基本上,我只想列出那些 windows 不会创建的文件。我如何列出 4.txt 和 4a.txt ...... s == s
是真的吗?
myList = '''
4.txt
4a.txt
spam
'''
myregex = 'a'
s = re.sub(myregex,'',myList)
if s == s:
print "got it!" # prints 'got it'
print s == s # prints 'true'
你不需要正则表达式。
只需检查文件是否存在os.path.exists
import os
lst = [your_fileNames....]
for arch in lst:
if not os.path.exists(arch):
print arch
代码:
import re
myList = '''
4.txt
4a.txt
spam
'''.split()
myList
reList = [
'4\w?\.txt', #4 followed by 0 or 1 alphanumeric then .txt
'4.*\.txt', #4 followed by anything and then .txt
'\d\w?\.txt', #a digit, then 0 or 1 alphanumeric then .txt
]
for name in myList:
for regex in reList:
if re.match(regex, name):
print ("%s matches %s" % (name, regex))
else:
print ("%s does not match %s" % (name, regex))
输出:
=================================重新开始================= ================
4.txt matches 4\w?\.txt 4.txt matches 4.*\.txt 4.txt matches \d\w?\.txt 4a.txt matches 4\w?\.txt 4a.txt matches 4.*\.txt 4a.txt matches \d\w?\.txt spam does not match 4\w?\.txt spam does not match 4.*\.txt spam does not match \d\w?\.txt