How to reverse every other sublist of a list? Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How can the lowest non-zero value in a nested list be found and its position in the sublist returned?Comparing elements of the $n^textth$ sublist in a ragged list with the $n^textth$ member of a sequenceSublist inside simple listHow to split list into disjoint lists?Selecting every last element of a nested listAdd elements of one list to sublists of another listEliminating elements from sublists under a global conditionRemove the right sublistList manipulation - adding last element of sublist to each sublistHow to partition a 1D list according to specified lengths?
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How to reverse every other sublist of a list?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How can the lowest non-zero value in a nested list be found and its position in the sublist returned?Comparing elements of the $n^textth$ sublist in a ragged list with the $n^textth$ member of a sequenceSublist inside simple listHow to split list into disjoint lists?Selecting every last element of a nested listAdd elements of one list to sublists of another listEliminating elements from sublists under a global conditionRemove the right sublistList manipulation - adding last element of sublist to each sublistHow to partition a 1D list according to specified lengths?
$begingroup$
I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?
list-manipulation
asked Apr 9 at 20:32
nanjunnanjun
44629
$endgroup$
add a comment |
$begingroup$
I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?
list-manipulation
asked Apr 9 at 20:32
nanjunnanjun
44629
$endgroup$
add a comment |
$begingroup$
I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?
list-manipulation
asked Apr 9 at 20:32
nanjunnanjun
44629
$endgroup$
I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?
list-manipulation
list-manipulation
asked Apr 9 at 20:32
nanjunnanjun
44629
asked Apr 9 at 20:32
nanjunnanjun
44629
asked Apr 9 at 20:32
nanjunnanjun
44629
asked Apr 9 at 20:32
nanjunnanjun
44629
asked Apr 9 at 20:32
nanjunnanjun
44629
44629
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
$endgroup$
add a comment |
$begingroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] = list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k582169
$endgroup$
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
$begingroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered Apr 9 at 21:14
Jason B.Jason B.
48.9k389197
$endgroup$
add a comment |
$begingroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered Apr 9 at 23:16
cphyscphys
815
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
$endgroup$
add a comment |
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
$endgroup$
add a comment |
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
$endgroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
answered Apr 9 at 20:33
marchmarch
17.6k22870
answered Apr 9 at 20:33
marchmarch
17.6k22870
answered Apr 9 at 20:33
marchmarch
17.6k22870
17.6k22870
add a comment |
add a comment |
$begingroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] = list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k582169
$endgroup$
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
$begingroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] = list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k582169
$endgroup$
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
$begingroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] = list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k582169
$endgroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] = list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k582169
edited Apr 10 at 6:12
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k582169
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k582169
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k582169
60.2k582169
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
2
2
$begingroup$
A bit cleaner, s/b comparable in speed:
Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]$endgroup$
– ciao
Apr 10 at 21:00
$begingroup$
A bit cleaner, s/b comparable in speed:
Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
$begingroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered Apr 9 at 21:14
Jason B.Jason B.
48.9k389197
$endgroup$
add a comment |
$begingroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered Apr 9 at 21:14
Jason B.Jason B.
48.9k389197
$endgroup$
add a comment |
$begingroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered Apr 9 at 21:14
Jason B.Jason B.
48.9k389197
$endgroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered Apr 9 at 21:14
Jason B.Jason B.
48.9k389197
answered Apr 9 at 21:14
Jason B.Jason B.
48.9k389197
answered Apr 9 at 21:14
Jason B.Jason B.
48.9k389197
answered Apr 9 at 21:14
Jason B.Jason B.
48.9k389197
48.9k389197
add a comment |
add a comment |
$begingroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered Apr 9 at 23:16
cphyscphys
815
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered Apr 9 at 23:16
cphyscphys
815
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered Apr 9 at 23:16
cphyscphys
815
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered Apr 9 at 23:16
cphyscphys
815
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 9 at 23:16
cphyscphys
815
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 9 at 23:16
cphyscphys
815
answered Apr 9 at 23:16
cphyscphys
815
815
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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