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Accepted
Alexey Sinitsky
Asked:2024-08-27 02:06:49 +0000 UTC2024-08-27 02:06:49 +0000 UTC

按相同元素合并子列表[重复]

  • 772

有一个列表输出：

[
    ['Сергей', '+71234567890'],
    ['Сергей', '+71234567810'],
    ['Михаил', '+51234567890'],
    ['Николай', '+72134567890']
]

怎么办？会变成这样吗？

[
    ['Сергей', ['+71234567890','+71234567810'],
    ['Михаил', '+51234567890'],
    ['Николай', '+72134567890']
]

那些。谁具有相同的姓名和多个电话号码，是否创建了嵌套列表？

python

2 个回答

  1. Best Answer

    C dict.setdefault：

    source = [
    ['Сергей', '+71234567890'],
    ['Сергей', '+71234567810'],
    ['Михаил', '+51234567890'],
    ['Николай', '+72134567890']
]

d = {}
for k, v in source:
    d.setdefault(k, []).append(v)
target = list(map(list, d.items()))

print(target)

    [['Сергей', ['+71234567890', '+71234567810']], ['Михаил', ['+51234567890']], ['Николай', ['+72134567890']]]

    没有它：

    source = [
    ['Сергей', '+71234567890'],
    ['Сергей', '+71234567810'],
    ['Михаил', '+51234567890'],
    ['Николай', '+72134567890']
]

d = {}
for k, v in source:
    if k not in d:
        d[k] = []
    d[k].append(v)

target = list(map(list, d.items()))

print(target)
    • 1

  2. 是的，我做了一切，这是完整的代码..

    lst = ['+71234567890 Сергей', '+71234567810 Сергей', '+51234567890 Михаил', '+72134567890 Николай']


lst_in = []

for x in lst:
    A = [x.split() for x in lst]
print(A)

for x in A:
    x[0], x[1] = x[1],x[0]
    print(x)
    lst_in.append(x)

print(lst_in)

d={}

for i, v in lst_in:
   if i not in d:
       d[i] = []
   d[i].append(v)

print(d)
    • 0

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