Python将多个可迭代对象转换为点列表[重复]

Question

假设我有三个列表：

a = [0, 1, 2]
b = [3, 4, 5]
c = [6, 7, 8]

我想要一个从每个方法中选择三个值的所有方法的列表，就像一个点网格。我的目标是这样的：

d = [[0, 3, 6], [0, 3, 7], [0, 3, 8], [0, 4, 6], [0, 4, 7], [0, 4, 8], [0, 5, 6], [0, 5, 7], [0, 5, 8], 
     [1, 3, 6], [1, 3, 7], [1, 3, 8], [1, 4, 6], [1, 4, 7], [1, 4, 8], [1, 5, 6], [1, 5, 7], [1, 5, 8], 
     [2, 3, 6], [2, 3, 7], [2, 3, 8], [2, 4, 6], [2, 4, 7], [2, 4, 8], [2, 5, 6], [2, 5, 7], [2, 5, 8]]

我想定义一个函数，它可以接受任意数量的迭代，并且可以做我上面所说的。有没有办法做到这一点？

Answer 1

我认为itertools.product 可以为您完成这项工作

from itertools import product

a = [0, 1, 2]
b = [3, 4, 5]
c = [6, 7, 8]

d = list(product(a, b, c))
d

Answer 2

你可以使用itertools.product:

>>> import itertools
>>> 
>>> a = [0, 1, 2]
>>> b = [3, 4, 5]
>>> c = [6, 7, 8]
>>> 
>>> list(itertools.product(a, b, c))
[(0, 3, 6), (0, 3, 7), (0, 3, 8),
 (0, 4, 6), (0, 4, 7), (0, 4, 8),
 (0, 5, 6), (0, 5, 7), (0, 5, 8),
 (1, 3, 6), (1, 3, 7), (1, 3, 8),
 (1, 4, 6), (1, 4, 7), (1, 4, 8),
 (1, 5, 6), (1, 5, 7), (1, 5, 8),
 (2, 3, 6), (2, 3, 7), (2, 3, 8),
 (2, 4, 6), (2, 4, 7), (2, 4, 8),
 (2, 5, 6), (2, 5, 7), (2, 5, 8)]
>>> 

要处理任意数量的列表，您可以：

>>> abc = [a, b, c]
>>> abc
[[0, 1, 2], [3, 4, 5], [6, 7, 8]]
>>> 
>>> list(itertools.product(*abc))
[(0, 3, 6), (0, 3, 7), (0, 3, 8),
 (0, 4, 6), (0, 4, 7), (0, 4, 8),
 (0, 5, 6), (0, 5, 7), (0, 5, 8),
 (1, 3, 6), (1, 3, 7), (1, 3, 8),
 (1, 4, 6), (1, 4, 7), (1, 4, 8),
 (1, 5, 6), (1, 5, 7), (1, 5, 8),
 (2, 3, 6), (2, 3, 7), (2, 3, 8),
 (2, 4, 6), (2, 4, 7), (2, 4, 8),
 (2, 5, 6), (2, 5, 7), (2, 5, 8)]
>>> 

Answer 3

a = [0, 1, 2]
b = [3, 4, 5]
c = [6, 7, 8]
s = []
for i in a :
    for j in b :
        for h in c :
            s.append([i,j,h])
print(s)