嵌套列表理解等价

Question

是否可以将以下代码替换为列表理解表达式？

input = ['1\t2,3\t4,5', '61\t7,8\t9,0']

res = []
li = [i.split() for i in input]
for i in li:
    l = [i[0]]
    l = l + [e.split(',') for e in i[1:]]
    res.append(l)

问题在于，每个子列表中的第一个元素应与其余元素区别对待。

Answer 1

我不得不说，考虑到可读性，这实际上并不是Pythonic。

>>> l = ['1\t2,3\t4,5', '61\t7,8\t9,0']
>>> [[i[0]]+[e.split(',') for e in i[1:]] for i in [x.split() for x in l]]
[['1', ['2', '3'], ['4', '5']], ['61', ['7', '8'], ['9', '0']]]

Answer 2

>>> input = ['1\t2,3\t4,5', '61\t7,8\t9,0']
>>> 
>>> [[a.split()[0]] + [b.split(',') for b in a.split()[1:]] for a in input]
[['1', ['2', '3'], ['4', '5']], ['61', ['7', '8'], ['9', '0']]]

Answer 3

>>> import csv
>>> data = ['1\t2,3\t4,5', '61\t7,8\t9,0']
>>> [x[:1] + list(csv.reader(x[1:], delimiter=','))
     for x in csv.reader(data, delimiter='\t')]
[['1', ['2', '3'], ['4', '5']], ['61', ['7', '8'], ['9', '0']]]