是否可以在熊猫中加入单词?我有一个单词列表,我想再次将它们变成短语
数据
0 [hello, she, can, seem, to, form, something, like, a, coherent,...
1 [not, any, more,...
2 [it, is, unclear, if, any, better, deal,...
3 [but, few, in, her, party, seem, inclined ...
4 [it, is, unclear, if, the, basic, conditions, for, any,...
Name: Data, dtype: object
stop_words = set(stopwords.words('english'))
#new words
new_stopwords = {'hello'}
new_list = stop_words.union(new_stopwords)
#remove from NLTK stop list
not_stopwords = {'no', 'not, 'any'}
stopwords_list = set([word for word in new_list if word not in not_stopwords])
df['Data'] = df['Data'].' '.join([wrd for wrd in Data if wrd not in stopwords_list])
输出:
File "<ipython-input-281-498b9daa386f>", line 1
df['Description_pretraites'] = df['Description_pretraites'].' '.join([wrd for wrd in replace_hour_token if wrd not in stopwords_list])
^
SyntaxError: invalid syntax
好的输出
0 [can seem form something like coherent...
1 [not any more...
2 [is unclear any better deal...
3 [few party seem inclined ...
4 [is unclear basic conditions any...
Name: Data, dtype: object
从我所见,在熊猫中,该联接可用于连接列。但是是否可以在一个列中进行联接?
答案 0 :(得分:2)
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与生成器一起使用:
.apply
或嵌套列表理解:
df['Data']=df['Data'].apply(lambda x: ' '.join(wrd for wrd in x if wrd not in stopwords_list))
示例:
df['Data'] = [' '.join(wrd for wrd in x if wrd not in stopwords_list) for x in df['Data']]