查找列表的笛卡尔积，根据条件过滤掉元素

Question

我最近遇到了这个符号

a = [1,2,3,4]
b = [2,4,6]

c = [(x,y,z) for x in a for y in b for z in a]

首先我不知道如何搜索创建c的符号，这种结构有名称吗？

另外，我相信c 可以更新为不允许x 等于z。请问你能帮我解决这个问题吗？

我已经尝试了各种方法

c = [(x,y,z) for x in a for y in b for z in a for x != z]

但到目前为止，我找不到任何有效的方法，甚至是有效的语法。

我想要完成的是找到(a,b,a) 的每个组合，其中a 在每一行中只能使用一次，所以结果是

[(1, 2, 2),
 (1, 2, 3),
 (1, 2, 4),
 (1, 4, 2),
 (1, 4, 3),
 (1, 4, 4),
 (1, 6, 2),
 (1, 6, 3),
 (1, 6, 4),
 (2, 2, 1),
 (2, 2, 3),
 (2, 2, 4),
 (2, 4, 1),
 (2, 4, 3),
 (2, 4, 4),
 (2, 6, 1),
 (2, 6, 3),
 (2, 6, 4),
 (3, 2, 1),
 (3, 2, 2),
 (3, 2, 4),
 (3, 4, 1),
 (3, 4, 2),
 (3, 4, 4),
 (3, 6, 1),
 (3, 6, 2),
 (3, 6, 4),
 (4, 2, 1),
 (4, 2, 2),
 (4, 2, 3),
 (4, 4, 1),
 (4, 4, 2),
 (4, 4, 3),
 (4, 6, 1),
 (4, 6, 2),
 (4, 6, 3)]

谢谢

Answer 1

它被称为 list comprehension，您可以在其中使用 logical if 将返回列表中的结果过滤为：

>>> a = [1,2,3,4]
>>> b = [2,4,6]
#     if condition to skip results where `x` equals `z` v
>>> c = [(x,y,z) for x in a for y in b for z in a if x != z]
>>> c   
[(1, 2, 2), (1, 2, 3), (1, 2, 4), (1, 4, 2), (1, 4, 3), (1, 4, 4), (1, 6, 2), (1, 6, 3), (1, 6, 4), (2, 2, 1), (2, 2, 3), (2, 2, 4), (2, 4, 1), (2, 4, 3), (2, 4, 4), (2, 6, 1), (2, 6, 3), (2, 6, 4), (3, 2, 1), (3, 2, 2), (3, 2, 4), (3, 4, 1), (3, 4, 2), (3, 4, 4), (3, 6, 1), (3, 6, 2), (3, 6, 4), (4, 2, 1), (4, 2, 2), (4, 2, 3), (4, 4, 1), (4, 4, 2), (4, 4, 3), (4, 6, 1), (4, 6, 2), (4, 6, 3)]

除了使用 嵌套列表理解，您也可以使用 itertools.product 获得相同的行为：

>>> from itertools import product

>>> [(x,y,z) for x, y, z in product(a, b, a) if x !=z]
[(1, 2, 2), (1, 2, 3), (1, 2, 4), (1, 4, 2), (1, 4, 3), (1, 4, 4), (1, 6, 2), (1, 6, 3), (1, 6, 4), (2, 2, 1), (2, 2, 3), (2, 2, 4), (2, 4, 1), (2, 4, 3), (2, 4, 4), (2, 6, 1), (2, 6, 3), (2, 6, 4), (3, 2, 1), (3, 2, 2), (3, 2, 4), (3, 4, 1), (3, 4, 2), (3, 4, 4), (3, 6, 1), (3, 6, 2), (3, 6, 4), (4, 2, 1), (4, 2, 2), (4, 2, 3), (4, 4, 1), (4, 4, 2), (4, 4, 3), (4, 6, 1), (4, 6, 2), (4, 6, 3)]

Answer 2

这是list comprehension，过滤条件的正确语法是：

c = [(x,y,z) for x in a for y in b for z in a if x != z]

List comprehensions

Answer 3

用于创建c 的语法称为list comprehension。几乎您的确切情况是这些文档中的第四个代码示例：

列表推导式由包含表达式的括号组成，后跟一个 for 子句，然后是零个或多个 for 或 if 子句。结果将是一个新列表，该列表是在其后面的 for 和 if 子句的上下文中评估表达式而产生的。例如，如果两个列表的元素不相等，则此 listcomp 会合并它们：

>>>[(x, y) for x in [1,2,3] for y in [3,1,4] if x != y]

[(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)]

您只需将最后一个 for 更改为 if：

c = [(x,y,z) for x in a for y in b for z in a if x != z]

Answer 4

为了多样化，我想添加以下基于itertools.product 和规避 if 检查的解决方案。

from itertools import product
a = [1,2,3,4]
b = [2,4,6]

c = []
for i, item in enumerate(a):
    c.extend((item, x, y) for x, y in product(b, a[:i] + a[i+1:]))
print(c)

制作：

[(1, 2, 2), (1, 2, 3), (1, 2, 4), (1, 4, 2), (1, 4, 3), (1, 4, 4), (1, 6, 2), (1, 6, 3), (1, 6, 4), (2, 2, 1), (2, 2, 3), (2, 2, 4), (2, 4, 1), (2, 4, 3), (2, 4, 4), (2, 6, 1), (2, 6, 3), (2, 6, 4), (3, 2, 1), (3, 2, 2), (3, 2, 4), (3, 4, 1), (3, 4, 2), (3, 4, 4), (3, 6, 1), (3, 6, 2), (3, 6, 4), (4, 2, 1), (4, 2, 2), (4, 2, 3), (4, 4, 1), (4, 4, 2), (4, 4, 3), (4, 6, 1), (4, 6, 2), (4, 6, 3)]

---

我也做了一些时间来比较不同的方法：

from timeit import timeit

setup = '''

from itertools import product


def list_comp(a, b):
    return [(x,y,z) for x in a for y in b for z in a if x != z]


def itertools_listComp(a, b):
    return [(x,y,z) for x, y, z in product(a, b, a) if x !=z]


def itertools_forLoop(a, b):
    c = []
    for i, item in enumerate(a):
        c.extend((item, x, y) for x, y in product(b, a[:i] + a[i + 1:]))
    return c

a = [1, 2, 3, 4]
b = [2, 4, 6]
'''

print('list_comp:', timeit(stmt="list_comp(a, b)", setup=setup, number=1000))
print('itertools_forLoop:', timeit(stmt="itertools_forLoop(a, b)", setup=setup, number=1000))
print('itertools_listComp:', timeit(stmt="itertools_listComp(a, b)", setup=setup, number=1000))

结果是：

list_comp:          0.0050
itertools_listComp: 0.0056
itertools_forLoop:  0.0086

所以看起来简单的列表理解是最快的。这种行为在更大的a 列表中也存在。