Python中列表的嵌套列表

Question

我有以下清单：

mylst = ['bla,bli', 'blula', 'blabla,bloblu,blibla', 'bla']

如何将其转换为嵌套列表，例如：

mylst = [['bla','bli'], ['blula'], ['blabla','bloblu','blibla'], ['bla']]

Answer 1

如果您想用逗号分开，只需在list comprehension中这样做：

mylst = [elem.split(',') for elem in mylst]

^{} method生成新列表：

>>> mylst = ['bla,bli', 'blula', 'blabla,bloblu,blibla', 'bla']
>>> [elem.split(',') for elem in mylst]
[['bla', 'bli'], ['blula'], ['blabla', 'bloblu', 'blibla'], ['bla']]

Answer 2

>>> mylst = ['bla,bli', 'blula', 'blabla,bloblu,blibla', 'bla']
>>> mylst = [item.split(',') for item in mylst]
>>> mylst
[['bla', 'bli'], ['blula'], ['blabla', 'bloblu', 'blibla'], ['bla']]

Answer 3

其实很简单！你知道吗

myLst2 = map(lambda x: x.split(','), mylst)

正如martijnpieters在python3中指出的，这将返回一个映射迭代器。要获取列表，请执行以下操作：

mapIter = map(lambda x: x.split(','), mylst)
myLst2 = list(mapIter)