嵌套列表<p>我有以下列表：</p> <pre><code>mylst = ['bla,bli', 'blula', 'blabla,bloblu,blibla', 'bla'] </code></pre> <p>怎樣纔可以轉化爲一個嵌套列表，例如：</p> <pre><code>mylst = [['bla','bli'], ['blula'], ['blabla','bloblu','blibla'], ['bla']] </code></pre>

Question

列表

Answer 1

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

Answer 2

如果你想拆就逗號，只是在list comprehension這樣做：

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

的str.split() method產生新的列表：

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

Answer 3

居然真的這麼簡單！

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

正如python3中Martijn Pieters指出的那樣，這將返回一個映射迭代器。爲了得到一個列表做：

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