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Accepted
J. Doe
Asked:2020-12-09 00:36:36 +0000 UTC2020-12-09 00:36:36 +0000 UTC

列表推导中循环的嵌套顺序

  • 772

对以下内容感兴趣：

a = [[[[],...],...],...] # некоторый 3-мерный массив
l = [a[x][y][z] for x in range(5) for y in range(10) for z in range(15)]

列表将如何初始化l？

所以：

l = []
for x in range(5):
    for y in range(10):
        for z in range(15):
            l.append(a[x][y][z])

或像这样：

l = []
for z in range(15):
    for y in range(10):
        for x in range(5):
            l.append(a[x][y][z])

?

python

1 个回答

  1. Best Answer
    a = [['x1', 'x2', 'x3'], ['y1', 'y2', 'y3'], ['z1', 'z2', 'z3']]
l=[]
l = [[x,y,z] for x in a[0] for y in a[1] for z in a[2]]

from pprint import pprint as pp
pp(l)
[['x1', 'y1', 'z1'],
 ['x1', 'y1', 'z2'],
 ['x1', 'y1', 'z3'],
 ['x1', 'y2', 'z1'],
 ['x1', 'y2', 'z2'],
 ['x1', 'y2', 'z3'],
 ['x1', 'y3', 'z1'],
 ['x1', 'y3', 'z2'],
 ['x1', 'y3', 'z3'],
 ['x2', 'y1', 'z1'],
 ['x2', 'y1', 'z2'],
 ['x2', 'y1', 'z3'],
 ['x2', 'y2', 'z1'],
 ['x2', 'y2', 'z2'],
 ['x2', 'y2', 'z3'],
 ['x2', 'y3', 'z1'],
 ['x2', 'y3', 'z2'],
 ['x2', 'y3', 'z3'],
 ['x3', 'y1', 'z1'],
 ['x3', 'y1', 'z2'],
 ['x3', 'y1', 'z3'],
 ['x3', 'y2', 'z1'],
 ['x3', 'y2', 'z2'],
 ['x3', 'y2', 'z3'],
 ['x3', 'y3', 'z1'],
 ['x3', 'y3', 'z2'],
 ['x3', 'y3', 'z3']]
    • 0

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