主页 / 问题 / 1433525
Accepted
Nommad
Asked:2022-07-27 17:22:16 +0000 UTC2022-07-27 17:22:16 +0000 UTC

检查元组列表中的元素以进行奇偶校验

  • 772
ListWithCortege = [(2, 4, 2, 1, 0), (4, 2, 1, 0), (1, 1, 0), (4, 1), (1, 4, 1),(1,)]

result = []
for cort in ListWithCortege:
    result.append(list(zip(cort, cort[1:])))
for i in result:
    print(i)

结论：

[(2, 4), (4, 2), (2, 1), (1, 0)]
[(4, 2), (2, 1), (1, 0)]
[(1, 1), (1, 0)]
[(4, 1)]
[(1, 4), (4, 1)]
[]

如何在这里进行类型检查：如果元组列表的元素中的元素数 = 1，则添加99999. 对于要(1,)成为的元素(1,99999)。

输出应该是：

[(2, 4), (4, 2), (2, 1), (1, 0)]
[(4, 2), (2, 1), (1, 0)]
[(1, 1), (1, 0)]
[(4, 1)]
[(1, 4), (4, 1)]
[(1,99999)]
python list

2 个回答

  1. Best Answer

    或像这样：

    ListWithCortege = [(2, 4, 2, 1, 0), (4, 2, 1, 0), (1, 1, 0), (4, 1), (1, 4, 1),(1,)]

result = []
for cort in ListWithCortege:
    result.append(list(zip(cort, cort[1:] or (99999,))))
                                         ^^^^^^^^^^^^
for i in result:
    print(i)
'''
[(2, 4), (4, 2), (2, 1), (1, 0)]
[(4, 2), (2, 1), (1, 0)]
[(1, 1), (1, 0)]
[(4, 1)]
[(1, 4), (4, 1)]
[(1, 99999)]
    • 5

  2. 例如，像这样：

    result.append(list(zip(cort, list(cort[1:]) + [99999] * (len(cort) == 1))))
                             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    • 1

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