例如，我有以下列表：

word1 = ['organization', 'community']

我有一个功能，可以从列表中的单词中获取同义词：

from nltk.corpus import wordnet as wn

def getSynonyms(word1):

synonymList1 = []

for data1 in word1:

wordnetSynset1 = wn.synsets(data1)

tempList1=[]

for synset1 in wordnetSynset1:

synLemmas = synset1.lemma_names()

for i in xrange(len(synLemmas)):

word = synLemmas[i].replace('_',' ')

if word not in tempList1:

tempList1.append(word)

synonymList1.append(tempList1)

return synonymList1

syn1 = getSynonyms(word1)

print syn1

这是输出：

[[u'organization', u'organisation', u'arrangement', u'system', u'administration', u'governance', u'governing body', u'establishment', u'brass', u'constitution', u'formation'], [u'community', u'community of interests', u'residential district', u'residential area', u'biotic community']]

^上面的输出显示了'organization'和的每个同义词集都'community'在列表内列出。然后我降低list的级别：

newlist1 = [val for sublist in syn1 for val in sublist]

这是输出：

[u'organization', u'organisation', u'arrangement', u'system', u'administration', u'governance', u'governing body', u'establishment', u'brass', u'constitution', u'formation', u'community', u'community of interests', u'residential district', u'residential area', u'biotic community']

^现在所有同义词集保持相同的字符串，没有子列表。而我现在想要做的是使所有的同义词newlist1集互相列出。我希望输出将是这样的：

[[u'organization'], [u'organisation'], [u'arrangement'], [u'system'], [u'administration'], [u'governance'], [u'governing body'], [u'establishment'], [u'brass'], [u'constitution'], [u'formation'], [u'community'], [u'community of interests'], [u'residential district'], [u'residential area'], [u'biotic community']]

我正在尝试此代码：

uplist1 = [[] for x in syn1]

uplist1.extend(syn1)

print uplist1

但是结果不是我所期望的：

[[], [], [u'organization', u'organisation', u'arrangement', u'system', u'administration', u'governance', u'governing body', u'establishment', u'brass', u'constitution', u'formation'], [u'community', u'community of interests', u'residential district', u'residential area', u'biotic community']]

它显示了两个空列表和两个同义集的两个列表'organization'和'community'

如何使每个同义词集的子列表？

解决方案

像这样吗

uplist1 = []

for i in syn1:

uplist1.append([i])

编辑：

与列表理解等效：

uplist1 = [[i] for i in syn1]