How can I get pairs of values, where the first is taken from one list and the second from another list?
Question:
I want something like the code below, but in a "Pythonic" style or using the standard library:
def combinations(a,b):
for i in a:
for j in b:
yield(i,j)
Answers:
These are not really "combinations" in the sense of combinatorics. These are rather elements from the Cartesian product of a and b. The function in the standard library to generate these pairs is itertools.product():
for i, j in itertools.product(a, b):
# Whatever
The itertools library has combinatorics functions. Like Sven stated, itertools.product would be the appropriate function in this case:
list(itertools.product('ab', 'cd'))
[('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd')]
A nested generator expression will work too:
product = ((i, j) for i in a for j in b)
for i, j in product:
# ...
>>>a=[1,2,3]
>>>b=[4,5,6]
>>>zip(a,b)
[(1, 4), (2, 5), (3, 6)]
As Sven said, your code is attempting to get all ordered pairs of elements of the lists a and b. In this case itertools.product(a,b) is what you want.
If instead you actually want "combinations", which are all unordered pairs of distinct elements of the list a, then you want itertools.combinations(a,2).
>>> for pair in itertools.combinations([1,2,3,4],2):
... print pair
...
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)
Create set of pairs (even,odd) combination
>>> a = { (i,j) for i in range(0,10,2) for j in range(1,10,2)}
>>> a
{(4, 7), (6, 9), (0, 7), (2, 1), (8, 9), (0, 3), (2, 5), (8, 5), (4, 9), (6, 7), (2, 9), (8, 1), (6, 3), (4, 1), (4, 5), (0, 5), (2, 3), (8, 7), (6, 5), (0, 1), (2, 7), (8, 3), (6, 1), (4, 3), (0, 9)}
def combinations(lista, listb):
return { (i,j) for i in lista for j in listb }
>>> combinations([1,3,5,6],[11,21,133,134,443])
{(1, 21), (5, 133), (5, 11), (5, 134), (6, 11), (6, 134), (1, 443), (3, 11), (6, 21), (3, 21), (1, 133), (1, 134), (5, 21), (3, 134), (5, 443), (6, 443), (1, 11), (3, 443), (6, 133), (3, 133)}
A question we might ask is whether you want to generate all ordered pairs or all unordered pairs. The nested generator expression provided in the answer by mhyfritz will give you all ordered pairs.
If you want all unordered pairs (that is, (1, 2) and (2, 1) counts as the same pair), then you need to filter out the duplicates. An easy way to do this is to add a conditional to the end of the generator expression like so:
myList= [1, 2, 3, 4, 5]
unorderedPairGenerator = ((x, y) for x in myList for y in myList if y > x)
for pair in unorderedPairGenerator:
print(pair)
#(1, 2)
#(1, 3)
#(1, 4)
#(1, 5)
#(2, 3)
#(2, 4)
#(2, 5)
#(3, 4)
#(3, 5)
#(4, 5)
I want something like the code below, but in a "Pythonic" style or using the standard library:
def combinations(a,b):
for i in a:
for j in b:
yield(i,j)
These are not really "combinations" in the sense of combinatorics. These are rather elements from the Cartesian product of a and b. The function in the standard library to generate these pairs is itertools.product():
for i, j in itertools.product(a, b):
# Whatever
The itertools library has combinatorics functions. Like Sven stated, itertools.product would be the appropriate function in this case:
list(itertools.product('ab', 'cd'))
[('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd')]
A nested generator expression will work too:
product = ((i, j) for i in a for j in b)
for i, j in product:
# ...
>>>a=[1,2,3]
>>>b=[4,5,6]
>>>zip(a,b)
[(1, 4), (2, 5), (3, 6)]
As Sven said, your code is attempting to get all ordered pairs of elements of the lists a and b. In this case itertools.product(a,b) is what you want.
If instead you actually want "combinations", which are all unordered pairs of distinct elements of the list a, then you want itertools.combinations(a,2).
>>> for pair in itertools.combinations([1,2,3,4],2):
... print pair
...
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)
Create set of pairs (even,odd) combination
>>> a = { (i,j) for i in range(0,10,2) for j in range(1,10,2)}
>>> a
{(4, 7), (6, 9), (0, 7), (2, 1), (8, 9), (0, 3), (2, 5), (8, 5), (4, 9), (6, 7), (2, 9), (8, 1), (6, 3), (4, 1), (4, 5), (0, 5), (2, 3), (8, 7), (6, 5), (0, 1), (2, 7), (8, 3), (6, 1), (4, 3), (0, 9)}
def combinations(lista, listb):
return { (i,j) for i in lista for j in listb }
>>> combinations([1,3,5,6],[11,21,133,134,443])
{(1, 21), (5, 133), (5, 11), (5, 134), (6, 11), (6, 134), (1, 443), (3, 11), (6, 21), (3, 21), (1, 133), (1, 134), (5, 21), (3, 134), (5, 443), (6, 443), (1, 11), (3, 443), (6, 133), (3, 133)}
A question we might ask is whether you want to generate all ordered pairs or all unordered pairs. The nested generator expression provided in the answer by mhyfritz will give you all ordered pairs.
If you want all unordered pairs (that is, (1, 2) and (2, 1) counts as the same pair), then you need to filter out the duplicates. An easy way to do this is to add a conditional to the end of the generator expression like so:
myList= [1, 2, 3, 4, 5]
unorderedPairGenerator = ((x, y) for x in myList for y in myList if y > x)
for pair in unorderedPairGenerator:
print(pair)
#(1, 2)
#(1, 3)
#(1, 4)
#(1, 5)
#(2, 3)
#(2, 4)
#(2, 5)
#(3, 4)
#(3, 5)
#(4, 5)