我想从文本文件中提取单引号之间的所有单词。文本文件如下所示:
u'MMA': 10,
=u'acrylic'= : 19,
== u'acting lessons': 2,
=u'aerobic': 141,
=u'alto': 2= 4,
=u&#= 39;art therapy': 4,
=u'ballet': 939,
=u'ballroom'= ;: 234,
= =u'banjo': 38,
理想情况下,我的输出看起来像这样:
MMA,
acrylic,
acting lessons,
...
从浏览帖子来看,似乎我应该使用一些NLTK / regex的组合来实现这一目标。我尝试过以下方法:
import re
file = open('artsplus_categories.txt', 'r').readlines()
for line in file:
list = re.search('^''$', file)
file.close()
并收到以下错误:
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/re.py", line 142, in search
return _compile(pattern, flags).search(string)
TypeError: expected string or buffer
我认为错误可能是由于我如何寻找模式造成的。我的逻辑是我在' ....'。
内搜索所有内容绊倒re.py?
谢谢!
遵循Ashwini的评论:
import re
file = open('artsplus_categories.txt', 'r').readlines()
for line in file:
list = re.search('^''$', line)
print list
#file.close()
但输出中没有任何内容:
Samuel-Finegolds-MacBook-Pro:~ samuelfinegold$ /var/folders/jv/9_sy0bn10mbdft1bk9t14qz40000gn/T/Cleanup\ At\ Startup/artsplus_categories_clean-393952531.278.py.command ; exit;
None
logout
@Rasco:这是我得到的错误:
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/re.py", line 177, in findall
return _compile(pattern, flags).findall(string)
TypeError: expected string or buffer
logout
我正在使用此代码:
file2 = open('artsplus_categories.txt', 'r').readlines()
list = re.findall("'[^']*'", file2)
for x in list:
print (x)
答案 0 :(得分:2)
您没有将line
传递给正则表达式,而是将整个列表(文件)传递给它。您应该将line
传递给re.search
而不是file
。
for line in file:
lis = re.search('^''$', line) # line not file
请勿使用list
,file
作为变量名称。它们是内置函数。
<强>更新强>
with open('artsplus_categories.txt') as f:
for line in f:
print re.search(r"'(.*)'", line).group(1)
...
MMA
acrylic
acting lessons
aerobic
alto
art therapy
ballet
ballroom
banjo
答案 1 :(得分:2)
试试这个代码示例:
import re
file = """u'MMA': 10,
=u'acrylic'= : 19,
== u'acting lessons': 2,
=u'aerobic': 141,
=u'alto': 2= 4,
=u&#= 39;art therapy': 4,
=u'ballet': 939,
=u'ballroom'= ;: 234,
= =u'banjo': 38,"""
list = re.findall("'[^']*'", file)
for x in list:
print (x)
显示正确的值。请记住,示例中的某个值无法正确打开引号,因此匹配将在那里打破。