我正在尝试清理我的句子以及我想在句子中删除这些标签的内容(它们是下划线的形式,后跟一个单词,例如" _UH")。 基本上我想删除后跟下划线的字符串(也删除下划线本身)
文本:
['hanks_NNS sir_VBP',
'Oh_UH thanks_NNS to_TO remember_VB']
需要输出:
['hanks sir',
'Oh thanks to remember']
以下是我尝试过的代码:
for i in text:
k= i.split(" ")
print (k)
for z in k:
if "_" in z:
j=z.replace("_",'')
print (j)
当前输出:
ThanksNNS
sirVBP
OhUH
thanksNNS
toTO
rememberVB
RemindVB
答案 0 :(得分:3)
您可以使用re.sub()
执行此操作。匹配字符串中的所需子字符串,并将子字符串替换为空字符串:
import re
text = ['hanks_NNS sir_VBP', 'Oh_UH thanks_NNS to_TO remember_VB']
curated_text = [re.sub(r'_\S*', r'', a) for a in text]
print curated_text
输出:
['hanks sir', 'Oh thanks to remember']
正则表达式:
_\S* - Underscore followed by 0 or more non space characters
text = ['hanks_NNS sir_VBP', 'Oh_UH thanks_NNS to_TO remember_VB']
curated_text = [] # Outer container for holding strings in text.
for i in text:
d = [] # Inner container for holding different parts of same string.
for b in i.split():
c = b.split('_')[0] # Discard second element after split
d.append(c) # Append first element to inner container.
curated_text.append(' '.join(d)) # Join the elements of inner container.
#Append the curated string to the outer container.
print curated_text
输出:
['hanks sir', 'Oh thanks to remember']
当你想要用空字符串替换'_'
和后面的字符时,你只是用空字符串替换'_'
。
for i in text:
k= i.split(" ")
print (k)
for z in k:
if "_" in z:
j=z.replace("_",'') # <--- 'hanks_NNS' becomes 'hanksNNS'
print (j)