Combinations of unique products with itertools
up vote
1
down vote
favorite
I have a nested list and would like to make a product of two items.
test = [[('juice', 'NOUN'), ('orange', 'FLAVOR')],
[('juice', 'NOUN'), ('orange', 'FLAVOR'), ('lemon', 'FLAVOR')],
[('orange', 'FLAVOR'), ('chip', 'NOUN')]]
What I expect is something like this:
[(('juice', 'NOUN'), ('lemon', 'FLAVOR')),
(('juice', 'NOUN'), ('chip', 'NOUN')),
(('orange', 'FLAVOR'), ('lemon', 'FLAVOR')),
(('orange', 'FLAVOR'), ('chip', 'NOUN')),
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))]
That is to say, I would like to get the permutation across lists but only for unique items. I prefer to use itertools. Previously, I tried list(itertools.product(*test)) But I realized it would produce the product of the length of a nested list...
My current code:
unique_list = list(set(itertools.chain(*test)))
list(itertools.combinations(unique_list, 2))
My thought process is to get the unique items in the nested list first, so the nested list will be [[('juice', 'NOUN'), ('orange', 'FLAVOR')], [('lemon', 'FLAVOR')], [('chip', 'NOUN')]] and then use the itertools.combinations to permute. Yet, it will permute within the list (i.e. juice and orange appear together), which I do not want in my results.
python list permutation itertools
edited Nov 21 at 7:43
Ev. Kounis
10.5k21545
asked Nov 20 at 15:52
Abbey
727
|
show 2 more comments
up vote
1
down vote
favorite
I have a nested list and would like to make a product of two items.
test = [[('juice', 'NOUN'), ('orange', 'FLAVOR')],
[('juice', 'NOUN'), ('orange', 'FLAVOR'), ('lemon', 'FLAVOR')],
[('orange', 'FLAVOR'), ('chip', 'NOUN')]]
What I expect is something like this:
[(('juice', 'NOUN'), ('lemon', 'FLAVOR')),
(('juice', 'NOUN'), ('chip', 'NOUN')),
(('orange', 'FLAVOR'), ('lemon', 'FLAVOR')),
(('orange', 'FLAVOR'), ('chip', 'NOUN')),
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))]
That is to say, I would like to get the permutation across lists but only for unique items. I prefer to use itertools. Previously, I tried list(itertools.product(*test)) But I realized it would produce the product of the length of a nested list...
My current code:
unique_list = list(set(itertools.chain(*test)))
list(itertools.combinations(unique_list, 2))
My thought process is to get the unique items in the nested list first, so the nested list will be [[('juice', 'NOUN'), ('orange', 'FLAVOR')], [('lemon', 'FLAVOR')], [('chip', 'NOUN')]] and then use the itertools.combinations to permute. Yet, it will permute within the list (i.e. juice and orange appear together), which I do not want in my results.
python list permutation itertools
edited Nov 21 at 7:43
Ev. Kounis
10.5k21545
asked Nov 20 at 15:52
Abbey
727
why do you not have(('lemon', 'FLAVOR'), ('chip', 'NOUN'))in your results?
– Ev. Kounis
Nov 20 at 15:59
@Ev. Kounis oh! Thanks for pointing out! I want that in my results. I just edited my question
– Abbey
Nov 20 at 16:00
Have you tried anything so far? Any code you could show?
– plocks
Nov 20 at 16:02
is the length of the original list fixed?
– Ev. Kounis
Nov 20 at 16:03
I don't understand why(('juice', 'NOUN'), ('orange', 'FLAVOR'))isn't part of the expected output. It (and its reverse) can be made a whole bunch of ways from your three input lists.
– Blckknght
Nov 21 at 7:54
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a nested list and would like to make a product of two items.
test = [[('juice', 'NOUN'), ('orange', 'FLAVOR')],
[('juice', 'NOUN'), ('orange', 'FLAVOR'), ('lemon', 'FLAVOR')],
[('orange', 'FLAVOR'), ('chip', 'NOUN')]]
What I expect is something like this:
[(('juice', 'NOUN'), ('lemon', 'FLAVOR')),
(('juice', 'NOUN'), ('chip', 'NOUN')),
(('orange', 'FLAVOR'), ('lemon', 'FLAVOR')),
(('orange', 'FLAVOR'), ('chip', 'NOUN')),
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))]
That is to say, I would like to get the permutation across lists but only for unique items. I prefer to use itertools. Previously, I tried list(itertools.product(*test)) But I realized it would produce the product of the length of a nested list...
My current code:
unique_list = list(set(itertools.chain(*test)))
list(itertools.combinations(unique_list, 2))
My thought process is to get the unique items in the nested list first, so the nested list will be [[('juice', 'NOUN'), ('orange', 'FLAVOR')], [('lemon', 'FLAVOR')], [('chip', 'NOUN')]] and then use the itertools.combinations to permute. Yet, it will permute within the list (i.e. juice and orange appear together), which I do not want in my results.
python list permutation itertools
edited Nov 21 at 7:43
Ev. Kounis
10.5k21545
asked Nov 20 at 15:52
Abbey
727
I have a nested list and would like to make a product of two items.
test = [[('juice', 'NOUN'), ('orange', 'FLAVOR')],
[('juice', 'NOUN'), ('orange', 'FLAVOR'), ('lemon', 'FLAVOR')],
[('orange', 'FLAVOR'), ('chip', 'NOUN')]]
What I expect is something like this:
[(('juice', 'NOUN'), ('lemon', 'FLAVOR')),
(('juice', 'NOUN'), ('chip', 'NOUN')),
(('orange', 'FLAVOR'), ('lemon', 'FLAVOR')),
(('orange', 'FLAVOR'), ('chip', 'NOUN')),
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))]
That is to say, I would like to get the permutation across lists but only for unique items. I prefer to use itertools. Previously, I tried list(itertools.product(*test)) But I realized it would produce the product of the length of a nested list...
My current code:
unique_list = list(set(itertools.chain(*test)))
list(itertools.combinations(unique_list, 2))
My thought process is to get the unique items in the nested list first, so the nested list will be [[('juice', 'NOUN'), ('orange', 'FLAVOR')], [('lemon', 'FLAVOR')], [('chip', 'NOUN')]] and then use the itertools.combinations to permute. Yet, it will permute within the list (i.e. juice and orange appear together), which I do not want in my results.
python list permutation itertools
python list permutation itertools
edited Nov 21 at 7:43
Ev. Kounis
10.5k21545
asked Nov 20 at 15:52
Abbey
727
edited Nov 21 at 7:43
Ev. Kounis
10.5k21545
asked Nov 20 at 15:52
Abbey
727
edited Nov 21 at 7:43
Ev. Kounis
10.5k21545
edited Nov 21 at 7:43
Ev. Kounis
10.5k21545
edited Nov 21 at 7:43
Ev. Kounis
10.5k21545
10.5k21545
asked Nov 20 at 15:52
Abbey
727
asked Nov 20 at 15:52
Abbey
727
asked Nov 20 at 15:52
Abbey
727
727
why do you not have(('lemon', 'FLAVOR'), ('chip', 'NOUN'))in your results?
– Ev. Kounis
Nov 20 at 15:59
@Ev. Kounis oh! Thanks for pointing out! I want that in my results. I just edited my question
– Abbey
Nov 20 at 16:00
Have you tried anything so far? Any code you could show?
– plocks
Nov 20 at 16:02
is the length of the original list fixed?
– Ev. Kounis
Nov 20 at 16:03
I don't understand why(('juice', 'NOUN'), ('orange', 'FLAVOR'))isn't part of the expected output. It (and its reverse) can be made a whole bunch of ways from your three input lists.
– Blckknght
Nov 21 at 7:54
|
show 2 more comments
why do you not have(('lemon', 'FLAVOR'), ('chip', 'NOUN'))in your results?
– Ev. Kounis
Nov 20 at 15:59
@Ev. Kounis oh! Thanks for pointing out! I want that in my results. I just edited my question
– Abbey
Nov 20 at 16:00
Have you tried anything so far? Any code you could show?
– plocks
Nov 20 at 16:02
is the length of the original list fixed?
– Ev. Kounis
Nov 20 at 16:03
I don't understand why(('juice', 'NOUN'), ('orange', 'FLAVOR'))isn't part of the expected output. It (and its reverse) can be made a whole bunch of ways from your three input lists.
– Blckknght
Nov 21 at 7:54
why do you not have
(('lemon', 'FLAVOR'), ('chip', 'NOUN')) in your results?– Ev. Kounis
Nov 20 at 15:59
why do you not have
(('lemon', 'FLAVOR'), ('chip', 'NOUN')) in your results?– Ev. Kounis
Nov 20 at 15:59
@Ev. Kounis oh! Thanks for pointing out! I want that in my results. I just edited my question
– Abbey
Nov 20 at 16:00
@Ev. Kounis oh! Thanks for pointing out! I want that in my results. I just edited my question
– Abbey
Nov 20 at 16:00
Have you tried anything so far? Any code you could show?
– plocks
Nov 20 at 16:02
Have you tried anything so far? Any code you could show?
– plocks
Nov 20 at 16:02
is the length of the original list fixed?
– Ev. Kounis
Nov 20 at 16:03
is the length of the original list fixed?
– Ev. Kounis
Nov 20 at 16:03
I don't understand why
(('juice', 'NOUN'), ('orange', 'FLAVOR')) isn't part of the expected output. It (and its reverse) can be made a whole bunch of ways from your three input lists.– Blckknght
Nov 21 at 7:54
I don't understand why
(('juice', 'NOUN'), ('orange', 'FLAVOR')) isn't part of the expected output. It (and its reverse) can be made a whole bunch of ways from your three input lists.– Blckknght
Nov 21 at 7:54
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
This does what you want without fixing the size of the original list to 3:
Input:
test = [[('juice', 'NOUN'), ('orange', 'FLAVOR')],
[('juice', 'NOUN'), ('orange', 'FLAVOR'), ('lemon', 'FLAVOR')],
[('juice', 'NOUN'), ('chip', 'NOUN')]]
First, reformat input to remove duplicates (see note 1):
test = [[x for x in sublist if x not in sum(test[:i], )] for i, sublist in enumerate(test)]
Finally, get the product of the combinations.
from itertools import combinations, product
for c in combinations(test, 2):
for x in product(*c):
print(x)
which produces:
(('juice', 'NOUN'), ('lemon', 'FLAVOR'))
(('orange', 'FLAVOR'), ('lemon', 'FLAVOR'))
(('juice', 'NOUN'), ('chip', 'NOUN'))
(('orange', 'FLAVOR'), ('chip', 'NOUN'))
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))
- removes inner tuples if they were seen in any of the previous sublists. The magic here is done by the
sum(test[:i], )which "adds" all the previous sublists together to perform one membership check only.
There is also a list-comprehension version of the above for compactness and style-points:
res = [x for c in combinations(test, 2) for x in product(*c)]
answered Nov 20 at 16:05
Ev. Kounis
10.5k21545
I tested all the solutions you suggested and they worked like magic! Thanks
– Abbey
Nov 20 at 16:17
1
@Abbey Cheers! it was an interesting problem.
– Ev. Kounis
Nov 20 at 16:19
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This does what you want without fixing the size of the original list to 3:
Input:
test = [[('juice', 'NOUN'), ('orange', 'FLAVOR')],
[('juice', 'NOUN'), ('orange', 'FLAVOR'), ('lemon', 'FLAVOR')],
[('juice', 'NOUN'), ('chip', 'NOUN')]]
First, reformat input to remove duplicates (see note 1):
test = [[x for x in sublist if x not in sum(test[:i], )] for i, sublist in enumerate(test)]
Finally, get the product of the combinations.
from itertools import combinations, product
for c in combinations(test, 2):
for x in product(*c):
print(x)
which produces:
(('juice', 'NOUN'), ('lemon', 'FLAVOR'))
(('orange', 'FLAVOR'), ('lemon', 'FLAVOR'))
(('juice', 'NOUN'), ('chip', 'NOUN'))
(('orange', 'FLAVOR'), ('chip', 'NOUN'))
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))
- removes inner tuples if they were seen in any of the previous sublists. The magic here is done by the
sum(test[:i], )which "adds" all the previous sublists together to perform one membership check only.
There is also a list-comprehension version of the above for compactness and style-points:
res = [x for c in combinations(test, 2) for x in product(*c)]
answered Nov 20 at 16:05
Ev. Kounis
10.5k21545
I tested all the solutions you suggested and they worked like magic! Thanks
– Abbey
Nov 20 at 16:17
1
@Abbey Cheers! it was an interesting problem.
– Ev. Kounis
Nov 20 at 16:19
add a comment |
up vote
1
down vote
accepted
This does what you want without fixing the size of the original list to 3:
Input:
test = [[('juice', 'NOUN'), ('orange', 'FLAVOR')],
[('juice', 'NOUN'), ('orange', 'FLAVOR'), ('lemon', 'FLAVOR')],
[('juice', 'NOUN'), ('chip', 'NOUN')]]
First, reformat input to remove duplicates (see note 1):
test = [[x for x in sublist if x not in sum(test[:i], )] for i, sublist in enumerate(test)]
Finally, get the product of the combinations.
from itertools import combinations, product
for c in combinations(test, 2):
for x in product(*c):
print(x)
which produces:
(('juice', 'NOUN'), ('lemon', 'FLAVOR'))
(('orange', 'FLAVOR'), ('lemon', 'FLAVOR'))
(('juice', 'NOUN'), ('chip', 'NOUN'))
(('orange', 'FLAVOR'), ('chip', 'NOUN'))
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))
- removes inner tuples if they were seen in any of the previous sublists. The magic here is done by the
sum(test[:i], )which "adds" all the previous sublists together to perform one membership check only.
There is also a list-comprehension version of the above for compactness and style-points:
res = [x for c in combinations(test, 2) for x in product(*c)]
answered Nov 20 at 16:05
Ev. Kounis
10.5k21545
I tested all the solutions you suggested and they worked like magic! Thanks
– Abbey
Nov 20 at 16:17
1
@Abbey Cheers! it was an interesting problem.
– Ev. Kounis
Nov 20 at 16:19
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This does what you want without fixing the size of the original list to 3:
Input:
test = [[('juice', 'NOUN'), ('orange', 'FLAVOR')],
[('juice', 'NOUN'), ('orange', 'FLAVOR'), ('lemon', 'FLAVOR')],
[('juice', 'NOUN'), ('chip', 'NOUN')]]
First, reformat input to remove duplicates (see note 1):
test = [[x for x in sublist if x not in sum(test[:i], )] for i, sublist in enumerate(test)]
Finally, get the product of the combinations.
from itertools import combinations, product
for c in combinations(test, 2):
for x in product(*c):
print(x)
which produces:
(('juice', 'NOUN'), ('lemon', 'FLAVOR'))
(('orange', 'FLAVOR'), ('lemon', 'FLAVOR'))
(('juice', 'NOUN'), ('chip', 'NOUN'))
(('orange', 'FLAVOR'), ('chip', 'NOUN'))
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))
- removes inner tuples if they were seen in any of the previous sublists. The magic here is done by the
sum(test[:i], )which "adds" all the previous sublists together to perform one membership check only.
There is also a list-comprehension version of the above for compactness and style-points:
res = [x for c in combinations(test, 2) for x in product(*c)]
answered Nov 20 at 16:05
Ev. Kounis
10.5k21545
This does what you want without fixing the size of the original list to 3:
Input:
test = [[('juice', 'NOUN'), ('orange', 'FLAVOR')],
[('juice', 'NOUN'), ('orange', 'FLAVOR'), ('lemon', 'FLAVOR')],
[('juice', 'NOUN'), ('chip', 'NOUN')]]
First, reformat input to remove duplicates (see note 1):
test = [[x for x in sublist if x not in sum(test[:i], )] for i, sublist in enumerate(test)]
Finally, get the product of the combinations.
from itertools import combinations, product
for c in combinations(test, 2):
for x in product(*c):
print(x)
which produces:
(('juice', 'NOUN'), ('lemon', 'FLAVOR'))
(('orange', 'FLAVOR'), ('lemon', 'FLAVOR'))
(('juice', 'NOUN'), ('chip', 'NOUN'))
(('orange', 'FLAVOR'), ('chip', 'NOUN'))
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))
- removes inner tuples if they were seen in any of the previous sublists. The magic here is done by the
sum(test[:i], )which "adds" all the previous sublists together to perform one membership check only.
There is also a list-comprehension version of the above for compactness and style-points:
res = [x for c in combinations(test, 2) for x in product(*c)]
answered Nov 20 at 16:05
Ev. Kounis
10.5k21545
edited Nov 20 at 16:11
answered Nov 20 at 16:05
Ev. Kounis
10.5k21545
answered Nov 20 at 16:05
Ev. Kounis
10.5k21545
answered Nov 20 at 16:05
Ev. Kounis
10.5k21545
10.5k21545
I tested all the solutions you suggested and they worked like magic! Thanks
– Abbey
Nov 20 at 16:17
1
@Abbey Cheers! it was an interesting problem.
– Ev. Kounis
Nov 20 at 16:19
add a comment |
I tested all the solutions you suggested and they worked like magic! Thanks
– Abbey
Nov 20 at 16:17
1
@Abbey Cheers! it was an interesting problem.
– Ev. Kounis
Nov 20 at 16:19
I tested all the solutions you suggested and they worked like magic! Thanks
– Abbey
Nov 20 at 16:17
I tested all the solutions you suggested and they worked like magic! Thanks
– Abbey
Nov 20 at 16:17
1
1
@Abbey Cheers! it was an interesting problem.
– Ev. Kounis
Nov 20 at 16:19
@Abbey Cheers! it was an interesting problem.
– Ev. Kounis
Nov 20 at 16:19
add a comment |
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why do you not have
(('lemon', 'FLAVOR'), ('chip', 'NOUN'))in your results?– Ev. Kounis
Nov 20 at 15:59
@Ev. Kounis oh! Thanks for pointing out! I want that in my results. I just edited my question
– Abbey
Nov 20 at 16:00
Have you tried anything so far? Any code you could show?
– plocks
Nov 20 at 16:02
is the length of the original list fixed?
– Ev. Kounis
Nov 20 at 16:03
I don't understand why
(('juice', 'NOUN'), ('orange', 'FLAVOR'))isn't part of the expected output. It (and its reverse) can be made a whole bunch of ways from your three input lists.– Blckknght
Nov 21 at 7:54