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Neighboring nodes in the network
Test if directed graph is connectedWhy is NeighborhoodGraph so slow?How to add new nodes to an existing graph with fixed (coordinates) nodes?How do I upload a graph as an adjacency list and find the betweenness centrality?Arranging “ranked” nodes of a graph symmetricallyVertexLabels with Graph PropertiesNetwork with Radial Gradient Fill NodesHow to format vertices and control placement in a directed graphHow to label a large number of vertices using a list of namesColor the nodes according to certain valuesHighlight all paths in a graph below some threshold lengthSandbox algorithm for multifractal analysis of complex networks
$begingroup$
Consider the graph:
graph = 1 <-> 2, 1 <-> 4, 1 <-> 5, 1 <-> 8, 1 <-> 10, 1 <-> 26, 1 <-> 37, 1 <-> 42, 1 <-> 62, 1 <-> 86, 1 <-> 93, 1 <-> 100, 2 <-> 3, 2 <-> 7, 2 <-> 9, 2 <-> 12, 2 <-> 14, 2 <-> 17, 2 <-> 18, 2 <-> 25, 2 <-> 36, 2 <-> 41, 2 <-> 46, 2 <-> 50, 2 <-> 55, 2 <-> 72, 2 <-> 75, 3 <-> 6, 3 <-> 28, 3 <-> 34, 3 <-> 63, 4 <-> 13, 4 <-> 21, 5 <-> 20, 5 <-> 35, 5 <-> 40, 5 <-> 45, 5 <-> 48, 5 <-> 74, 6 <-> 31, 6 <-> 70, 9 <-> 11, 9 <-> 54, 9 <-> 67, 11 <-> 16, 11 <-> 24, 11 <-> 58, 11 <-> 60, 11 <-> 61, 11 <-> 65, 11 <-> 69, 12 <-> 27, 13 <-> 15, 13 <-> 33, 13 <-> 76, 14 <-> 30, 15 <-> 19, 15 <-> 96, 15 <-> 98, 16 <-> 57, 16 <-> 90, 19 <-> 22, 19 <-> 23, 19 <-> 39, 19 <-> 80, 19 <-> 83, 21 <-> 38, 22 <-> 59, 22 <-> 82, 25 <-> 29, 25 <-> 56, 25 <-> 94, 26 <-> 32, 26 <-> 43, 26 <-> 71, 27 <-> 47, 30 <-> 77, 30 <-> 78, 33 <-> 79, 33 <-> 97, 39 <-> 49, 39 <-> 51, 40 <-> 44, 40 <-> 73, 42 <-> 68, 48 <-> 52, 48 <-> 81, 50 <-> 53, 50 <-> 64, 50 <-> 89, 56 <-> 66, 56 <-> 92, 59 <-> 91, 62 <-> 88, 67 <-> 87, 74 <-> 95, 82 <-> 84, 82 <-> 85, 82 <-> 99;
net = Graph[graph, VertexShapeFunction -> "Name"]
Let's choose any node 'g' in the graph:
g=19;
Let 'r' denote the distance (counted in the number of nodes) from the node 'g':
d = GraphDiameter[net]
r = Range[1, d]
How to count all neighboring nodes within radius 'r' from the node 'g' ?
For example for node g=19 we have 6 nodes for r=1 (nodes: 80,83,22,39,23,15). For r=2 we have 7 nodes: 59,82,49,51,98,96,13.
graphs-and-networks
edited Apr 4 at 8:26
J. M. is away♦
98.9k10311467
asked Apr 4 at 8:25
ralphralph
1687
$endgroup$
add a comment |
$begingroup$
Consider the graph:
graph = 1 <-> 2, 1 <-> 4, 1 <-> 5, 1 <-> 8, 1 <-> 10, 1 <-> 26, 1 <-> 37, 1 <-> 42, 1 <-> 62, 1 <-> 86, 1 <-> 93, 1 <-> 100, 2 <-> 3, 2 <-> 7, 2 <-> 9, 2 <-> 12, 2 <-> 14, 2 <-> 17, 2 <-> 18, 2 <-> 25, 2 <-> 36, 2 <-> 41, 2 <-> 46, 2 <-> 50, 2 <-> 55, 2 <-> 72, 2 <-> 75, 3 <-> 6, 3 <-> 28, 3 <-> 34, 3 <-> 63, 4 <-> 13, 4 <-> 21, 5 <-> 20, 5 <-> 35, 5 <-> 40, 5 <-> 45, 5 <-> 48, 5 <-> 74, 6 <-> 31, 6 <-> 70, 9 <-> 11, 9 <-> 54, 9 <-> 67, 11 <-> 16, 11 <-> 24, 11 <-> 58, 11 <-> 60, 11 <-> 61, 11 <-> 65, 11 <-> 69, 12 <-> 27, 13 <-> 15, 13 <-> 33, 13 <-> 76, 14 <-> 30, 15 <-> 19, 15 <-> 96, 15 <-> 98, 16 <-> 57, 16 <-> 90, 19 <-> 22, 19 <-> 23, 19 <-> 39, 19 <-> 80, 19 <-> 83, 21 <-> 38, 22 <-> 59, 22 <-> 82, 25 <-> 29, 25 <-> 56, 25 <-> 94, 26 <-> 32, 26 <-> 43, 26 <-> 71, 27 <-> 47, 30 <-> 77, 30 <-> 78, 33 <-> 79, 33 <-> 97, 39 <-> 49, 39 <-> 51, 40 <-> 44, 40 <-> 73, 42 <-> 68, 48 <-> 52, 48 <-> 81, 50 <-> 53, 50 <-> 64, 50 <-> 89, 56 <-> 66, 56 <-> 92, 59 <-> 91, 62 <-> 88, 67 <-> 87, 74 <-> 95, 82 <-> 84, 82 <-> 85, 82 <-> 99;
net = Graph[graph, VertexShapeFunction -> "Name"]
Let's choose any node 'g' in the graph:
g=19;
Let 'r' denote the distance (counted in the number of nodes) from the node 'g':
d = GraphDiameter[net]
r = Range[1, d]
How to count all neighboring nodes within radius 'r' from the node 'g' ?
For example for node g=19 we have 6 nodes for r=1 (nodes: 80,83,22,39,23,15). For r=2 we have 7 nodes: 59,82,49,51,98,96,13.
graphs-and-networks
edited Apr 4 at 8:26
J. M. is away♦
98.9k10311467
asked Apr 4 at 8:25
ralphralph
1687
$endgroup$
add a comment |
$begingroup$
Consider the graph:
graph = 1 <-> 2, 1 <-> 4, 1 <-> 5, 1 <-> 8, 1 <-> 10, 1 <-> 26, 1 <-> 37, 1 <-> 42, 1 <-> 62, 1 <-> 86, 1 <-> 93, 1 <-> 100, 2 <-> 3, 2 <-> 7, 2 <-> 9, 2 <-> 12, 2 <-> 14, 2 <-> 17, 2 <-> 18, 2 <-> 25, 2 <-> 36, 2 <-> 41, 2 <-> 46, 2 <-> 50, 2 <-> 55, 2 <-> 72, 2 <-> 75, 3 <-> 6, 3 <-> 28, 3 <-> 34, 3 <-> 63, 4 <-> 13, 4 <-> 21, 5 <-> 20, 5 <-> 35, 5 <-> 40, 5 <-> 45, 5 <-> 48, 5 <-> 74, 6 <-> 31, 6 <-> 70, 9 <-> 11, 9 <-> 54, 9 <-> 67, 11 <-> 16, 11 <-> 24, 11 <-> 58, 11 <-> 60, 11 <-> 61, 11 <-> 65, 11 <-> 69, 12 <-> 27, 13 <-> 15, 13 <-> 33, 13 <-> 76, 14 <-> 30, 15 <-> 19, 15 <-> 96, 15 <-> 98, 16 <-> 57, 16 <-> 90, 19 <-> 22, 19 <-> 23, 19 <-> 39, 19 <-> 80, 19 <-> 83, 21 <-> 38, 22 <-> 59, 22 <-> 82, 25 <-> 29, 25 <-> 56, 25 <-> 94, 26 <-> 32, 26 <-> 43, 26 <-> 71, 27 <-> 47, 30 <-> 77, 30 <-> 78, 33 <-> 79, 33 <-> 97, 39 <-> 49, 39 <-> 51, 40 <-> 44, 40 <-> 73, 42 <-> 68, 48 <-> 52, 48 <-> 81, 50 <-> 53, 50 <-> 64, 50 <-> 89, 56 <-> 66, 56 <-> 92, 59 <-> 91, 62 <-> 88, 67 <-> 87, 74 <-> 95, 82 <-> 84, 82 <-> 85, 82 <-> 99;
net = Graph[graph, VertexShapeFunction -> "Name"]
Let's choose any node 'g' in the graph:
g=19;
Let 'r' denote the distance (counted in the number of nodes) from the node 'g':
d = GraphDiameter[net]
r = Range[1, d]
How to count all neighboring nodes within radius 'r' from the node 'g' ?
For example for node g=19 we have 6 nodes for r=1 (nodes: 80,83,22,39,23,15). For r=2 we have 7 nodes: 59,82,49,51,98,96,13.
graphs-and-networks
edited Apr 4 at 8:26
J. M. is away♦
98.9k10311467
asked Apr 4 at 8:25
ralphralph
1687
$endgroup$
Consider the graph:
graph = 1 <-> 2, 1 <-> 4, 1 <-> 5, 1 <-> 8, 1 <-> 10, 1 <-> 26, 1 <-> 37, 1 <-> 42, 1 <-> 62, 1 <-> 86, 1 <-> 93, 1 <-> 100, 2 <-> 3, 2 <-> 7, 2 <-> 9, 2 <-> 12, 2 <-> 14, 2 <-> 17, 2 <-> 18, 2 <-> 25, 2 <-> 36, 2 <-> 41, 2 <-> 46, 2 <-> 50, 2 <-> 55, 2 <-> 72, 2 <-> 75, 3 <-> 6, 3 <-> 28, 3 <-> 34, 3 <-> 63, 4 <-> 13, 4 <-> 21, 5 <-> 20, 5 <-> 35, 5 <-> 40, 5 <-> 45, 5 <-> 48, 5 <-> 74, 6 <-> 31, 6 <-> 70, 9 <-> 11, 9 <-> 54, 9 <-> 67, 11 <-> 16, 11 <-> 24, 11 <-> 58, 11 <-> 60, 11 <-> 61, 11 <-> 65, 11 <-> 69, 12 <-> 27, 13 <-> 15, 13 <-> 33, 13 <-> 76, 14 <-> 30, 15 <-> 19, 15 <-> 96, 15 <-> 98, 16 <-> 57, 16 <-> 90, 19 <-> 22, 19 <-> 23, 19 <-> 39, 19 <-> 80, 19 <-> 83, 21 <-> 38, 22 <-> 59, 22 <-> 82, 25 <-> 29, 25 <-> 56, 25 <-> 94, 26 <-> 32, 26 <-> 43, 26 <-> 71, 27 <-> 47, 30 <-> 77, 30 <-> 78, 33 <-> 79, 33 <-> 97, 39 <-> 49, 39 <-> 51, 40 <-> 44, 40 <-> 73, 42 <-> 68, 48 <-> 52, 48 <-> 81, 50 <-> 53, 50 <-> 64, 50 <-> 89, 56 <-> 66, 56 <-> 92, 59 <-> 91, 62 <-> 88, 67 <-> 87, 74 <-> 95, 82 <-> 84, 82 <-> 85, 82 <-> 99;
net = Graph[graph, VertexShapeFunction -> "Name"]
Let's choose any node 'g' in the graph:
g=19;
Let 'r' denote the distance (counted in the number of nodes) from the node 'g':
d = GraphDiameter[net]
r = Range[1, d]
How to count all neighboring nodes within radius 'r' from the node 'g' ?
For example for node g=19 we have 6 nodes for r=1 (nodes: 80,83,22,39,23,15). For r=2 we have 7 nodes: 59,82,49,51,98,96,13.
graphs-and-networks
graphs-and-networks
edited Apr 4 at 8:26
J. M. is away♦
98.9k10311467
asked Apr 4 at 8:25
ralphralph
1687
edited Apr 4 at 8:26
J. M. is away♦
98.9k10311467
asked Apr 4 at 8:25
ralphralph
1687
edited Apr 4 at 8:26
J. M. is away♦
98.9k10311467
edited Apr 4 at 8:26
J. M. is away♦
98.9k10311467
edited Apr 4 at 8:26
J. M. is away♦
98.9k10311467
98.9k10311467
asked Apr 4 at 8:25
ralphralph
1687
asked Apr 4 at 8:25
ralphralph
1687
asked Apr 4 at 8:25
ralphralph
1687
1687
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
I will choose a bit better GraphLayout for a tree:
net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];
I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.
nei[v_, d_] := NeighborhoodGraph[net, v, d]
Take distance 1:
nei[19, 1]
and see it is right:
HighlightGraph[net, nei[19, 1]]
Now you can compute whatever you need:
VertexList[nei[19, 1]]
Length[%] - 1
19, 15, 22, 23, 39, 80, 83
6
For the distance 2:
VertexList[nei[19, 1]]
VertexList[nei[19, 2]]
Complement[%, %%]
Length[%]
19, 15, 22, 23, 39, 80, 83
19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98
13, 49, 51, 59, 82, 96, 98
7
Timings for large graphs
net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];
nei[v_, d_] := NeighborhoodGraph[net, v, d]
dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]
Table[AbsoluteTiming[dist15;][[1]], 5]
0.097359, 0.094737, 0.092589, 0.08872, 0.087478
answered Apr 4 at 9:11
Vitaliy KaurovVitaliy Kaurov
57.6k6162282
$endgroup$
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15).
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
@ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 12:41
$begingroup$
Please forgive me. I meant about 200,000 no 20,000 nodes.
$endgroup$
– ralph
Apr 4 at 12:57
$begingroup$
@Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 13:51
$begingroup$
@VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
$endgroup$
– Szabolcs
Apr 4 at 16:01
|
show 2 more comments
$begingroup$
You could build it using BreadthFirstScan:
net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];
distance =
GroupBy[Reap[
BreadthFirstScan[net,
19, "DiscoverVertex" -> (Sow[#3 -> #1] &)]][[2, 1]],
First -> Last, Association["length" -> Length[#], "set" -> #] &];
Get length:
distance[3, "length"]
1194
distance[[All, "length"]]
<|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
-> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>
and set
distance[21, "set"]
182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
159530, 196846, 144772
For weighted graphs:
SeedRandom[123];net2 = Graph[net, EdgeWeight -> RandomInteger[1, 20, EdgeCount[net]]];
edgeWeight[g_, x_, y_] :=
With[weight = PropertyValue[g, UndirectedEdge[x, y],EdgeWeight],
If[NumericQ[weight], weight, 0]]
Clear[dist]; dist[_] := 0;
weights =
Reap[BreadthFirstScan[net2,
9, "DiscoverVertex" -> ((dist[#1] =
dist[#2] + edgeWeight[net2, #1, #2];
Sow[#1 -> dist[#1]]) &)]][[2, 1]];
set = Select[weights, #[[2]] <= 5 &];
set[[;; 10]]
9 -> 0, 66 -> 4, 126 -> 5, 160 -> 5, 190 -> 3, 274 -> 3, 283 -> 4,
312 -> 4, 519 -> 5, 537 -> 4
set // Length
105
Note that BreadthFirstScan approach might not work in general (non tree graphs).
answered Apr 4 at 14:29
halmirhalmir
10.6k2544
$endgroup$
1
$begingroup$
Amazingly fast, halmir! Any idea why this solution is so much faster thanGraphDistance, which I would have thought works byBreadthFirstScaninternally?
$endgroup$
– Roman
Apr 4 at 16:05
1
$begingroup$
@Roman I had the conviction thatGraphDistancecompute the entireGraphDistanceMatrixeven if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
$endgroup$
– Szabolcs
Apr 4 at 16:14
$begingroup$
@Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity ofGraphDistanceis quadratic in the graph size even when given just one vertex. That should not be so.
$endgroup$
– Szabolcs
Apr 4 at 16:17
$begingroup$
@halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
$endgroup$
– Szabolcs
Apr 4 at 16:20
$begingroup$
@Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
$endgroup$
– Szabolcs
Apr 4 at 16:30
|
show 7 more comments
$begingroup$
To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:
Counts@GraphDistance[net, g]
<|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>
Look them all up in order:
BinCounts[GraphDistance[net, g], 0, d, 1]
1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0
answered Apr 4 at 9:04
RomanRoman
4,53011127
$endgroup$
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15)
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
Yes if you want only short distances then @szabolcs has better tools available. ThisGraphDistancesolution is only good if you want the distances to all nodes in the graph.
$endgroup$
– Roman
Apr 4 at 13:50
add a comment |
$begingroup$
How to count all neighboring nodes within radius 'r' from the node 'g' ?
Use IGraph/M.
IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.
If you want to do it for multiple distances in one go, use IGDistanceCounts,
IGDistanceCounts[graph, vertex]
This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.
For weighted distances, use IGDistanceHistogram.
answered Apr 4 at 13:40
SzabolcsSzabolcs
163k14448945
$endgroup$
$begingroup$
Thanks. And how to count the same as the 'IGDistanceCounts[graph, vertex]' formula but for weighted networks?
$endgroup$
– ralph
Apr 4 at 14:15
$begingroup$
@ralph As I said above, useIGDistanceHistogram
$endgroup$
– Szabolcs
Apr 4 at 16:01
$begingroup$
Mr=IGDistanceHistogram[net1, ??] (*for weighted graph *) ???
$endgroup$
– ralph
2 days ago
$begingroup$
@ralph Did you check the documentation? If you checked the documentation and you found it to be unclear, you are very welcome to suggest improvements.
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@ralph The syntax isIGDistanceHistogram[graph, binSize, vertex]wherebinSizeis the bin size used for constructing the distance histogram. You must put the vertex in a list as the syntax also accepts multiple vertices.
$endgroup$
– Szabolcs
2 days ago
|
show 5 more comments
$begingroup$
For weighted network:
g1 = 4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9;
w1 = 10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1;
w2=Table[1, 29];
net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
s = RandomSample[VertexList[net1], 15];
Mr = Table[IGDistanceCounts[net1, s[[i]]], i, 1, Length[s]] (*for non weighted*)
Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)
Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)
answered Apr 4 at 17:40
ralphralph
1687
$endgroup$
add a comment |
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5 Answers
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5 Answers
5
active
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active
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votes
$begingroup$
I will choose a bit better GraphLayout for a tree:
net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];
I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.
nei[v_, d_] := NeighborhoodGraph[net, v, d]
Take distance 1:
nei[19, 1]
and see it is right:
HighlightGraph[net, nei[19, 1]]
Now you can compute whatever you need:
VertexList[nei[19, 1]]
Length[%] - 1
19, 15, 22, 23, 39, 80, 83
6
For the distance 2:
VertexList[nei[19, 1]]
VertexList[nei[19, 2]]
Complement[%, %%]
Length[%]
19, 15, 22, 23, 39, 80, 83
19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98
13, 49, 51, 59, 82, 96, 98
7
Timings for large graphs
net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];
nei[v_, d_] := NeighborhoodGraph[net, v, d]
dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]
Table[AbsoluteTiming[dist15;][[1]], 5]
0.097359, 0.094737, 0.092589, 0.08872, 0.087478
answered Apr 4 at 9:11
Vitaliy KaurovVitaliy Kaurov
57.6k6162282
$endgroup$
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15).
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
@ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 12:41
$begingroup$
Please forgive me. I meant about 200,000 no 20,000 nodes.
$endgroup$
– ralph
Apr 4 at 12:57
$begingroup$
@Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 13:51
$begingroup$
@VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
$endgroup$
– Szabolcs
Apr 4 at 16:01
|
show 2 more comments
$begingroup$
I will choose a bit better GraphLayout for a tree:
net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];
I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.
nei[v_, d_] := NeighborhoodGraph[net, v, d]
Take distance 1:
nei[19, 1]
and see it is right:
HighlightGraph[net, nei[19, 1]]
Now you can compute whatever you need:
VertexList[nei[19, 1]]
Length[%] - 1
19, 15, 22, 23, 39, 80, 83
6
For the distance 2:
VertexList[nei[19, 1]]
VertexList[nei[19, 2]]
Complement[%, %%]
Length[%]
19, 15, 22, 23, 39, 80, 83
19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98
13, 49, 51, 59, 82, 96, 98
7
Timings for large graphs
net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];
nei[v_, d_] := NeighborhoodGraph[net, v, d]
dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]
Table[AbsoluteTiming[dist15;][[1]], 5]
0.097359, 0.094737, 0.092589, 0.08872, 0.087478
answered Apr 4 at 9:11
Vitaliy KaurovVitaliy Kaurov
57.6k6162282
$endgroup$
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15).
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
@ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 12:41
$begingroup$
Please forgive me. I meant about 200,000 no 20,000 nodes.
$endgroup$
– ralph
Apr 4 at 12:57
$begingroup$
@Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 13:51
$begingroup$
@VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
$endgroup$
– Szabolcs
Apr 4 at 16:01
|
show 2 more comments
$begingroup$
I will choose a bit better GraphLayout for a tree:
net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];
I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.
nei[v_, d_] := NeighborhoodGraph[net, v, d]
Take distance 1:
nei[19, 1]
and see it is right:
HighlightGraph[net, nei[19, 1]]
Now you can compute whatever you need:
VertexList[nei[19, 1]]
Length[%] - 1
19, 15, 22, 23, 39, 80, 83
6
For the distance 2:
VertexList[nei[19, 1]]
VertexList[nei[19, 2]]
Complement[%, %%]
Length[%]
19, 15, 22, 23, 39, 80, 83
19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98
13, 49, 51, 59, 82, 96, 98
7
Timings for large graphs
net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];
nei[v_, d_] := NeighborhoodGraph[net, v, d]
dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]
Table[AbsoluteTiming[dist15;][[1]], 5]
0.097359, 0.094737, 0.092589, 0.08872, 0.087478
answered Apr 4 at 9:11
Vitaliy KaurovVitaliy Kaurov
57.6k6162282
$endgroup$
I will choose a bit better GraphLayout for a tree:
net = Graph[graph, VertexLabels -> "Name", GraphLayout -> "RadialEmbedding"];
I suggest don't just count directly - get an object - a subgraph - of your query, so you can then run various computations on it and don't need count all over again based on different criteria w/ a different code.
nei[v_, d_] := NeighborhoodGraph[net, v, d]
Take distance 1:
nei[19, 1]
and see it is right:
HighlightGraph[net, nei[19, 1]]
Now you can compute whatever you need:
VertexList[nei[19, 1]]
Length[%] - 1
19, 15, 22, 23, 39, 80, 83
6
For the distance 2:
VertexList[nei[19, 1]]
VertexList[nei[19, 2]]
Complement[%, %%]
Length[%]
19, 15, 22, 23, 39, 80, 83
19, 13, 15, 22, 23, 39, 49, 51, 59, 80, 82, 83, 96, 98
13, 49, 51, 59, 82, 96, 98
7
Timings for large graphs
net = RandomGraph[BarabasiAlbertGraphDistribution[20000, 1]];
nei[v_, d_] := NeighborhoodGraph[net, v, d]
dist15:=Length[Complement[VertexList[nei[#,15]],VertexList[nei[#,14]]]&@RandomInteger[1000]]
Table[AbsoluteTiming[dist15;][[1]], 5]
0.097359, 0.094737, 0.092589, 0.08872, 0.087478
answered Apr 4 at 9:11
Vitaliy KaurovVitaliy Kaurov
57.6k6162282
edited Apr 4 at 12:44
answered Apr 4 at 9:11
Vitaliy KaurovVitaliy Kaurov
57.6k6162282
answered Apr 4 at 9:11
Vitaliy KaurovVitaliy Kaurov
57.6k6162282
answered Apr 4 at 9:11
Vitaliy KaurovVitaliy Kaurov
57.6k6162282
57.6k6162282
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15).
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
@ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 12:41
$begingroup$
Please forgive me. I meant about 200,000 no 20,000 nodes.
$endgroup$
– ralph
Apr 4 at 12:57
$begingroup$
@Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 13:51
$begingroup$
@VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
$endgroup$
– Szabolcs
Apr 4 at 16:01
|
show 2 more comments
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15).
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
@ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 12:41
$begingroup$
Please forgive me. I meant about 200,000 no 20,000 nodes.
$endgroup$
– ralph
Apr 4 at 12:57
$begingroup$
@Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 13:51
$begingroup$
@VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
$endgroup$
– Szabolcs
Apr 4 at 16:01
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15).
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15).
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
@ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 12:41
$begingroup$
@ralph is 0.1 seconds is slow? What timings do you need? No criteria for timings is mentioned in your original post.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 12:41
$begingroup$
Please forgive me. I meant about 200,000 no 20,000 nodes.
$endgroup$
– ralph
Apr 4 at 12:57
$begingroup$
Please forgive me. I meant about 200,000 no 20,000 nodes.
$endgroup$
– ralph
Apr 4 at 12:57
$begingroup$
@Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 13:51
$begingroup$
@Szabolcs i was just answering question without performance consideration as it was not asked in the OP, which had a tiny graph. I added benchmark after he made a comment, and then he changed his comment again.
$endgroup$
– Vitaliy Kaurov
Apr 4 at 13:51
$begingroup$
@VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
$endgroup$
– Szabolcs
Apr 4 at 16:01
$begingroup$
@VitaliyKaurov Sorry about the comments, I was wrong: this was actually fixed in 12.0. That is why I deleted them.
$endgroup$
– Szabolcs
Apr 4 at 16:01
|
show 2 more comments
$begingroup$
You could build it using BreadthFirstScan:
net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];
distance =
GroupBy[Reap[
BreadthFirstScan[net,
19, "DiscoverVertex" -> (Sow[#3 -> #1] &)]][[2, 1]],
First -> Last, Association["length" -> Length[#], "set" -> #] &];
Get length:
distance[3, "length"]
1194
distance[[All, "length"]]
<|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
-> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>
and set
distance[21, "set"]
182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
159530, 196846, 144772
For weighted graphs:
SeedRandom[123];net2 = Graph[net, EdgeWeight -> RandomInteger[1, 20, EdgeCount[net]]];
edgeWeight[g_, x_, y_] :=
With[weight = PropertyValue[g, UndirectedEdge[x, y],EdgeWeight],
If[NumericQ[weight], weight, 0]]
Clear[dist]; dist[_] := 0;
weights =
Reap[BreadthFirstScan[net2,
9, "DiscoverVertex" -> ((dist[#1] =
dist[#2] + edgeWeight[net2, #1, #2];
Sow[#1 -> dist[#1]]) &)]][[2, 1]];
set = Select[weights, #[[2]] <= 5 &];
set[[;; 10]]
9 -> 0, 66 -> 4, 126 -> 5, 160 -> 5, 190 -> 3, 274 -> 3, 283 -> 4,
312 -> 4, 519 -> 5, 537 -> 4
set // Length
105
Note that BreadthFirstScan approach might not work in general (non tree graphs).
answered Apr 4 at 14:29
halmirhalmir
10.6k2544
$endgroup$
1
$begingroup$
Amazingly fast, halmir! Any idea why this solution is so much faster thanGraphDistance, which I would have thought works byBreadthFirstScaninternally?
$endgroup$
– Roman
Apr 4 at 16:05
1
$begingroup$
@Roman I had the conviction thatGraphDistancecompute the entireGraphDistanceMatrixeven if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
$endgroup$
– Szabolcs
Apr 4 at 16:14
$begingroup$
@Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity ofGraphDistanceis quadratic in the graph size even when given just one vertex. That should not be so.
$endgroup$
– Szabolcs
Apr 4 at 16:17
$begingroup$
@halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
$endgroup$
– Szabolcs
Apr 4 at 16:20
$begingroup$
@Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
$endgroup$
– Szabolcs
Apr 4 at 16:30
|
show 7 more comments
$begingroup$
You could build it using BreadthFirstScan:
net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];
distance =
GroupBy[Reap[
BreadthFirstScan[net,
19, "DiscoverVertex" -> (Sow[#3 -> #1] &)]][[2, 1]],
First -> Last, Association["length" -> Length[#], "set" -> #] &];
Get length:
distance[3, "length"]
1194
distance[[All, "length"]]
<|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
-> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>
and set
distance[21, "set"]
182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
159530, 196846, 144772
For weighted graphs:
SeedRandom[123];net2 = Graph[net, EdgeWeight -> RandomInteger[1, 20, EdgeCount[net]]];
edgeWeight[g_, x_, y_] :=
With[weight = PropertyValue[g, UndirectedEdge[x, y],EdgeWeight],
If[NumericQ[weight], weight, 0]]
Clear[dist]; dist[_] := 0;
weights =
Reap[BreadthFirstScan[net2,
9, "DiscoverVertex" -> ((dist[#1] =
dist[#2] + edgeWeight[net2, #1, #2];
Sow[#1 -> dist[#1]]) &)]][[2, 1]];
set = Select[weights, #[[2]] <= 5 &];
set[[;; 10]]
9 -> 0, 66 -> 4, 126 -> 5, 160 -> 5, 190 -> 3, 274 -> 3, 283 -> 4,
312 -> 4, 519 -> 5, 537 -> 4
set // Length
105
Note that BreadthFirstScan approach might not work in general (non tree graphs).
answered Apr 4 at 14:29
halmirhalmir
10.6k2544
$endgroup$
1
$begingroup$
Amazingly fast, halmir! Any idea why this solution is so much faster thanGraphDistance, which I would have thought works byBreadthFirstScaninternally?
$endgroup$
– Roman
Apr 4 at 16:05
1
$begingroup$
@Roman I had the conviction thatGraphDistancecompute the entireGraphDistanceMatrixeven if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
$endgroup$
– Szabolcs
Apr 4 at 16:14
$begingroup$
@Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity ofGraphDistanceis quadratic in the graph size even when given just one vertex. That should not be so.
$endgroup$
– Szabolcs
Apr 4 at 16:17
$begingroup$
@halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
$endgroup$
– Szabolcs
Apr 4 at 16:20
$begingroup$
@Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
$endgroup$
– Szabolcs
Apr 4 at 16:30
|
show 7 more comments
$begingroup$
You could build it using BreadthFirstScan:
net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];
distance =
GroupBy[Reap[
BreadthFirstScan[net,
19, "DiscoverVertex" -> (Sow[#3 -> #1] &)]][[2, 1]],
First -> Last, Association["length" -> Length[#], "set" -> #] &];
Get length:
distance[3, "length"]
1194
distance[[All, "length"]]
<|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
-> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>
and set
distance[21, "set"]
182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
159530, 196846, 144772
For weighted graphs:
SeedRandom[123];net2 = Graph[net, EdgeWeight -> RandomInteger[1, 20, EdgeCount[net]]];
edgeWeight[g_, x_, y_] :=
With[weight = PropertyValue[g, UndirectedEdge[x, y],EdgeWeight],
If[NumericQ[weight], weight, 0]]
Clear[dist]; dist[_] := 0;
weights =
Reap[BreadthFirstScan[net2,
9, "DiscoverVertex" -> ((dist[#1] =
dist[#2] + edgeWeight[net2, #1, #2];
Sow[#1 -> dist[#1]]) &)]][[2, 1]];
set = Select[weights, #[[2]] <= 5 &];
set[[;; 10]]
9 -> 0, 66 -> 4, 126 -> 5, 160 -> 5, 190 -> 3, 274 -> 3, 283 -> 4,
312 -> 4, 519 -> 5, 537 -> 4
set // Length
105
Note that BreadthFirstScan approach might not work in general (non tree graphs).
answered Apr 4 at 14:29
halmirhalmir
10.6k2544
$endgroup$
You could build it using BreadthFirstScan:
net = RandomGraph[BarabasiAlbertGraphDistribution[200000, 1]];
distance =
GroupBy[Reap[
BreadthFirstScan[net,
19, "DiscoverVertex" -> (Sow[#3 -> #1] &)]][[2, 1]],
First -> Last, Association["length" -> Length[#], "set" -> #] &];
Get length:
distance[3, "length"]
1194
distance[[All, "length"]]
<|0 -> 1, 1 -> 214, 2 -> 1194, 3 -> 3058, 4 -> 5826, 5 -> 10069, 6
-> 15110, 7 -> 19992, 8 -> 23821, 9 -> 24910, 10 -> 24767, 11 -> 21459, 12 -> 17869, 13 -> 13525, 14 -> 9119, 15 -> 5146, 16 -> 2406,
17 -> 1025, 18 -> 337, 19 -> 106, 20 -> 34, 21 -> 11, 22 -> 1|>
and set
distance[21, "set"]
182224, 145742, 171910, 124658, 125540, 128520, 196392, 166986,
159530, 196846, 144772
For weighted graphs:
SeedRandom[123];net2 = Graph[net, EdgeWeight -> RandomInteger[1, 20, EdgeCount[net]]];
edgeWeight[g_, x_, y_] :=
With[weight = PropertyValue[g, UndirectedEdge[x, y],EdgeWeight],
If[NumericQ[weight], weight, 0]]
Clear[dist]; dist[_] := 0;
weights =
Reap[BreadthFirstScan[net2,
9, "DiscoverVertex" -> ((dist[#1] =
dist[#2] + edgeWeight[net2, #1, #2];
Sow[#1 -> dist[#1]]) &)]][[2, 1]];
set = Select[weights, #[[2]] <= 5 &];
set[[;; 10]]
9 -> 0, 66 -> 4, 126 -> 5, 160 -> 5, 190 -> 3, 274 -> 3, 283 -> 4,
312 -> 4, 519 -> 5, 537 -> 4
set // Length
105
Note that BreadthFirstScan approach might not work in general (non tree graphs).
answered Apr 4 at 14:29
halmirhalmir
10.6k2544
edited 2 days ago
answered Apr 4 at 14:29
halmirhalmir
10.6k2544
answered Apr 4 at 14:29
halmirhalmir
10.6k2544
answered Apr 4 at 14:29
halmirhalmir
10.6k2544
10.6k2544
1
$begingroup$
Amazingly fast, halmir! Any idea why this solution is so much faster thanGraphDistance, which I would have thought works byBreadthFirstScaninternally?
$endgroup$
– Roman
Apr 4 at 16:05
1
$begingroup$
@Roman I had the conviction thatGraphDistancecompute the entireGraphDistanceMatrixeven if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
$endgroup$
– Szabolcs
Apr 4 at 16:14
$begingroup$
@Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity ofGraphDistanceis quadratic in the graph size even when given just one vertex. That should not be so.
$endgroup$
– Szabolcs
Apr 4 at 16:17
$begingroup$
@halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
$endgroup$
– Szabolcs
Apr 4 at 16:20
$begingroup$
@Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
$endgroup$
– Szabolcs
Apr 4 at 16:30
|
show 7 more comments
1
$begingroup$
Amazingly fast, halmir! Any idea why this solution is so much faster thanGraphDistance, which I would have thought works byBreadthFirstScaninternally?
$endgroup$
– Roman
Apr 4 at 16:05
1
$begingroup$
@Roman I had the conviction thatGraphDistancecompute the entireGraphDistanceMatrixeven if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.
$endgroup$
– Szabolcs
Apr 4 at 16:14
$begingroup$
@Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity ofGraphDistanceis quadratic in the graph size even when given just one vertex. That should not be so.
$endgroup$
– Szabolcs
Apr 4 at 16:17
$begingroup$
@halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
$endgroup$
– Szabolcs
Apr 4 at 16:20
$begingroup$
@Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
$endgroup$
– Szabolcs
Apr 4 at 16:30
1
1
$begingroup$
Amazingly fast, halmir! Any idea why this solution is so much faster than
GraphDistance, which I would have thought works by BreadthFirstScan internally?$endgroup$
– Roman
Apr 4 at 16:05
$begingroup$
Amazingly fast, halmir! Any idea why this solution is so much faster than
GraphDistance, which I would have thought works by BreadthFirstScan internally?$endgroup$
– Roman
Apr 4 at 16:05
1
1
$begingroup$
@Roman I had the conviction that
GraphDistance compute the entire GraphDistanceMatrix even if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.$endgroup$
– Szabolcs
Apr 4 at 16:14
$begingroup$
@Roman I had the conviction that
GraphDistance compute the entire GraphDistanceMatrix even if you gave it only one vertex. I do not remember what led me to this conclusion though. I do remember that I put a lot of effort into this functionality area in IGraph/M as I could not use M's built-ins for large graphs.$endgroup$
– Szabolcs
Apr 4 at 16:14
$begingroup$
@Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity of
GraphDistance is quadratic in the graph size even when given just one vertex. That should not be so.$endgroup$
– Szabolcs
Apr 4 at 16:17
$begingroup$
@Roman A qucik test tells me that on a tree (which is being benchmarked here) the complexity of
GraphDistance is quadratic in the graph size even when given just one vertex. That should not be so.$endgroup$
– Szabolcs
Apr 4 at 16:17
$begingroup$
@halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
$endgroup$
– Szabolcs
Apr 4 at 16:20
$begingroup$
@halmir Can you tell us whether this is a bug and if it is fixable? The quadratic complexity looks like a bug.
$endgroup$
– Szabolcs
Apr 4 at 16:20
$begingroup$
@Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
$endgroup$
– Szabolcs
Apr 4 at 16:30
$begingroup$
@Roman I strongly suspect that I may have reported this issue to Wolfram in the past. See e.g. this post I wrote 3 years ago, where I mention it: mathematica.stackexchange.com/a/109408/12
$endgroup$
– Szabolcs
Apr 4 at 16:30
|
show 7 more comments
$begingroup$
To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:
Counts@GraphDistance[net, g]
<|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>
Look them all up in order:
BinCounts[GraphDistance[net, g], 0, d, 1]
1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0
answered Apr 4 at 9:04
RomanRoman
4,53011127
$endgroup$
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15)
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
Yes if you want only short distances then @szabolcs has better tools available. ThisGraphDistancesolution is only good if you want the distances to all nodes in the graph.
$endgroup$
– Roman
Apr 4 at 13:50
add a comment |
$begingroup$
To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:
Counts@GraphDistance[net, g]
<|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>
Look them all up in order:
BinCounts[GraphDistance[net, g], 0, d, 1]
1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0
answered Apr 4 at 9:04
RomanRoman
4,53011127
$endgroup$
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15)
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
Yes if you want only short distances then @szabolcs has better tools available. ThisGraphDistancesolution is only good if you want the distances to all nodes in the graph.
$endgroup$
– Roman
Apr 4 at 13:50
add a comment |
$begingroup$
To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:
Counts@GraphDistance[net, g]
<|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>
Look them all up in order:
BinCounts[GraphDistance[net, g], 0, d, 1]
1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0
answered Apr 4 at 9:04
RomanRoman
4,53011127
$endgroup$
To count how many nodes there are at every distance (unsorted Association): use this if you want to Lookup a particular distance:
Counts@GraphDistance[net, g]
<|4 -> 4, 5 -> 12, 3 -> 7, 6 -> 26, 7 -> 20, 2 -> 7, 8 -> 15, 1 -> 6, 0 -> 1, 9 -> 2|>
Look them all up in order:
BinCounts[GraphDistance[net, g], 0, d, 1]
1, 6, 7, 7, 4, 12, 26, 20, 15, 2, 0, 0
answered Apr 4 at 9:04
RomanRoman
4,53011127
edited Apr 4 at 12:19
answered Apr 4 at 9:04
RomanRoman
4,53011127
answered Apr 4 at 9:04
RomanRoman
4,53011127
answered Apr 4 at 9:04
RomanRoman
4,53011127
4,53011127
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15)
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
Yes if you want only short distances then @szabolcs has better tools available. ThisGraphDistancesolution is only good if you want the distances to all nodes in the graph.
$endgroup$
– Roman
Apr 4 at 13:50
add a comment |
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15)
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
Yes if you want only short distances then @szabolcs has better tools available. ThisGraphDistancesolution is only good if you want the distances to all nodes in the graph.
$endgroup$
– Roman
Apr 4 at 13:50
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15)
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
Thank you. The code gives correct results but is memory-consuming for large networks (around 200,000 nodes: net = RandomGraph [BarabasiAlbertGraphDistribution [20,000, 1] and d = 1,2,3,4, ..., 15)
$endgroup$
– ralph
Apr 4 at 12:24
$begingroup$
Yes if you want only short distances then @szabolcs has better tools available. This
GraphDistance solution is only good if you want the distances to all nodes in the graph.$endgroup$
– Roman
Apr 4 at 13:50
$begingroup$
Yes if you want only short distances then @szabolcs has better tools available. This
GraphDistance solution is only good if you want the distances to all nodes in the graph.$endgroup$
– Roman
Apr 4 at 13:50
add a comment |
$begingroup$
How to count all neighboring nodes within radius 'r' from the node 'g' ?
Use IGraph/M.
IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.
If you want to do it for multiple distances in one go, use IGDistanceCounts,
IGDistanceCounts[graph, vertex]
This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.
For weighted distances, use IGDistanceHistogram.
answered Apr 4 at 13:40
SzabolcsSzabolcs
163k14448945
$endgroup$
$begingroup$
Thanks. And how to count the same as the 'IGDistanceCounts[graph, vertex]' formula but for weighted networks?
$endgroup$
– ralph
Apr 4 at 14:15
$begingroup$
@ralph As I said above, useIGDistanceHistogram
$endgroup$
– Szabolcs
Apr 4 at 16:01
$begingroup$
Mr=IGDistanceHistogram[net1, ??] (*for weighted graph *) ???
$endgroup$
– ralph
2 days ago
$begingroup$
@ralph Did you check the documentation? If you checked the documentation and you found it to be unclear, you are very welcome to suggest improvements.
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@ralph The syntax isIGDistanceHistogram[graph, binSize, vertex]wherebinSizeis the bin size used for constructing the distance histogram. You must put the vertex in a list as the syntax also accepts multiple vertices.
$endgroup$
– Szabolcs
2 days ago
|
show 5 more comments
$begingroup$
How to count all neighboring nodes within radius 'r' from the node 'g' ?
Use IGraph/M.
IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.
If you want to do it for multiple distances in one go, use IGDistanceCounts,
IGDistanceCounts[graph, vertex]
This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.
For weighted distances, use IGDistanceHistogram.
answered Apr 4 at 13:40
SzabolcsSzabolcs
163k14448945
$endgroup$
$begingroup$
Thanks. And how to count the same as the 'IGDistanceCounts[graph, vertex]' formula but for weighted networks?
$endgroup$
– ralph
Apr 4 at 14:15
$begingroup$
@ralph As I said above, useIGDistanceHistogram
$endgroup$
– Szabolcs
Apr 4 at 16:01
$begingroup$
Mr=IGDistanceHistogram[net1, ??] (*for weighted graph *) ???
$endgroup$
– ralph
2 days ago
$begingroup$
@ralph Did you check the documentation? If you checked the documentation and you found it to be unclear, you are very welcome to suggest improvements.
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@ralph The syntax isIGDistanceHistogram[graph, binSize, vertex]wherebinSizeis the bin size used for constructing the distance histogram. You must put the vertex in a list as the syntax also accepts multiple vertices.
$endgroup$
– Szabolcs
2 days ago
|
show 5 more comments
$begingroup$
How to count all neighboring nodes within radius 'r' from the node 'g' ?
Use IGraph/M.
IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.
If you want to do it for multiple distances in one go, use IGDistanceCounts,
IGDistanceCounts[graph, vertex]
This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.
For weighted distances, use IGDistanceHistogram.
answered Apr 4 at 13:40
SzabolcsSzabolcs
163k14448945
$endgroup$
How to count all neighboring nodes within radius 'r' from the node 'g' ?
Use IGraph/M.
IGNeighborhoodSize does precisely this and is probably your fastest bet, but I do not have time to benchmark it against other solutions right now.
If you want to do it for multiple distances in one go, use IGDistanceCounts,
IGDistanceCounts[graph, vertex]
This gives you the counts of other vertices found at all (unweighted) distances. You can then simply Accumulate that list to get the result for all r at the same time.
For weighted distances, use IGDistanceHistogram.
answered Apr 4 at 13:40
SzabolcsSzabolcs
163k14448945
answered Apr 4 at 13:40
SzabolcsSzabolcs
163k14448945
answered Apr 4 at 13:40
SzabolcsSzabolcs
163k14448945
answered Apr 4 at 13:40
SzabolcsSzabolcs
163k14448945
163k14448945
$begingroup$
Thanks. And how to count the same as the 'IGDistanceCounts[graph, vertex]' formula but for weighted networks?
$endgroup$
– ralph
Apr 4 at 14:15
$begingroup$
@ralph As I said above, useIGDistanceHistogram
$endgroup$
– Szabolcs
Apr 4 at 16:01
$begingroup$
Mr=IGDistanceHistogram[net1, ??] (*for weighted graph *) ???
$endgroup$
– ralph
2 days ago
$begingroup$
@ralph Did you check the documentation? If you checked the documentation and you found it to be unclear, you are very welcome to suggest improvements.
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@ralph The syntax isIGDistanceHistogram[graph, binSize, vertex]wherebinSizeis the bin size used for constructing the distance histogram. You must put the vertex in a list as the syntax also accepts multiple vertices.
$endgroup$
– Szabolcs
2 days ago
|
show 5 more comments
$begingroup$
Thanks. And how to count the same as the 'IGDistanceCounts[graph, vertex]' formula but for weighted networks?
$endgroup$
– ralph
Apr 4 at 14:15
$begingroup$
@ralph As I said above, useIGDistanceHistogram
$endgroup$
– Szabolcs
Apr 4 at 16:01
$begingroup$
Mr=IGDistanceHistogram[net1, ??] (*for weighted graph *) ???
$endgroup$
– ralph
2 days ago
$begingroup$
@ralph Did you check the documentation? If you checked the documentation and you found it to be unclear, you are very welcome to suggest improvements.
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@ralph The syntax isIGDistanceHistogram[graph, binSize, vertex]wherebinSizeis the bin size used for constructing the distance histogram. You must put the vertex in a list as the syntax also accepts multiple vertices.
$endgroup$
– Szabolcs
2 days ago
$begingroup$
Thanks. And how to count the same as the 'IGDistanceCounts[graph, vertex]' formula but for weighted networks?
$endgroup$
– ralph
Apr 4 at 14:15
$begingroup$
Thanks. And how to count the same as the 'IGDistanceCounts[graph, vertex]' formula but for weighted networks?
$endgroup$
– ralph
Apr 4 at 14:15
$begingroup$
@ralph As I said above, use
IGDistanceHistogram$endgroup$
– Szabolcs
Apr 4 at 16:01
$begingroup$
@ralph As I said above, use
IGDistanceHistogram$endgroup$
– Szabolcs
Apr 4 at 16:01
$begingroup$
Mr=IGDistanceHistogram[net1, ??] (*for weighted graph *) ???
$endgroup$
– ralph
2 days ago
$begingroup$
Mr=IGDistanceHistogram[net1, ??] (*for weighted graph *) ???
$endgroup$
– ralph
2 days ago
$begingroup$
@ralph Did you check the documentation? If you checked the documentation and you found it to be unclear, you are very welcome to suggest improvements.
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@ralph Did you check the documentation? If you checked the documentation and you found it to be unclear, you are very welcome to suggest improvements.
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@ralph The syntax is
IGDistanceHistogram[graph, binSize, vertex] where binSize is the bin size used for constructing the distance histogram. You must put the vertex in a list as the syntax also accepts multiple vertices.$endgroup$
– Szabolcs
2 days ago
$begingroup$
@ralph The syntax is
IGDistanceHistogram[graph, binSize, vertex] where binSize is the bin size used for constructing the distance histogram. You must put the vertex in a list as the syntax also accepts multiple vertices.$endgroup$
– Szabolcs
2 days ago
|
show 5 more comments
$begingroup$
For weighted network:
g1 = 4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9;
w1 = 10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1;
w2=Table[1, 29];
net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
s = RandomSample[VertexList[net1], 15];
Mr = Table[IGDistanceCounts[net1, s[[i]]], i, 1, Length[s]] (*for non weighted*)
Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)
Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)
answered Apr 4 at 17:40
ralphralph
1687
$endgroup$
add a comment |
$begingroup$
For weighted network:
g1 = 4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9;
w1 = 10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1;
w2=Table[1, 29];
net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
s = RandomSample[VertexList[net1], 15];
Mr = Table[IGDistanceCounts[net1, s[[i]]], i, 1, Length[s]] (*for non weighted*)
Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)
Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)
answered Apr 4 at 17:40
ralphralph
1687
$endgroup$
add a comment |
$begingroup$
For weighted network:
g1 = 4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9;
w1 = 10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1;
w2=Table[1, 29];
net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
s = RandomSample[VertexList[net1], 15];
Mr = Table[IGDistanceCounts[net1, s[[i]]], i, 1, Length[s]] (*for non weighted*)
Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)
Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)
answered Apr 4 at 17:40
ralphralph
1687
$endgroup$
For weighted network:
g1 = 4798 <-> 2641, 4798 <-> 2310, 4798 <-> 4721, 2310 <-> 1942,2310 <-> 961, 4721 <-> 4507, 4721 <-> 4779, 4779 <-> 4336, 4779 <-> 3238, 4336 <-> 3277, 4336 <-> 3514, 3277 <-> 2923, 2923 <-> 2772, 2923 <-> 2401, 2772 <-> 2, 2772 <-> 2771, 3514 <-> 3042, 3514 <-> 2739, 3042 <-> 3007, 3042 <-> 1655, 2739 <-> 2277, 2739 <-> 1895, 2 <-> 5, 2 <-> 3, 3277 <-> 100, 5 <-> 6, 5 <-> 7, 5 <-> 8, 5 <-> 9;
w1 = 10, 20, 20, 4, 35, 3, 4, 6, 17, 7, 13, 2, 2, 7, 2, 1, 3, 5, 3, 6,4, 6, 2, 1, 1, 1, 1, 1, 1;
w2=Table[1, 29];
net1 = Graph[g1, EdgeWeight -> w1, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
net2 = Graph[g1, EdgeWeight -> w2, EdgeLabels -> "EdgeWeight", VertexShapeFunction -> "Name"]
s = RandomSample[VertexList[net1], 15];
Mr = Table[IGDistanceCounts[net1, s[[i]]], i, 1, Length[s]] (*for non weighted*)
Mr2 = IGDistanceHistogram[net1, 9] (*for weighted graph ?*)
Mr3 = IGDistanceHistogram[net2, 9] (*for non weighted graph ? Mr3==Mr *)
answered Apr 4 at 17:40
ralphralph
1687
answered Apr 4 at 17:40
ralphralph
1687
answered Apr 4 at 17:40
ralphralph
1687
answered Apr 4 at 17:40
ralphralph
1687
1687
add a comment |
add a comment |
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