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Python “triplet” dictionary?
How to merge two dictionaries in a single expression?Calling an external command in PythonWhat are metaclasses in Python?How can I safely create a nested directory in Python?Does Python have a ternary conditional operator?How do I sort a dictionary by value?Add new keys to a dictionary?Check if a given key already exists in a dictionaryIterating over dictionaries using 'for' loopsDoes Python have a string 'contains' substring method?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
If we have (a1, b1)
and (a2, b2)
it's easy to use a dictionary to store the correspondences:
dict[a1] = b1
dict[a2] = b2
And we can get (a1, b1)
and (a2, b2)
back no problem.
But, if we have (a1, b1, c1)
and (a2, b2, c2)
, is it possible to get something like:
dict[a1] = (b1, c1)
dict[b1] = (a1, c1)
Where we can use either a1
or b1
to get the triplet (a1, b1, c2)
back? Does that make sense? I'm not quite sure which datatype to use for this problem. The above would work but there would be duplicate data.
Basically, if I have a triplet, which data type could I use such that I can use either the first or second value to get the triplet back?
python
|
show 4 more comments
If we have (a1, b1)
and (a2, b2)
it's easy to use a dictionary to store the correspondences:
dict[a1] = b1
dict[a2] = b2
And we can get (a1, b1)
and (a2, b2)
back no problem.
But, if we have (a1, b1, c1)
and (a2, b2, c2)
, is it possible to get something like:
dict[a1] = (b1, c1)
dict[b1] = (a1, c1)
Where we can use either a1
or b1
to get the triplet (a1, b1, c2)
back? Does that make sense? I'm not quite sure which datatype to use for this problem. The above would work but there would be duplicate data.
Basically, if I have a triplet, which data type could I use such that I can use either the first or second value to get the triplet back?
python
1
You want to be able to recover the triplet from any of the elements?
– Olivier Melançon
Apr 28 at 16:59
1
perfect time to learn how to make your own data types. see docs.python.org/3/library/stdtypes.html#typesmapping
– user633183
Apr 28 at 17:03
2
Quick hint: you would either need a custom mapping type (e.g. adict
subclass with__getitem__
overridden, at the very least) or you would need some custom container type that can wrap each item of your tuples and the tuples themselves such that they have the same__hash__
. I suspect you might have an X-Y problem though but hard to be sure.
– Iguananaut
Apr 28 at 17:05
6
I think it's an interesting problem. The answers given below are the most obvious but if you wanted to completely avoid duplication you'd have to do some tricks with hashes. I have to say, while it's a neat idea, in 25 years of programming I have never needed a data structure quite like this which is why I suspect an X-Y problem, but who knows stranger things have happened.
– Iguananaut
Apr 28 at 17:25
1
Apparently, this type of question needs a basic relational database as been pointed out here codereview.stackexchange.com/q/85842/127240 Consider usingsqlite
module for that. This is basicallySELECT * FROM table_foo WHERE b_value = 'b1';
which would probably be the same asSELECT * FROM table_foo WHERE a_value = 'a1';
And if I'm not mistaken a1 and b1 would be composite key for c1
– Sergiy Kolodyazhnyy
Apr 29 at 4:37
|
show 4 more comments
If we have (a1, b1)
and (a2, b2)
it's easy to use a dictionary to store the correspondences:
dict[a1] = b1
dict[a2] = b2
And we can get (a1, b1)
and (a2, b2)
back no problem.
But, if we have (a1, b1, c1)
and (a2, b2, c2)
, is it possible to get something like:
dict[a1] = (b1, c1)
dict[b1] = (a1, c1)
Where we can use either a1
or b1
to get the triplet (a1, b1, c2)
back? Does that make sense? I'm not quite sure which datatype to use for this problem. The above would work but there would be duplicate data.
Basically, if I have a triplet, which data type could I use such that I can use either the first or second value to get the triplet back?
python
If we have (a1, b1)
and (a2, b2)
it's easy to use a dictionary to store the correspondences:
dict[a1] = b1
dict[a2] = b2
And we can get (a1, b1)
and (a2, b2)
back no problem.
But, if we have (a1, b1, c1)
and (a2, b2, c2)
, is it possible to get something like:
dict[a1] = (b1, c1)
dict[b1] = (a1, c1)
Where we can use either a1
or b1
to get the triplet (a1, b1, c2)
back? Does that make sense? I'm not quite sure which datatype to use for this problem. The above would work but there would be duplicate data.
Basically, if I have a triplet, which data type could I use such that I can use either the first or second value to get the triplet back?
python
python
asked Apr 28 at 16:58
JustinJustin
3,30922550
3,30922550
1
You want to be able to recover the triplet from any of the elements?
– Olivier Melançon
Apr 28 at 16:59
1
perfect time to learn how to make your own data types. see docs.python.org/3/library/stdtypes.html#typesmapping
– user633183
Apr 28 at 17:03
2
Quick hint: you would either need a custom mapping type (e.g. adict
subclass with__getitem__
overridden, at the very least) or you would need some custom container type that can wrap each item of your tuples and the tuples themselves such that they have the same__hash__
. I suspect you might have an X-Y problem though but hard to be sure.
– Iguananaut
Apr 28 at 17:05
6
I think it's an interesting problem. The answers given below are the most obvious but if you wanted to completely avoid duplication you'd have to do some tricks with hashes. I have to say, while it's a neat idea, in 25 years of programming I have never needed a data structure quite like this which is why I suspect an X-Y problem, but who knows stranger things have happened.
– Iguananaut
Apr 28 at 17:25
1
Apparently, this type of question needs a basic relational database as been pointed out here codereview.stackexchange.com/q/85842/127240 Consider usingsqlite
module for that. This is basicallySELECT * FROM table_foo WHERE b_value = 'b1';
which would probably be the same asSELECT * FROM table_foo WHERE a_value = 'a1';
And if I'm not mistaken a1 and b1 would be composite key for c1
– Sergiy Kolodyazhnyy
Apr 29 at 4:37
|
show 4 more comments
1
You want to be able to recover the triplet from any of the elements?
– Olivier Melançon
Apr 28 at 16:59
1
perfect time to learn how to make your own data types. see docs.python.org/3/library/stdtypes.html#typesmapping
– user633183
Apr 28 at 17:03
2
Quick hint: you would either need a custom mapping type (e.g. adict
subclass with__getitem__
overridden, at the very least) or you would need some custom container type that can wrap each item of your tuples and the tuples themselves such that they have the same__hash__
. I suspect you might have an X-Y problem though but hard to be sure.
– Iguananaut
Apr 28 at 17:05
6
I think it's an interesting problem. The answers given below are the most obvious but if you wanted to completely avoid duplication you'd have to do some tricks with hashes. I have to say, while it's a neat idea, in 25 years of programming I have never needed a data structure quite like this which is why I suspect an X-Y problem, but who knows stranger things have happened.
– Iguananaut
Apr 28 at 17:25
1
Apparently, this type of question needs a basic relational database as been pointed out here codereview.stackexchange.com/q/85842/127240 Consider usingsqlite
module for that. This is basicallySELECT * FROM table_foo WHERE b_value = 'b1';
which would probably be the same asSELECT * FROM table_foo WHERE a_value = 'a1';
And if I'm not mistaken a1 and b1 would be composite key for c1
– Sergiy Kolodyazhnyy
Apr 29 at 4:37
1
1
You want to be able to recover the triplet from any of the elements?
– Olivier Melançon
Apr 28 at 16:59
You want to be able to recover the triplet from any of the elements?
– Olivier Melançon
Apr 28 at 16:59
1
1
perfect time to learn how to make your own data types. see docs.python.org/3/library/stdtypes.html#typesmapping
– user633183
Apr 28 at 17:03
perfect time to learn how to make your own data types. see docs.python.org/3/library/stdtypes.html#typesmapping
– user633183
Apr 28 at 17:03
2
2
Quick hint: you would either need a custom mapping type (e.g. a
dict
subclass with __getitem__
overridden, at the very least) or you would need some custom container type that can wrap each item of your tuples and the tuples themselves such that they have the same __hash__
. I suspect you might have an X-Y problem though but hard to be sure.– Iguananaut
Apr 28 at 17:05
Quick hint: you would either need a custom mapping type (e.g. a
dict
subclass with __getitem__
overridden, at the very least) or you would need some custom container type that can wrap each item of your tuples and the tuples themselves such that they have the same __hash__
. I suspect you might have an X-Y problem though but hard to be sure.– Iguananaut
Apr 28 at 17:05
6
6
I think it's an interesting problem. The answers given below are the most obvious but if you wanted to completely avoid duplication you'd have to do some tricks with hashes. I have to say, while it's a neat idea, in 25 years of programming I have never needed a data structure quite like this which is why I suspect an X-Y problem, but who knows stranger things have happened.
– Iguananaut
Apr 28 at 17:25
I think it's an interesting problem. The answers given below are the most obvious but if you wanted to completely avoid duplication you'd have to do some tricks with hashes. I have to say, while it's a neat idea, in 25 years of programming I have never needed a data structure quite like this which is why I suspect an X-Y problem, but who knows stranger things have happened.
– Iguananaut
Apr 28 at 17:25
1
1
Apparently, this type of question needs a basic relational database as been pointed out here codereview.stackexchange.com/q/85842/127240 Consider using
sqlite
module for that. This is basically SELECT * FROM table_foo WHERE b_value = 'b1';
which would probably be the same as SELECT * FROM table_foo WHERE a_value = 'a1';
And if I'm not mistaken a1 and b1 would be composite key for c1– Sergiy Kolodyazhnyy
Apr 29 at 4:37
Apparently, this type of question needs a basic relational database as been pointed out here codereview.stackexchange.com/q/85842/127240 Consider using
sqlite
module for that. This is basically SELECT * FROM table_foo WHERE b_value = 'b1';
which would probably be the same as SELECT * FROM table_foo WHERE a_value = 'a1';
And if I'm not mistaken a1 and b1 would be composite key for c1– Sergiy Kolodyazhnyy
Apr 29 at 4:37
|
show 4 more comments
5 Answers
5
active
oldest
votes
Solution
You can write your own mapping data-structure which allows to add
triples, or groups of any size, and recover the group with __getitem__
.
class GroupMap:
def __init__(self):
self.data =
def add(self, group):
for item in group:
self.data[item] = group
def __getitem__(self, item):
return self.data[item]
group = (1, 2, 3)
group_map = GroupMap()
group_map.add(group)
print(group_map[1]) # (1, 2, 3)
Notice that this GroupMap
can be used for groups of any size, not only triples.
The next step in the above would be to extend the class to avoid collisions according to the behaviour you want when a collision occurs.
Theory
You might wonder if there is a better way to represent groups of connected objects. The answer is not really.
Suppose you have a graph containing n vertices. Then for the graph to be connected, you must have at least n - 1 edges. In the above data-structure, I used n entry in the dict
, meaning the solution is nearly optimal.
Why not use n - 1 entries if it can be done? Because you would then need to traverse all your graph to recover the whole group. Thus using one more edge allows for O(1) lookup, which is a trade-off you probably want to take.
Wouldn't this still duplicate the information in the same way?
– MyNameIsCaleb
Apr 28 at 17:09
1
@GiraffeMan91 Yes, but there is no other way. A connexion between two items must be stored one way of another. Another way to visualize this is that a graph of n items must have at least (n-1) vertices to be connected. Here I use two more than that, but this is to allow fast lookup.
– Olivier Melançon
Apr 28 at 17:12
Yea I understand I just wanted to be clear that we agreed. I thought the primary element of the question was to reduce duplication which I don't believe can be done either, I just thought your answer turned what the post already said they could do into a class with methods, which is written nicely, but I felt the post already explained they knew how to do what your answer provided and they were looking for other methods.
– MyNameIsCaleb
Apr 28 at 17:22
@GiraffeMan91 Fair point, I edited the answer to explain why we cannot do better than that.
– Olivier Melançon
Apr 28 at 17:26
1
@GiraffeMan91 The problem I see with traversal is that it requires implementing some kind of graph data-structure which 1) slows down everything and 2) is a bit overly complex in my opinion for what we want to achieve.
– Olivier Melançon
Apr 28 at 17:33
|
show 5 more comments
An alternative if you want to subclass dict
(to get all other methods associated with dict
like .get
and whatnot) and only get the other elements when asked (for some reason). You can make a new dictionary that is all your own
class TupleDict(dict):
def __setitem__(self, key, value):
assert isinstance(key, tuple)
for i, e in enumerate(key):
dict.__setitem__(self, e, key[:i] + key[i+1:] + (value,))
dict.__setitem__(self, value, key)
and then assign any key that is a tuple to a single value (not sure I like this syntax, but we can make it different or use a stand-alone method)
d = TriDict()
d[(1,2)] = 4
and you will have the result of __getitem__
return the rest of the tuple not present.
>>> print(d[1])
(2, 4)
>>> print(d[2])
(1, 4)
print(d[4])
>>> (1, 2)
add a comment |
Building on Olivier Melançons answer, I came up with this - in case the position of the value in the tuple matters:
class GroupMap:
def __init__(self, data=None):
self.data =
if data:
self.add(data)
def add(self, data):
for idx, key in enumerate(data):
self.data.setdefault(idx, )[key] = data
def __getitem__(self, key):
# lookup in first index
return self.getby(0, key)
def getby(self, idx, key):
return self.data[idx].get(key)
data = ('a', 'b', 'c')
g = GroupMap(data)
more_data = ('b', 'a', 'z')
g.add(more_data)
assert g['a'] == data
assert g.getby(0, 'a') == data
assert g.getby(0, 'b') == more_data
assert g.getby(0, 'c') is None
assert g.getby(1, 'a') == more_data
assert g.getby(1, 'b') == data
assert g.getby(2, 'c') == data
assert g.getby(2, 'z') == more_data
assert id(data) == id(g['a']) == id(g.getby(1, 'b'))
add a comment |
Dictionaries can store key value pairs only.
You could make your own triplet dictionary however using operator overloading so that when you index with any member of the triplets you get back the other two, maybe something like this:
class trictionary:
def __init__(self):
self.data = []
def add(self, group):
self.data.append(group)
def __getitem__(self, key):
for group in data: #Find the set the key belongs to.
if key in group:
return tuple(group)
This avoids replicating the data and has the functionality you're looking for at the expense of performance. There may be a better way of doing the same.
3
I love "trictionary"
– modesitt
Apr 28 at 17:16
That was probably the part of my answer with the most merit...
– Eden Trainor
Apr 28 at 17:34
add a comment |
Your question has examples that deviate from the main question:
Basically, if I have a triplet, which data type could I use such that I can use either the first or second value to get the triplet back?
A dict. Assign key-value pairs of element
, triplet
(see @Olivier Melançon's answer):
Code
d =
for x in triplet:
d[x] = triplet
Demo
d["a"]
# ('a', 'b', 'c')
d["b"]
# ('a', 'b', 'c')
d["c"]
# ('a', 'b', 'c')
The OP requires clarity on preferred behavior in:
- adding elements, e.g.
d[a1] = (b1, c1)
vs.f((a1, b1, c1))
- element ordering, e.g.
(a1, b1, c1)
vs.(b1, a1, c1)
- duplicate data, e.g. hold
(a1, b1, c1)
three times or combinations of 2-tuple subsets
With these items addressed, more elaborate solutions are possible.
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Solution
You can write your own mapping data-structure which allows to add
triples, or groups of any size, and recover the group with __getitem__
.
class GroupMap:
def __init__(self):
self.data =
def add(self, group):
for item in group:
self.data[item] = group
def __getitem__(self, item):
return self.data[item]
group = (1, 2, 3)
group_map = GroupMap()
group_map.add(group)
print(group_map[1]) # (1, 2, 3)
Notice that this GroupMap
can be used for groups of any size, not only triples.
The next step in the above would be to extend the class to avoid collisions according to the behaviour you want when a collision occurs.
Theory
You might wonder if there is a better way to represent groups of connected objects. The answer is not really.
Suppose you have a graph containing n vertices. Then for the graph to be connected, you must have at least n - 1 edges. In the above data-structure, I used n entry in the dict
, meaning the solution is nearly optimal.
Why not use n - 1 entries if it can be done? Because you would then need to traverse all your graph to recover the whole group. Thus using one more edge allows for O(1) lookup, which is a trade-off you probably want to take.
Wouldn't this still duplicate the information in the same way?
– MyNameIsCaleb
Apr 28 at 17:09
1
@GiraffeMan91 Yes, but there is no other way. A connexion between two items must be stored one way of another. Another way to visualize this is that a graph of n items must have at least (n-1) vertices to be connected. Here I use two more than that, but this is to allow fast lookup.
– Olivier Melançon
Apr 28 at 17:12
Yea I understand I just wanted to be clear that we agreed. I thought the primary element of the question was to reduce duplication which I don't believe can be done either, I just thought your answer turned what the post already said they could do into a class with methods, which is written nicely, but I felt the post already explained they knew how to do what your answer provided and they were looking for other methods.
– MyNameIsCaleb
Apr 28 at 17:22
@GiraffeMan91 Fair point, I edited the answer to explain why we cannot do better than that.
– Olivier Melançon
Apr 28 at 17:26
1
@GiraffeMan91 The problem I see with traversal is that it requires implementing some kind of graph data-structure which 1) slows down everything and 2) is a bit overly complex in my opinion for what we want to achieve.
– Olivier Melançon
Apr 28 at 17:33
|
show 5 more comments
Solution
You can write your own mapping data-structure which allows to add
triples, or groups of any size, and recover the group with __getitem__
.
class GroupMap:
def __init__(self):
self.data =
def add(self, group):
for item in group:
self.data[item] = group
def __getitem__(self, item):
return self.data[item]
group = (1, 2, 3)
group_map = GroupMap()
group_map.add(group)
print(group_map[1]) # (1, 2, 3)
Notice that this GroupMap
can be used for groups of any size, not only triples.
The next step in the above would be to extend the class to avoid collisions according to the behaviour you want when a collision occurs.
Theory
You might wonder if there is a better way to represent groups of connected objects. The answer is not really.
Suppose you have a graph containing n vertices. Then for the graph to be connected, you must have at least n - 1 edges. In the above data-structure, I used n entry in the dict
, meaning the solution is nearly optimal.
Why not use n - 1 entries if it can be done? Because you would then need to traverse all your graph to recover the whole group. Thus using one more edge allows for O(1) lookup, which is a trade-off you probably want to take.
Wouldn't this still duplicate the information in the same way?
– MyNameIsCaleb
Apr 28 at 17:09
1
@GiraffeMan91 Yes, but there is no other way. A connexion between two items must be stored one way of another. Another way to visualize this is that a graph of n items must have at least (n-1) vertices to be connected. Here I use two more than that, but this is to allow fast lookup.
– Olivier Melançon
Apr 28 at 17:12
Yea I understand I just wanted to be clear that we agreed. I thought the primary element of the question was to reduce duplication which I don't believe can be done either, I just thought your answer turned what the post already said they could do into a class with methods, which is written nicely, but I felt the post already explained they knew how to do what your answer provided and they were looking for other methods.
– MyNameIsCaleb
Apr 28 at 17:22
@GiraffeMan91 Fair point, I edited the answer to explain why we cannot do better than that.
– Olivier Melançon
Apr 28 at 17:26
1
@GiraffeMan91 The problem I see with traversal is that it requires implementing some kind of graph data-structure which 1) slows down everything and 2) is a bit overly complex in my opinion for what we want to achieve.
– Olivier Melançon
Apr 28 at 17:33
|
show 5 more comments
Solution
You can write your own mapping data-structure which allows to add
triples, or groups of any size, and recover the group with __getitem__
.
class GroupMap:
def __init__(self):
self.data =
def add(self, group):
for item in group:
self.data[item] = group
def __getitem__(self, item):
return self.data[item]
group = (1, 2, 3)
group_map = GroupMap()
group_map.add(group)
print(group_map[1]) # (1, 2, 3)
Notice that this GroupMap
can be used for groups of any size, not only triples.
The next step in the above would be to extend the class to avoid collisions according to the behaviour you want when a collision occurs.
Theory
You might wonder if there is a better way to represent groups of connected objects. The answer is not really.
Suppose you have a graph containing n vertices. Then for the graph to be connected, you must have at least n - 1 edges. In the above data-structure, I used n entry in the dict
, meaning the solution is nearly optimal.
Why not use n - 1 entries if it can be done? Because you would then need to traverse all your graph to recover the whole group. Thus using one more edge allows for O(1) lookup, which is a trade-off you probably want to take.
Solution
You can write your own mapping data-structure which allows to add
triples, or groups of any size, and recover the group with __getitem__
.
class GroupMap:
def __init__(self):
self.data =
def add(self, group):
for item in group:
self.data[item] = group
def __getitem__(self, item):
return self.data[item]
group = (1, 2, 3)
group_map = GroupMap()
group_map.add(group)
print(group_map[1]) # (1, 2, 3)
Notice that this GroupMap
can be used for groups of any size, not only triples.
The next step in the above would be to extend the class to avoid collisions according to the behaviour you want when a collision occurs.
Theory
You might wonder if there is a better way to represent groups of connected objects. The answer is not really.
Suppose you have a graph containing n vertices. Then for the graph to be connected, you must have at least n - 1 edges. In the above data-structure, I used n entry in the dict
, meaning the solution is nearly optimal.
Why not use n - 1 entries if it can be done? Because you would then need to traverse all your graph to recover the whole group. Thus using one more edge allows for O(1) lookup, which is a trade-off you probably want to take.
edited Apr 28 at 17:36
answered Apr 28 at 17:07
Olivier MelançonOlivier Melançon
14.7k32144
14.7k32144
Wouldn't this still duplicate the information in the same way?
– MyNameIsCaleb
Apr 28 at 17:09
1
@GiraffeMan91 Yes, but there is no other way. A connexion between two items must be stored one way of another. Another way to visualize this is that a graph of n items must have at least (n-1) vertices to be connected. Here I use two more than that, but this is to allow fast lookup.
– Olivier Melançon
Apr 28 at 17:12
Yea I understand I just wanted to be clear that we agreed. I thought the primary element of the question was to reduce duplication which I don't believe can be done either, I just thought your answer turned what the post already said they could do into a class with methods, which is written nicely, but I felt the post already explained they knew how to do what your answer provided and they were looking for other methods.
– MyNameIsCaleb
Apr 28 at 17:22
@GiraffeMan91 Fair point, I edited the answer to explain why we cannot do better than that.
– Olivier Melançon
Apr 28 at 17:26
1
@GiraffeMan91 The problem I see with traversal is that it requires implementing some kind of graph data-structure which 1) slows down everything and 2) is a bit overly complex in my opinion for what we want to achieve.
– Olivier Melançon
Apr 28 at 17:33
|
show 5 more comments
Wouldn't this still duplicate the information in the same way?
– MyNameIsCaleb
Apr 28 at 17:09
1
@GiraffeMan91 Yes, but there is no other way. A connexion between two items must be stored one way of another. Another way to visualize this is that a graph of n items must have at least (n-1) vertices to be connected. Here I use two more than that, but this is to allow fast lookup.
– Olivier Melançon
Apr 28 at 17:12
Yea I understand I just wanted to be clear that we agreed. I thought the primary element of the question was to reduce duplication which I don't believe can be done either, I just thought your answer turned what the post already said they could do into a class with methods, which is written nicely, but I felt the post already explained they knew how to do what your answer provided and they were looking for other methods.
– MyNameIsCaleb
Apr 28 at 17:22
@GiraffeMan91 Fair point, I edited the answer to explain why we cannot do better than that.
– Olivier Melançon
Apr 28 at 17:26
1
@GiraffeMan91 The problem I see with traversal is that it requires implementing some kind of graph data-structure which 1) slows down everything and 2) is a bit overly complex in my opinion for what we want to achieve.
– Olivier Melançon
Apr 28 at 17:33
Wouldn't this still duplicate the information in the same way?
– MyNameIsCaleb
Apr 28 at 17:09
Wouldn't this still duplicate the information in the same way?
– MyNameIsCaleb
Apr 28 at 17:09
1
1
@GiraffeMan91 Yes, but there is no other way. A connexion between two items must be stored one way of another. Another way to visualize this is that a graph of n items must have at least (n-1) vertices to be connected. Here I use two more than that, but this is to allow fast lookup.
– Olivier Melançon
Apr 28 at 17:12
@GiraffeMan91 Yes, but there is no other way. A connexion between two items must be stored one way of another. Another way to visualize this is that a graph of n items must have at least (n-1) vertices to be connected. Here I use two more than that, but this is to allow fast lookup.
– Olivier Melançon
Apr 28 at 17:12
Yea I understand I just wanted to be clear that we agreed. I thought the primary element of the question was to reduce duplication which I don't believe can be done either, I just thought your answer turned what the post already said they could do into a class with methods, which is written nicely, but I felt the post already explained they knew how to do what your answer provided and they were looking for other methods.
– MyNameIsCaleb
Apr 28 at 17:22
Yea I understand I just wanted to be clear that we agreed. I thought the primary element of the question was to reduce duplication which I don't believe can be done either, I just thought your answer turned what the post already said they could do into a class with methods, which is written nicely, but I felt the post already explained they knew how to do what your answer provided and they were looking for other methods.
– MyNameIsCaleb
Apr 28 at 17:22
@GiraffeMan91 Fair point, I edited the answer to explain why we cannot do better than that.
– Olivier Melançon
Apr 28 at 17:26
@GiraffeMan91 Fair point, I edited the answer to explain why we cannot do better than that.
– Olivier Melançon
Apr 28 at 17:26
1
1
@GiraffeMan91 The problem I see with traversal is that it requires implementing some kind of graph data-structure which 1) slows down everything and 2) is a bit overly complex in my opinion for what we want to achieve.
– Olivier Melançon
Apr 28 at 17:33
@GiraffeMan91 The problem I see with traversal is that it requires implementing some kind of graph data-structure which 1) slows down everything and 2) is a bit overly complex in my opinion for what we want to achieve.
– Olivier Melançon
Apr 28 at 17:33
|
show 5 more comments
An alternative if you want to subclass dict
(to get all other methods associated with dict
like .get
and whatnot) and only get the other elements when asked (for some reason). You can make a new dictionary that is all your own
class TupleDict(dict):
def __setitem__(self, key, value):
assert isinstance(key, tuple)
for i, e in enumerate(key):
dict.__setitem__(self, e, key[:i] + key[i+1:] + (value,))
dict.__setitem__(self, value, key)
and then assign any key that is a tuple to a single value (not sure I like this syntax, but we can make it different or use a stand-alone method)
d = TriDict()
d[(1,2)] = 4
and you will have the result of __getitem__
return the rest of the tuple not present.
>>> print(d[1])
(2, 4)
>>> print(d[2])
(1, 4)
print(d[4])
>>> (1, 2)
add a comment |
An alternative if you want to subclass dict
(to get all other methods associated with dict
like .get
and whatnot) and only get the other elements when asked (for some reason). You can make a new dictionary that is all your own
class TupleDict(dict):
def __setitem__(self, key, value):
assert isinstance(key, tuple)
for i, e in enumerate(key):
dict.__setitem__(self, e, key[:i] + key[i+1:] + (value,))
dict.__setitem__(self, value, key)
and then assign any key that is a tuple to a single value (not sure I like this syntax, but we can make it different or use a stand-alone method)
d = TriDict()
d[(1,2)] = 4
and you will have the result of __getitem__
return the rest of the tuple not present.
>>> print(d[1])
(2, 4)
>>> print(d[2])
(1, 4)
print(d[4])
>>> (1, 2)
add a comment |
An alternative if you want to subclass dict
(to get all other methods associated with dict
like .get
and whatnot) and only get the other elements when asked (for some reason). You can make a new dictionary that is all your own
class TupleDict(dict):
def __setitem__(self, key, value):
assert isinstance(key, tuple)
for i, e in enumerate(key):
dict.__setitem__(self, e, key[:i] + key[i+1:] + (value,))
dict.__setitem__(self, value, key)
and then assign any key that is a tuple to a single value (not sure I like this syntax, but we can make it different or use a stand-alone method)
d = TriDict()
d[(1,2)] = 4
and you will have the result of __getitem__
return the rest of the tuple not present.
>>> print(d[1])
(2, 4)
>>> print(d[2])
(1, 4)
print(d[4])
>>> (1, 2)
An alternative if you want to subclass dict
(to get all other methods associated with dict
like .get
and whatnot) and only get the other elements when asked (for some reason). You can make a new dictionary that is all your own
class TupleDict(dict):
def __setitem__(self, key, value):
assert isinstance(key, tuple)
for i, e in enumerate(key):
dict.__setitem__(self, e, key[:i] + key[i+1:] + (value,))
dict.__setitem__(self, value, key)
and then assign any key that is a tuple to a single value (not sure I like this syntax, but we can make it different or use a stand-alone method)
d = TriDict()
d[(1,2)] = 4
and you will have the result of __getitem__
return the rest of the tuple not present.
>>> print(d[1])
(2, 4)
>>> print(d[2])
(1, 4)
print(d[4])
>>> (1, 2)
edited Apr 28 at 17:26
answered Apr 28 at 17:11
modesittmodesitt
3,17721745
3,17721745
add a comment |
add a comment |
Building on Olivier Melançons answer, I came up with this - in case the position of the value in the tuple matters:
class GroupMap:
def __init__(self, data=None):
self.data =
if data:
self.add(data)
def add(self, data):
for idx, key in enumerate(data):
self.data.setdefault(idx, )[key] = data
def __getitem__(self, key):
# lookup in first index
return self.getby(0, key)
def getby(self, idx, key):
return self.data[idx].get(key)
data = ('a', 'b', 'c')
g = GroupMap(data)
more_data = ('b', 'a', 'z')
g.add(more_data)
assert g['a'] == data
assert g.getby(0, 'a') == data
assert g.getby(0, 'b') == more_data
assert g.getby(0, 'c') is None
assert g.getby(1, 'a') == more_data
assert g.getby(1, 'b') == data
assert g.getby(2, 'c') == data
assert g.getby(2, 'z') == more_data
assert id(data) == id(g['a']) == id(g.getby(1, 'b'))
add a comment |
Building on Olivier Melançons answer, I came up with this - in case the position of the value in the tuple matters:
class GroupMap:
def __init__(self, data=None):
self.data =
if data:
self.add(data)
def add(self, data):
for idx, key in enumerate(data):
self.data.setdefault(idx, )[key] = data
def __getitem__(self, key):
# lookup in first index
return self.getby(0, key)
def getby(self, idx, key):
return self.data[idx].get(key)
data = ('a', 'b', 'c')
g = GroupMap(data)
more_data = ('b', 'a', 'z')
g.add(more_data)
assert g['a'] == data
assert g.getby(0, 'a') == data
assert g.getby(0, 'b') == more_data
assert g.getby(0, 'c') is None
assert g.getby(1, 'a') == more_data
assert g.getby(1, 'b') == data
assert g.getby(2, 'c') == data
assert g.getby(2, 'z') == more_data
assert id(data) == id(g['a']) == id(g.getby(1, 'b'))
add a comment |
Building on Olivier Melançons answer, I came up with this - in case the position of the value in the tuple matters:
class GroupMap:
def __init__(self, data=None):
self.data =
if data:
self.add(data)
def add(self, data):
for idx, key in enumerate(data):
self.data.setdefault(idx, )[key] = data
def __getitem__(self, key):
# lookup in first index
return self.getby(0, key)
def getby(self, idx, key):
return self.data[idx].get(key)
data = ('a', 'b', 'c')
g = GroupMap(data)
more_data = ('b', 'a', 'z')
g.add(more_data)
assert g['a'] == data
assert g.getby(0, 'a') == data
assert g.getby(0, 'b') == more_data
assert g.getby(0, 'c') is None
assert g.getby(1, 'a') == more_data
assert g.getby(1, 'b') == data
assert g.getby(2, 'c') == data
assert g.getby(2, 'z') == more_data
assert id(data) == id(g['a']) == id(g.getby(1, 'b'))
Building on Olivier Melançons answer, I came up with this - in case the position of the value in the tuple matters:
class GroupMap:
def __init__(self, data=None):
self.data =
if data:
self.add(data)
def add(self, data):
for idx, key in enumerate(data):
self.data.setdefault(idx, )[key] = data
def __getitem__(self, key):
# lookup in first index
return self.getby(0, key)
def getby(self, idx, key):
return self.data[idx].get(key)
data = ('a', 'b', 'c')
g = GroupMap(data)
more_data = ('b', 'a', 'z')
g.add(more_data)
assert g['a'] == data
assert g.getby(0, 'a') == data
assert g.getby(0, 'b') == more_data
assert g.getby(0, 'c') is None
assert g.getby(1, 'a') == more_data
assert g.getby(1, 'b') == data
assert g.getby(2, 'c') == data
assert g.getby(2, 'z') == more_data
assert id(data) == id(g['a']) == id(g.getby(1, 'b'))
answered Apr 28 at 18:40
CloudomationCloudomation
1,0101111
1,0101111
add a comment |
add a comment |
Dictionaries can store key value pairs only.
You could make your own triplet dictionary however using operator overloading so that when you index with any member of the triplets you get back the other two, maybe something like this:
class trictionary:
def __init__(self):
self.data = []
def add(self, group):
self.data.append(group)
def __getitem__(self, key):
for group in data: #Find the set the key belongs to.
if key in group:
return tuple(group)
This avoids replicating the data and has the functionality you're looking for at the expense of performance. There may be a better way of doing the same.
3
I love "trictionary"
– modesitt
Apr 28 at 17:16
That was probably the part of my answer with the most merit...
– Eden Trainor
Apr 28 at 17:34
add a comment |
Dictionaries can store key value pairs only.
You could make your own triplet dictionary however using operator overloading so that when you index with any member of the triplets you get back the other two, maybe something like this:
class trictionary:
def __init__(self):
self.data = []
def add(self, group):
self.data.append(group)
def __getitem__(self, key):
for group in data: #Find the set the key belongs to.
if key in group:
return tuple(group)
This avoids replicating the data and has the functionality you're looking for at the expense of performance. There may be a better way of doing the same.
3
I love "trictionary"
– modesitt
Apr 28 at 17:16
That was probably the part of my answer with the most merit...
– Eden Trainor
Apr 28 at 17:34
add a comment |
Dictionaries can store key value pairs only.
You could make your own triplet dictionary however using operator overloading so that when you index with any member of the triplets you get back the other two, maybe something like this:
class trictionary:
def __init__(self):
self.data = []
def add(self, group):
self.data.append(group)
def __getitem__(self, key):
for group in data: #Find the set the key belongs to.
if key in group:
return tuple(group)
This avoids replicating the data and has the functionality you're looking for at the expense of performance. There may be a better way of doing the same.
Dictionaries can store key value pairs only.
You could make your own triplet dictionary however using operator overloading so that when you index with any member of the triplets you get back the other two, maybe something like this:
class trictionary:
def __init__(self):
self.data = []
def add(self, group):
self.data.append(group)
def __getitem__(self, key):
for group in data: #Find the set the key belongs to.
if key in group:
return tuple(group)
This avoids replicating the data and has the functionality you're looking for at the expense of performance. There may be a better way of doing the same.
answered Apr 28 at 17:16
Eden TrainorEden Trainor
246
246
3
I love "trictionary"
– modesitt
Apr 28 at 17:16
That was probably the part of my answer with the most merit...
– Eden Trainor
Apr 28 at 17:34
add a comment |
3
I love "trictionary"
– modesitt
Apr 28 at 17:16
That was probably the part of my answer with the most merit...
– Eden Trainor
Apr 28 at 17:34
3
3
I love "trictionary"
– modesitt
Apr 28 at 17:16
I love "trictionary"
– modesitt
Apr 28 at 17:16
That was probably the part of my answer with the most merit...
– Eden Trainor
Apr 28 at 17:34
That was probably the part of my answer with the most merit...
– Eden Trainor
Apr 28 at 17:34
add a comment |
Your question has examples that deviate from the main question:
Basically, if I have a triplet, which data type could I use such that I can use either the first or second value to get the triplet back?
A dict. Assign key-value pairs of element
, triplet
(see @Olivier Melançon's answer):
Code
d =
for x in triplet:
d[x] = triplet
Demo
d["a"]
# ('a', 'b', 'c')
d["b"]
# ('a', 'b', 'c')
d["c"]
# ('a', 'b', 'c')
The OP requires clarity on preferred behavior in:
- adding elements, e.g.
d[a1] = (b1, c1)
vs.f((a1, b1, c1))
- element ordering, e.g.
(a1, b1, c1)
vs.(b1, a1, c1)
- duplicate data, e.g. hold
(a1, b1, c1)
three times or combinations of 2-tuple subsets
With these items addressed, more elaborate solutions are possible.
add a comment |
Your question has examples that deviate from the main question:
Basically, if I have a triplet, which data type could I use such that I can use either the first or second value to get the triplet back?
A dict. Assign key-value pairs of element
, triplet
(see @Olivier Melançon's answer):
Code
d =
for x in triplet:
d[x] = triplet
Demo
d["a"]
# ('a', 'b', 'c')
d["b"]
# ('a', 'b', 'c')
d["c"]
# ('a', 'b', 'c')
The OP requires clarity on preferred behavior in:
- adding elements, e.g.
d[a1] = (b1, c1)
vs.f((a1, b1, c1))
- element ordering, e.g.
(a1, b1, c1)
vs.(b1, a1, c1)
- duplicate data, e.g. hold
(a1, b1, c1)
three times or combinations of 2-tuple subsets
With these items addressed, more elaborate solutions are possible.
add a comment |
Your question has examples that deviate from the main question:
Basically, if I have a triplet, which data type could I use such that I can use either the first or second value to get the triplet back?
A dict. Assign key-value pairs of element
, triplet
(see @Olivier Melançon's answer):
Code
d =
for x in triplet:
d[x] = triplet
Demo
d["a"]
# ('a', 'b', 'c')
d["b"]
# ('a', 'b', 'c')
d["c"]
# ('a', 'b', 'c')
The OP requires clarity on preferred behavior in:
- adding elements, e.g.
d[a1] = (b1, c1)
vs.f((a1, b1, c1))
- element ordering, e.g.
(a1, b1, c1)
vs.(b1, a1, c1)
- duplicate data, e.g. hold
(a1, b1, c1)
three times or combinations of 2-tuple subsets
With these items addressed, more elaborate solutions are possible.
Your question has examples that deviate from the main question:
Basically, if I have a triplet, which data type could I use such that I can use either the first or second value to get the triplet back?
A dict. Assign key-value pairs of element
, triplet
(see @Olivier Melançon's answer):
Code
d =
for x in triplet:
d[x] = triplet
Demo
d["a"]
# ('a', 'b', 'c')
d["b"]
# ('a', 'b', 'c')
d["c"]
# ('a', 'b', 'c')
The OP requires clarity on preferred behavior in:
- adding elements, e.g.
d[a1] = (b1, c1)
vs.f((a1, b1, c1))
- element ordering, e.g.
(a1, b1, c1)
vs.(b1, a1, c1)
- duplicate data, e.g. hold
(a1, b1, c1)
three times or combinations of 2-tuple subsets
With these items addressed, more elaborate solutions are possible.
edited Apr 29 at 18:27
answered Apr 29 at 18:08
pylangpylang
14.8k24558
14.8k24558
add a comment |
add a comment |
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1
You want to be able to recover the triplet from any of the elements?
– Olivier Melançon
Apr 28 at 16:59
1
perfect time to learn how to make your own data types. see docs.python.org/3/library/stdtypes.html#typesmapping
– user633183
Apr 28 at 17:03
2
Quick hint: you would either need a custom mapping type (e.g. a
dict
subclass with__getitem__
overridden, at the very least) or you would need some custom container type that can wrap each item of your tuples and the tuples themselves such that they have the same__hash__
. I suspect you might have an X-Y problem though but hard to be sure.– Iguananaut
Apr 28 at 17:05
6
I think it's an interesting problem. The answers given below are the most obvious but if you wanted to completely avoid duplication you'd have to do some tricks with hashes. I have to say, while it's a neat idea, in 25 years of programming I have never needed a data structure quite like this which is why I suspect an X-Y problem, but who knows stranger things have happened.
– Iguananaut
Apr 28 at 17:25
1
Apparently, this type of question needs a basic relational database as been pointed out here codereview.stackexchange.com/q/85842/127240 Consider using
sqlite
module for that. This is basicallySELECT * FROM table_foo WHERE b_value = 'b1';
which would probably be the same asSELECT * FROM table_foo WHERE a_value = 'a1';
And if I'm not mistaken a1 and b1 would be composite key for c1– Sergiy Kolodyazhnyy
Apr 29 at 4:37