不含Itertools的笛卡尔积非Pythonic方法

我这里有这个代码：

import itertools

col_groups = [c1+c2 for c1, c2 in itertools.product(d1_columns, d2_columns)]

没有itertools我怎么能做这样的事情？一种非蟒蛇式的方式

def colGroups(d1, d2):
  answer = []
  for c1 in d1:
    for c2 in d2:
      answer.append(c1+c2)
  return answer

测试：

>>> d1 = list(range(10))
>>> d2 = list(range(10,20))
>>> d1
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> d2
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> def colGroups(d1, d2):
...   answer = []
...   for c1 in d1:
...     for c2 in d2:
...       answer.append(c1+c2)
...   return answer
... 
>>> colGroups(d1, d2)
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28]

。。。或嵌套的listcomp等效项：

colGroups = [c1+c2 for c1 in d1 for c2 in d2]

测试：

>>> d1 = list(range(10))
>>> d2 = list(range(10,20))
>>> d1
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> d2
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> def colGroups(d1, d2):
...   answer = []
...   for c1 in d1:
...     for c2 in d2:
...       answer.append(c1+c2)
...   return answer
... 
>>> colGroups(d1, d2)
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28]

。。。或嵌套的listcomp等效项：

colGroups = [c1+c2 for c1 in d1 for c2 in d2]

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如果您只需要两个列表的笛卡尔乘积，则可以使用嵌套列表理解

[ i + j for i in d1_columns for j in d2_columns ]

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如果您只需要两个列表的笛卡尔乘积，则可以使用嵌套列表理解

[ i + j for i in d1_columns for j in d2_columns ]

---

您始终可以使用中给出的代码

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您始终可以使用中给出的代码

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“你为什么要这么做？”彼得德格洛普只是好奇。为什么你要这么做？“彼得德格洛普只是好奇。哼哼，这是避免使用迭代工具的有趣方式。@ USER 309969.实际上：请考虑任何答案，如果它们帮助你，我得到这个错误：不支持的操作数类型（s）。对于*：'map'和'int'您要传递的参数是什么？我有这样的内容：我创建了上面的函数。。。然后，我做一个函数调用并在另一个函数中传递这样的函数：CyLyGys= [C1+C2为C1，C2在产品（D1x列，D2x列）]，这是一个有趣的避免使用迭代工具的方法。@ USER 309969.实际上：请考虑任何答案，如果它们帮助您，我得到这个错误To:不支持的操作数类型（s）对于*：'map'和'int'您要传递的参数是什么？我有这样的内容：我创建了上面的函数。。。然后我执行一个函数调用，并在另一个函数中传递如下函数：col_groups=[c1+c2表示c1，c2表示乘积（d1_列，d2_列）]