我想要类似下面的代码,但是使用“Pythonic”风格或标准库:
def combinations(a,b):
for i in a:
for j in b:
yield(i,j)
def combinations(a,b):
for i in a:
for j in b:
yield(i,j)
从组合学的角度来看,这些并不是真正的“组合”。它们更像是来自于a
和b
的笛卡尔积中的元素。标准库中生成这些对的函数是itertools.product()
:
for i, j in itertools.product(a, b):
# Whatever
Sven所说,你的代码试图获取列表a
和b
的所有有序元素对。在这种情况下,你需要使用itertools.product(a,b)
。
如果你实际上想要的是“组合”,即列表a
中所有不同元素的无序对,则需要使用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)
嵌套生成器表达式也可以使用:
product = ((i, j) for i in a for j in b)
for i, j in product:
# ...
yield
;) - mhyfritz>>>a=[1,2,3]
>>>b=[4,5,6]
>>>zip(a,b)
[(1, 4), (2, 5), (3, 6)]
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)
>>> 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
和b
中的每个元素创建一对。这真的是你想要的吗? - Felix Kling