如何從句子列表中創建單詞列表？

Question

如果我有一個字符串列表，如：

lst = ['aa bb', 'cc dd', 'cc aa'] 

我怎樣才能進入的唯一字列表這個像這樣：

['aa', 'bb', 'cc', 'dd'] 

使用的理解？下面是據我已經得到了，無濟於事：

wordList = [x.split() for row in lst for x in row] 

Answer 1

我認爲最簡單的方法可能是這樣，雖然不是最有效的。

set(' '.join(lst).split()) 

如果你真的想要一個列表，然後只是包裝，在給list()

Answer 2

你要循環拆分值：

wordList = [word for row in lst for word in row.split()] 

然後用一組，使獨特的整個列表：

wordList = list({word for row in lst for word in row.split()}) 

或只使用一組，並用它做：

wordList = {word for row in lst for word in row.split()} 

演示：

>>> lst = ['aa bb', 'cc dd', 'cc aa'] 
>>> list({word for row in lst for word in row.split()}) 
['aa', 'cc', 'dd', 'bb'] 
>>> {word for row in lst for word in row.split()} 
set(['aa', 'cc', 'dd', 'bb']) 

如果順序事項（上面的代碼返回詞語任意順序，排序順序是巧合憑藉的CPython的實現細節），使用一個單獨的組來跟蹤重複值：

seen = set() 
wordList = [word for row in lst for word in row.split() if word not in seen and not seen.add(word)] 

爲了說明的差別，一個更好的輸入樣本：

>>> lst = ['the quick brown fox', 'brown speckled hen', 'the hen and the fox'] 
>>> seen = set() 
>>> [word for row in lst for word in row.split() if word not in seen and not seen.add(word)] 
['the', 'quick', 'brown', 'fox', 'speckled', 'hen', 'and'] 
>>> {word for row in lst for word in row.split()} 
set(['and', 'brown', 'fox', 'speckled', 'quick', 'the', 'hen']) 

Answer 3

In [25]: list({y for x in lst for y in x.split()}) 
Out[25]: ['aa', 'cc', 'dd', 'bb'] 

要保持它的順序，意味着從列表中刪除重複項，您可以參考http://www.peterbe.com/plog/uniqifiers-benchmark。

Answer 4

維持秩序，你可以這樣做：

>>> from collections import OrderedDict 
>>> lst = ['aa bb', 'cc dd', 'cc aa'] 
>>> new = [] 
>>> for i in lst: 
...  new.extend(i.split()) 
... 
>>> list(OrderedDict.fromkeys(new)) 
['aa', 'bb', 'cc', 'dd'] 

注意使用set()是最有可能更快，因爲馬亭指出。