备用答案（不使用设置）

Question

如何转换此列表列表：

[['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]

到这个元组列表：

[(0, [1, 2, 3]), (1, [0, 4, 6, 7, 9]), (2, [0, 3, 6, 8, 9])]

我不确定如何实施下一步？（我不能用字典，集，双端，平分模块。但是，你可以，实际上应该使用.sort或sort函数。）

这是我的尝试：

network= [['10'], ['0 1'], ['0 2'], ['0 3'], ['1 4'], ['1 6'], ['1 7'], ['1 9'], ['2 3'], ['2 6'], ['2 8'], ['2 9']]
network.remove(network[0])
friends=[]

for i in range(len(network)):
    element= (network[i][0]).split(' ')
    friends.append(element)


t=len(friends)
s= len(friends[0])
lst=[]

for i in range(t):
    a= (friends[i][0])
    if a not in lst:
        lst.append(int(a))
        for i in range(t):
            if a == friends[i][0]:
                b=(friends[i][1])
                lst.append([b])
print(tuple(lst))

输出：

(0, ['1'], ['2'], ['3'], 0, ['1'], ['2'], ['3'], 0, ['1'], ['2'], ['3'], 1, ['4'], ['6'], ['7'], ['9'], 1, ['4'], ['6'], ['7'], ['9'], 1, ['4'], ['6'], ['7'], ['9'], 1, ['4'], ['6'], ['7'], ['9'], 2, ['3'], ['6'], ['8'], ['9'], 2, ['3'], ['6'], ['8'], ['9'], 2, ['3'], ['6'], ['8'], ['9'], 2, ['3'], ['6'], ['8'], ['9'])

我似乎非常接近，不知道该怎么办？

Answer 1

一种更简单的方法：

l = [['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
a=set(i[0] for i in l)
b=list( (i,[]) for i in a)
[b[int(i[0])][1].append(i[1]) for i in l]
print(b)

输出：

[('0', ['1', '2', '3']), ('1', ['4', '6', '7', '9']), ('2', ['3', '6', '8', '9'])]

备用答案（不使用设置）

l = [['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
a=[]
for i in l:
    if i[0] not in a:
        a.append(i[0])
b=list( (i,[]) for i in a)
[b[int(i[0])][1].append(i[1]) for i in l]
print(b)

还输出

[('0', ['1', '2', '3']), ('1', ['4', '6', '7', '9']), ('2', ['3', '6', '8', '9'])]

Answer 2

您可以使用Pandas：

import pandas as pd
import numpy as np
l = [['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
df = pd.DataFrame(l, dtype=np.int)
s = df.groupby(0)[1].apply(list)
list(zip(s.index, s))

输出：

[(0, [1, 2, 3]), (1, [4, 6, 7, 9]), (2, [3, 6, 8, 9])]

如何从列表列表中创建元组列表？

2 个答案:

备用答案（不使用设置）