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jokop
Asked:2020-10-26 07:41:33 +0000 UTC2020-10-26 07:41:33 +0000 UTC

比较嵌套列表

  • 772

我有两个长度为 45 的嵌套列表，如下所示：

a = [['a','d','c','f'], ['w','r','t','c'], ['w','r','t','y'], ['r','t','t']]
b = [['q','w','e'], ['r','t','z'], ['t','y','t'], ['t','t','y']]

但是不仅嵌套了4个元素，而且嵌套了45个。您需要创建一个列表，其中将计算列表中的元素c与列表中的元素匹配的数量。аb

结果，我想得到一个这样的列表：

с = [[0], [2], [2], [2]]
python

3 个回答

  1. Best Answer

    你可以拧计数器：

    from collections import Counter
a = [['a','d','c','f'], ['w','r','t','c'], ['w','r','t','y'], ['r','t','t']]
b = [['q','w','e'], ['r','t','z'], ['t','y','t'], ['t', 't','y']]
r = ([sum((Counter(a) & Counter(b)).values())] for a, b in zip(a, b))
print(list(r))

    结论：

    [[0], [2], [2], [2]]

    演示。

    • 4 
  2. a = [['a','d','c','f'], ['w','r','t','c'], ['w','r','t','y'    ], ['r','t','t']]
b = [['q','w','e'    ], [    'r','t','z'], [        't','y','t'], [    't','t','y']]
#                             +   +                  +   +              +   +

c = []

for i, v in enumerate(a):
    n = 0
    for j in v:
        if j in b[i]:
            n += 1
            ind = b[i].index(j)
            b[i].pop(ind)
    c.append([n])

print(c)

    [[0], [2], [2], [2]]
    • 1

  3. 从答案来看，结果毕竟应该是[[0], [2], [2], [2]]，而不是[[0], [1], [1], [2]]，这是令人困惑的。如果答案是数据a并且b确实应该是[[0], [2], [2], [2]]，那么不使用任何模块的单行解决方案将如下所示：

    c = [[sum(e in y for e in x)] for x, y in zip(a, b)]
    • 1

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