如何在列表理解中添加列表列表 [英] How to add list of lists in a list comprehension

问题描述

我有一个列表推导式生成的列表，它通过查找哪些字符串的长度为3来按组对stripped中的数据进行排序，并且我想将它们合并，从而将它们与单个列表分开单长度字符串.

I have a list which is produced by a list comprehension and it sorts the data in stripped according to groups by finding which strings have a length of 3 and I want to merge them so that are in a single list separately from single length strings.

stripped = ['a,b', 'c,d', 'e', '', 'f,g', 'h', '', '']
lst = [[i.split(',')] if len(i) is 3 else i for i in stripped]
print(lst)
#[[['a', 'b']], [['c', 'd']], 'e', '', [['f', 'g']], 'h', '', '']

我想改为生成[[['a', 'b'], ['c', 'd'],['f', 'g']], 'e', '','h', '', '']

如果可能，我如何使用单列表理解来实现这一目标?

How can I achieve this with a Single list-comprehension if possible?

因为高效和简单而被 @HennyH的答案接受了

accepted @HennyH's answer because of its high efficiency and simplicity

推荐答案

l = [[]]
for e in stripped:
    (l[0] if len(e) == 3 else l).append(e)
>>> 
[['a,b', 'c,d', 'f,g'], 'e', '', 'h', '', '']

或者匹配OP的3个长字符串的输出:

Or to match the OP's output for the 3 long strings:

for e in stripped:
    l[0].append(e.split(',')) if len(e) == 3 else l.append(e)
>>> 
[[['a', 'b'], ['c', 'd'], ['f', 'g']], 'e', '', 'h', '', '']

这样，就不会像Inbar的解决方案那样将两个列表A，B进行附加连接.您还可以将stripped转换为生成器表达式，这样就不必将两个列表保存在内存中.

This way there is no additional concatenation of two lists A,B as Inbar's solution provided. You can also turn stripped into a generator expression so you don't need to hold the two lists in memory.

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