我正在努力创造
[ x
for x in [1,2,3]
for y in [3,1,4] ]
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输出:
[1, 1, 1, 2, 2, 2, 3, 3, 3]
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但我想要的是创造
预期产出:
[1, 1, 1, 2, 3, 3, 3, 3]
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是否可以在列表理解中执行此操作?
使用此zip()功能将您的号码与其计数配对:
numbers = [1, 2, 3]
counts = [3, 1, 4]
output = [n for n, c in zip(numbers, counts) for _ in range(c)]
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