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[英]Python regex to match alpha characters only ([A-Za-z] but for Unicode)
[英]python regular expression $[a-zA-Z ][0-9a-zA-Z ]* with forbidding
我正在尋找Python中的正則表達式,在字符串中找到了我:
$[a-zA-Z_][0-9a-zA-Z_]*
可以有更多這些,並且可以用空格(\\ s)分隔
這一切都將非常容易,但是如果有任何與模式不匹配的內容,我還需要禁止整個字符串。 (+空字符串也可以)
我給你舉一些例子:
$x$y0123 => OK, gives me [$x, $y0123]
$ => BAD (only $)
"" or " \t" => OK, gives me []
$x @hi => BAD, cause @hi, does not match the pattern
它可以是更多的正則表達式,而不必只是一個。
regex = re.compile("(\\$[a-zA-Z_][0-9a-zA-Z_]*)") regex.findall(string)
如果我也不必檢查那些事情,那沒關系。
要檢查整個字符串,最好使用re.match函數而不是re.findall,並且還允許使用空格的模式將像這樣^((\\$[a-zA-Z_][0-9a-zA-Z_])|(\\s))*$
嗯,我不太確定您要做什么,但是也許您需要2個正則表達式:第一個用於檢查格式的有效性,第二個用於檢索匹配項。
import re
stuff = ["$x$y0123", "$", "", " \t", "$x @hi"]
p1 = re.compile(r'(?:\$[A-Z_]\w*|\s)*$', re.IGNORECASE)
p2 = re.compile(r'\$[A-Z_]\w*|\s+', re.IGNORECASE)
for thing in stuff:
if p1.match(thing):
print(p2.findall(thing))
將打印:
['$x', '$y0123']
[]
[' \t']
嘗試這個:
import re
s1 = '$x$y0123 $_xyz1$B0dR_'
s2 = '$x$y0123 $_xyz1$B0dR_ @1'
s3 = '$'
s4 = ' \t'
s5 = ''
def process(s, pattern):
'''Find substrings in s that match pattern
if string is not completely composed of substings that match pattern
raises AttributeError
s --> str
pattern --> str
returns list
'''
rex = re.compile(pattern)
matches = list()
while s:
## print '*'*8
## print s1
m = rex.match(s)
matches.append(m)
## print '\t', m.group(), m.span(), m.endpos
s = s[m.end():]
return matches
pattern = '\$[a-zA-Z_][0-9a-zA-Z_]*'
for s in [s1, s2, s3, s4, s5]:
print '*'*8
# remove whitespace
s = re.sub('\s', '', s)
if not s:
print 'empty string'
continue
try:
matches = process(s, pattern)
except AttributeError:
print 'this string has bad stuff in it'
print s
continue
print '\n'.join(m.group() for m in matches)
>>>
********
$x
$y0123
$_xyz1
$B0dR_
********
this string has bad stuff in it
$x$y0123$_xyz1$B0dR_@1
********
this string has bad stuff in it
$
********
empty string
********
empty string
>>>
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