Python: Regular expression re

#re正则的用法: Match match matches from the beginning  search takes one back, findout take so match, slit split  Sub Replace

#-*-Coding:utf8-*-#Auth:fulimei#re regular usage: match match matches from the beginning search takes one back, findout fetch so match, slit split sub ReplaceImportReres=re.match ('^chen','Chenronghua')#^ Match character startResa=res.group ()#matching characters are printed outPrint(RES)Print(RESA)#If you write it dead , you don't have to.#\d matches a number, \d+ matches one or more digitsRe.search#Search this uses the mostA=re.search ('R[a-z]+a','112ronghua44ronghua')#[A-z] take A to Z of a character, more to match the last to take the firstPrint(a)#span= (3, ten), match= ' Ronghua ' >B=re.search ('R[a-za-z]+a','112ronaghua44ronghua')Print(b)#span= (3, one), match= ' Ronaghua ' >#match the previous character 0 or 1 times? Usage ofC=re.search ('AAL?','AADBD')Print(c)#L can not, matched 0 times match= ' AA 'D=re.search ('AAL?','AALDBD')Print(d)#matched 10 times match= ' AAL 'F=re.search ('[0-9]{3}','a1bc234')#Match 3 digits match= ' 234 ' >Print(f)#match one to three times, take one and returnG=re.search ('[0-9]{1,3}','a1bc234')#take one and return to match= ' 1 ' >Print(g)#match all, note findall No group methodH=re.findall ('[0-9]{1,3}','A1B2C345DE')Print(h)#[' 1 ', ' 2 ', ' 345 ']L=re.search ('abc| ABC','ABCDABC')#match= ' ABC 'Print(l) J=re.findall ('abc| ABC','ABCDABC')#match= ' ABC 'Print(J)#[' abc ', ' abc ']K=re.search ('abc{2}','FLMABCCC')Print(k)#match= ' ABCC ' >#match two pipe characters two times as | | =M=re.search ('(\|\|=) {2}','abc| | =|| =')Print(m)#match= ' | | =|| = '#\a match from the last \zN=re.search ('\a[0-9]+[a-z]\z','123b')#the end must be a lowercase letterPrint(n)#match= ' 123b '#\d matches non-numericO=re.search ('\d+','123$%a\r\n')Print(o)#match= ' $%a\r\n ' >#match numbers + lettersP=re.search ('\w+','12a3b$%')Print(p)#match= ' 12a3b ' >#Match Special charactersQ=re.search ('\w+','12a3b$%')Print(q)#match= ' $% ' >#\s Match space wrap tab (\ t)R=re.search ('\s+','123$%a \ r \ n')Print(R)#' (? P<name>, ...) ' Group matching provinces and countiesS=re.search ('(? P<province>[0-9]{3}) (? P<city>[0-9]{3}) (? P<BIRTHDAY>[0-9]{4})','220723198901021625'). Groupdict ()Print(s)Print(s[' City'])#divide split by whatT=re.split ('[0-9]+','ABC12DF44GG456DF')Print(t)#[' abc ', ' DF ', ' GG ', ' DF ']#Sub Substitution replace all numbers with |Y=re.sub ('[0-9]+','|','A1w234fg2ff2gg')Print(y)#a|w|fg|ff| GGZ=re.sub ('[0-9]+','|','A1w234fg2ff2gg', count=2)#Replace the top 2Print(z)#A|w|fg2ff2gg

the use of re regularization

# \s Match Space wrap tab (\ t)

Python: Regular expression re