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Combinations of multiple lists
How do I check if a list is empty?Finding the index of an item given a list containing it in PythonDifference between append vs. extend list methods in PythonHow to return multiple values from a function?How to make a flat list out of list of lists?How do I concatenate two lists in Python?How to clone or copy a list?How do I list all files of a directory?How to read a file line-by-line into a list?Catch multiple exceptions in one line (except block)
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
Suppose I have three lists:
list1 --> [a, b, c, d, e, f, g, h]
list2 --> [i, j, k]
list3 --> [l, m, n, o, p]
I wish to generate all combinations where I take five elements from list1, two elements from list2 and three elements from list3.
eg.
a, b, c, d, e, i, j, l, m, n
a, b, c, d, e, i, j, l, m, o
etc.
I tried to use itertools.combinations.
l1_combinations = itertools.combinations(list1, 5)
l2_combinations = itertools.combinations(list2, 2)
l3_combinations = itertools.combinations(list3, 3)
for l1_iterator in list(l1_combinations):
for l2_iterator in list(l2_combinations): #added a missing )
for l3_iterator in list(l3_combinations):
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
But I am getting output with iterations happening only on list3. In all the output, only first five elements from list1 and first two elements from list2 are present. Combinations with other elements from those two lists aren't present.
Can someone help me here and also explain what exactly did i miss ?
python
edited Apr 4 at 20:02
amchugh89
595621
asked Apr 4 at 19:44
Arun AranganathanArun Aranganathan
525
add a comment |
Suppose I have three lists:
list1 --> [a, b, c, d, e, f, g, h]
list2 --> [i, j, k]
list3 --> [l, m, n, o, p]
I wish to generate all combinations where I take five elements from list1, two elements from list2 and three elements from list3.
eg.
a, b, c, d, e, i, j, l, m, n
a, b, c, d, e, i, j, l, m, o
etc.
I tried to use itertools.combinations.
l1_combinations = itertools.combinations(list1, 5)
l2_combinations = itertools.combinations(list2, 2)
l3_combinations = itertools.combinations(list3, 3)
for l1_iterator in list(l1_combinations):
for l2_iterator in list(l2_combinations): #added a missing )
for l3_iterator in list(l3_combinations):
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
But I am getting output with iterations happening only on list3. In all the output, only first five elements from list1 and first two elements from list2 are present. Combinations with other elements from those two lists aren't present.
Can someone help me here and also explain what exactly did i miss ?
python
edited Apr 4 at 20:02
amchugh89
595621
asked Apr 4 at 19:44
Arun AranganathanArun Aranganathan
525
Well I'm sure the missing)infor l2_iterator in list(l2_combinations:doesn't help...
– Reedinationer
Apr 4 at 19:51
add a comment |
Suppose I have three lists:
list1 --> [a, b, c, d, e, f, g, h]
list2 --> [i, j, k]
list3 --> [l, m, n, o, p]
I wish to generate all combinations where I take five elements from list1, two elements from list2 and three elements from list3.
eg.
a, b, c, d, e, i, j, l, m, n
a, b, c, d, e, i, j, l, m, o
etc.
I tried to use itertools.combinations.
l1_combinations = itertools.combinations(list1, 5)
l2_combinations = itertools.combinations(list2, 2)
l3_combinations = itertools.combinations(list3, 3)
for l1_iterator in list(l1_combinations):
for l2_iterator in list(l2_combinations): #added a missing )
for l3_iterator in list(l3_combinations):
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
But I am getting output with iterations happening only on list3. In all the output, only first five elements from list1 and first two elements from list2 are present. Combinations with other elements from those two lists aren't present.
Can someone help me here and also explain what exactly did i miss ?
python
edited Apr 4 at 20:02
amchugh89
595621
asked Apr 4 at 19:44
Arun AranganathanArun Aranganathan
525
Suppose I have three lists:
list1 --> [a, b, c, d, e, f, g, h]
list2 --> [i, j, k]
list3 --> [l, m, n, o, p]
I wish to generate all combinations where I take five elements from list1, two elements from list2 and three elements from list3.
eg.
a, b, c, d, e, i, j, l, m, n
a, b, c, d, e, i, j, l, m, o
etc.
I tried to use itertools.combinations.
l1_combinations = itertools.combinations(list1, 5)
l2_combinations = itertools.combinations(list2, 2)
l3_combinations = itertools.combinations(list3, 3)
for l1_iterator in list(l1_combinations):
for l2_iterator in list(l2_combinations): #added a missing )
for l3_iterator in list(l3_combinations):
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
But I am getting output with iterations happening only on list3. In all the output, only first five elements from list1 and first two elements from list2 are present. Combinations with other elements from those two lists aren't present.
Can someone help me here and also explain what exactly did i miss ?
python
python
edited Apr 4 at 20:02
amchugh89
595621
asked Apr 4 at 19:44
Arun AranganathanArun Aranganathan
525
edited Apr 4 at 20:02
amchugh89
595621
asked Apr 4 at 19:44
Arun AranganathanArun Aranganathan
525
edited Apr 4 at 20:02
amchugh89
595621
edited Apr 4 at 20:02
amchugh89
595621
edited Apr 4 at 20:02
amchugh89
595621
595621
asked Apr 4 at 19:44
Arun AranganathanArun Aranganathan
525
asked Apr 4 at 19:44
Arun AranganathanArun Aranganathan
525
asked Apr 4 at 19:44
Arun AranganathanArun Aranganathan
525
525
Well I'm sure the missing)infor l2_iterator in list(l2_combinations:doesn't help...
– Reedinationer
Apr 4 at 19:51
add a comment |
Well I'm sure the missing)infor l2_iterator in list(l2_combinations:doesn't help...
– Reedinationer
Apr 4 at 19:51
Well I'm sure the missing
) in for l2_iterator in list(l2_combinations: doesn't help...– Reedinationer
Apr 4 at 19:51
Well I'm sure the missing
) in for l2_iterator in list(l2_combinations: doesn't help...– Reedinationer
Apr 4 at 19:51
add a comment |
2 Answers
2
active
oldest
votes
As an alternative to regenerating the list of combinations, compute the product of the combinations up front; this also saves you from nesting for loops.
from itertools import combinations, product
list1 = list("abcdefgh")
list2 = list("ijk")
list3 = list("lmnop")
l1 = combinations(list1, 5)
l2 = combinations(list2, 2)
l3 = combinations(list3, 3)
for c1, c2, c3 in product(l1, l2, l3):
sample = c1 + c2 + c3
print(sample)
answered Apr 4 at 19:56
chepnerchepner
261k35251344
add a comment |
Don't iterate over the same iterator multiple times, after the first time it's exhausted. Iterate over a fresh iterator each time:
for l1_iterator in itertools.combinations(list1, 5):
for l2_iterator in itertools.combinations(list2, 2):
for l3_iterator in itertools.combinations(list3, 3):
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
Or make lists of each one in advance to avoid recomputation:
l1_combinations = list(itertools.combinations(list1, 5))
l2_combinations = list(itertools.combinations(list2, 2))
l3_combinations = list(itertools.combinations(list3, 3))
for l1_iterator in l1_combinations:
for l2_iterator in l2_combinations:
for l3_iterator in l3_combinations:
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
answered Apr 4 at 19:52
Alex HallAlex Hall
23.1k32053
What would be a solution to not regenerate the combinations every time?
– ritlew
Apr 4 at 19:55
2
@ritlew see edit.
– Alex Hall
Apr 4 at 19:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
As an alternative to regenerating the list of combinations, compute the product of the combinations up front; this also saves you from nesting for loops.
from itertools import combinations, product
list1 = list("abcdefgh")
list2 = list("ijk")
list3 = list("lmnop")
l1 = combinations(list1, 5)
l2 = combinations(list2, 2)
l3 = combinations(list3, 3)
for c1, c2, c3 in product(l1, l2, l3):
sample = c1 + c2 + c3
print(sample)
answered Apr 4 at 19:56
chepnerchepner
261k35251344
add a comment |
As an alternative to regenerating the list of combinations, compute the product of the combinations up front; this also saves you from nesting for loops.
from itertools import combinations, product
list1 = list("abcdefgh")
list2 = list("ijk")
list3 = list("lmnop")
l1 = combinations(list1, 5)
l2 = combinations(list2, 2)
l3 = combinations(list3, 3)
for c1, c2, c3 in product(l1, l2, l3):
sample = c1 + c2 + c3
print(sample)
answered Apr 4 at 19:56
chepnerchepner
261k35251344
add a comment |
As an alternative to regenerating the list of combinations, compute the product of the combinations up front; this also saves you from nesting for loops.
from itertools import combinations, product
list1 = list("abcdefgh")
list2 = list("ijk")
list3 = list("lmnop")
l1 = combinations(list1, 5)
l2 = combinations(list2, 2)
l3 = combinations(list3, 3)
for c1, c2, c3 in product(l1, l2, l3):
sample = c1 + c2 + c3
print(sample)
answered Apr 4 at 19:56
chepnerchepner
261k35251344
As an alternative to regenerating the list of combinations, compute the product of the combinations up front; this also saves you from nesting for loops.
from itertools import combinations, product
list1 = list("abcdefgh")
list2 = list("ijk")
list3 = list("lmnop")
l1 = combinations(list1, 5)
l2 = combinations(list2, 2)
l3 = combinations(list3, 3)
for c1, c2, c3 in product(l1, l2, l3):
sample = c1 + c2 + c3
print(sample)
answered Apr 4 at 19:56
chepnerchepner
261k35251344
answered Apr 4 at 19:56
chepnerchepner
261k35251344
answered Apr 4 at 19:56
chepnerchepner
261k35251344
answered Apr 4 at 19:56
chepnerchepner
261k35251344
261k35251344
add a comment |
add a comment |
Don't iterate over the same iterator multiple times, after the first time it's exhausted. Iterate over a fresh iterator each time:
for l1_iterator in itertools.combinations(list1, 5):
for l2_iterator in itertools.combinations(list2, 2):
for l3_iterator in itertools.combinations(list3, 3):
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
Or make lists of each one in advance to avoid recomputation:
l1_combinations = list(itertools.combinations(list1, 5))
l2_combinations = list(itertools.combinations(list2, 2))
l3_combinations = list(itertools.combinations(list3, 3))
for l1_iterator in l1_combinations:
for l2_iterator in l2_combinations:
for l3_iterator in l3_combinations:
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
answered Apr 4 at 19:52
Alex HallAlex Hall
23.1k32053
What would be a solution to not regenerate the combinations every time?
– ritlew
Apr 4 at 19:55
2
@ritlew see edit.
– Alex Hall
Apr 4 at 19:59
add a comment |
Don't iterate over the same iterator multiple times, after the first time it's exhausted. Iterate over a fresh iterator each time:
for l1_iterator in itertools.combinations(list1, 5):
for l2_iterator in itertools.combinations(list2, 2):
for l3_iterator in itertools.combinations(list3, 3):
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
Or make lists of each one in advance to avoid recomputation:
l1_combinations = list(itertools.combinations(list1, 5))
l2_combinations = list(itertools.combinations(list2, 2))
l3_combinations = list(itertools.combinations(list3, 3))
for l1_iterator in l1_combinations:
for l2_iterator in l2_combinations:
for l3_iterator in l3_combinations:
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
answered Apr 4 at 19:52
Alex HallAlex Hall
23.1k32053
What would be a solution to not regenerate the combinations every time?
– ritlew
Apr 4 at 19:55
2
@ritlew see edit.
– Alex Hall
Apr 4 at 19:59
add a comment |
Don't iterate over the same iterator multiple times, after the first time it's exhausted. Iterate over a fresh iterator each time:
for l1_iterator in itertools.combinations(list1, 5):
for l2_iterator in itertools.combinations(list2, 2):
for l3_iterator in itertools.combinations(list3, 3):
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
Or make lists of each one in advance to avoid recomputation:
l1_combinations = list(itertools.combinations(list1, 5))
l2_combinations = list(itertools.combinations(list2, 2))
l3_combinations = list(itertools.combinations(list3, 3))
for l1_iterator in l1_combinations:
for l2_iterator in l2_combinations:
for l3_iterator in l3_combinations:
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
answered Apr 4 at 19:52
Alex HallAlex Hall
23.1k32053
Don't iterate over the same iterator multiple times, after the first time it's exhausted. Iterate over a fresh iterator each time:
for l1_iterator in itertools.combinations(list1, 5):
for l2_iterator in itertools.combinations(list2, 2):
for l3_iterator in itertools.combinations(list3, 3):
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
Or make lists of each one in advance to avoid recomputation:
l1_combinations = list(itertools.combinations(list1, 5))
l2_combinations = list(itertools.combinations(list2, 2))
l3_combinations = list(itertools.combinations(list3, 3))
for l1_iterator in l1_combinations:
for l2_iterator in l2_combinations:
for l3_iterator in l3_combinations:
sample = l1_iterator + l2_iterator + l3_iterator
print(sample)
answered Apr 4 at 19:52
Alex HallAlex Hall
23.1k32053
edited Apr 4 at 19:58
answered Apr 4 at 19:52
Alex HallAlex Hall
23.1k32053
answered Apr 4 at 19:52
Alex HallAlex Hall
23.1k32053
answered Apr 4 at 19:52
Alex HallAlex Hall
23.1k32053
23.1k32053
What would be a solution to not regenerate the combinations every time?
– ritlew
Apr 4 at 19:55
2
@ritlew see edit.
– Alex Hall
Apr 4 at 19:59
add a comment |
What would be a solution to not regenerate the combinations every time?
– ritlew
Apr 4 at 19:55
2
@ritlew see edit.
– Alex Hall
Apr 4 at 19:59
What would be a solution to not regenerate the combinations every time?
– ritlew
Apr 4 at 19:55
What would be a solution to not regenerate the combinations every time?
– ritlew
Apr 4 at 19:55
2
2
@ritlew see edit.
– Alex Hall
Apr 4 at 19:59
@ritlew see edit.
– Alex Hall
Apr 4 at 19:59
add a comment |
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Well I'm sure the missing
)infor l2_iterator in list(l2_combinations:doesn't help...– Reedinationer
Apr 4 at 19:51