Python Regular Expressions

Python uses the RE module to provide the ability to handle regular expressions

Constant

常量                                     说明re.M、re.MULTILINE          多行模式  re.S、re.DOTALL                单行模式re.I、re.IGNORECASE       忽略大小写re.X、re.VERBOSE            忽略表达式中的空白字符使用 | 位或 运算开启多种选项

Method

Compile

re.compile(pattern,flags =0)设定flags，编译模式，返回正则表达式对象regex。pattern就是正则表达式字符串，flags是选项。正则表达式需要被编译，为了提高效率，这些被编译后的结果被保存，下次使用同样的pattern的时候，就不需要再次编译。re的其他方法为了提高效率都调用了编译方法，就是为了提速

One-time match

Re.match (pattern,string,flags=0) Regex.match (String[,pos[,endpos]) match match matches from the beginning of the string, The Regex object Match method can reset the starting position and end position, returning the match object Re.search (pattern,string,flags=0) Regex.search (String[,pos[,endpos]) Searching from scratch until the first match, the Regex object Search method can reset the start and end positions, returning the match object Re.fullmatch (pattern,string,flags=0) REGEX.FULLMATC (string[, Pos[,endpos]]) the entire string and regular expression matches the import re s = ' bottle\nbag\nbig\napple ' for i,c in Enumerate (s,1): Print ((i-1,c), end  = ' \ n ' if i%8==0 Else ') print () (0, ' B ') (1, ' O ') (2, ' t ') (3, ' t ') (4, ' L ') (5, ' E ') (6, ' \ n ') (7, ' B ') (8, ' a ') (9, ' G ') (Ten, ' \ n ') (One, ' B ') (The ' I ') (The ' G ') (' \ n ') (In ' a ') (+, ' P ') (+, ' P ') (at ' l ') (d, ' e ') #match method print ('--match--') result = Re.match (' B ', s) # find one without looking for print (1,result) result = Re.match (' A ', s) print (2, Result) #没找到返回Noneresult = Re.match (' ^a ', s,re. M) # Still looking from the beginning, multiline mode does not use print (3,result) result = Re.match (' ^a ', s,re. S) #依然从头开始找print (4,result) #先编译, and then use regular expressions to express regex = Re.compile (' a ') result = Regex.match (s) #依然从头开始找print (5,result)  result = Regex.match (s,15)# put index 15 as start looking for print (6,result) print () Search method print ('--search--') result = Re.search (' A ', s) # Scan to find the first position matching print (7,result Regex = Re.compile (' b ') result = Regex.search (s,1) print (8,result) # Bagregex = Re.compile (' ^b ', re. M) result = Regex.search (s) # no matter if it is not multiple lines, find it back to print (8.5,result) #bootleresult = Regex.search (s,8) print (9,result) # Bigfullmatch method result = Re.fullmatch (' bag ', s) print (10,result) regex = re.compile (' bag ') result = Regex.fullmatch (s) Print (11,result) result = Regex.fullmatch (s,7) print (12,result) result = Regex.fullmatch (s,7,10) print (13,result) # To exactly match, the less is not enough, [7,10) Full-text method Re.findall (pattern,string,flags=0) Regex.findall (String[,pos[,endpos]]) to the entire string, from left to right, Returns a list of all occurrences Re.finditer (pattern,string,flags=0) regex.fingiter (String[,pos[,endpos]) that matches the entire string, from left to right, and returns all occurrences. Returns an iterator note that each iteration returns a match object. Import re s = ' bottle\nbag\nbig\nable ' for i,c in Enumerate (s,1): Print ((i-1,c), end = ' \ n ' if i%8==0 Else ') pr Int () (0, ' B ') (1, ' O ') (2, ' t ') (3, ' t ') (4, ' L ') (5, ' E ') (6, ' \ n ') (7, ' B ') (8, ' a ') (9, ' G ') (, ' \ n ') (one, ' B ') (A, ' I ') (A, ' I ') (+, ' \ n ') (+, ' a ') (+, ' B ') (+, ' l ') (+, ' e ') FindAll method result = Re.findall (' b , ', s) print (1,result) regex = Re.compile (' ^b ') result = Regex.findall (s) print (2,result) regex = Re.compile (' ^b ', re. M) result = Regex.findall (s,7) print (3,result) #bag Bigregex = Re.compile (' ^b ', re. s) result = Regex.findall (s) print (4,result) # Bottleregex = Re.compile (' ^b ', re. M) result = Regex.findall (s,7,10) print (5,result) # Bagfiditer Method result = Regex.finditer (s) print (type result) print ( Next (Result) print (next result)

match Substitution

Re.sub (pattern,replacement,string,count=0,flags=0) regex.sub (replacement,string,count=0) uses pattern to match string strings , use repl substitution for matches. The replacement can be string, Bytes, functionre.subn (pattern,replace,string,count=0,flags=0) regex.subn (replance,string, string,count=0) with sub returns a tuple (New_string,number_of_subs_made) import re s = "' Bottle\nbag\nbig\napple" ' for i,c in  Enumerate (s,1): Print ((i-1,c), end = ' \ n ' if i%8==0 Else ') print () (0, ' B ') (1, ' O ') (2, ' t ') (3, ' t ') (4, ' L ') (5,  ' E ') (6, ' \ n ') (7, ' B ') (8, ' a ') (9, ' G ') (ten, ' \ n ') (one, ' B ') (A, ' I ') (+, ' g ') (+, ' \ n ') (+, ' a ') (+, ' P ') (+, ' P ') (at ' l ') (+, ' e ') Replacement method regex = Re.compile (' b\wg ') result = Regex.sub (' magedu ', s) print (1,result) result = Regex.sub (' magedu ', s,1) Print (2,result) regex = Re.compile (' \s+ ') result = Regex.subn (' \ t ', s) print (3,result) # The substituted string and the number of substitutions of the tuple split string Re.split ( pattren,string,maxsplit,flags=0) Re.split (split string) Import res = "Bottle02 bag03 big1100 Able" "for i,c in enumerate (s,1): Print ((i-1,c), end= ' \ n ' if i%8==0 Else ') print () extracts each line of words out of print (S.split ()) #做不到 [' big1 ' ...] result = Re.split (' [\s\d]+ ', s) print (1,result) regex = Re.compile (' ^[\s\d]+ ') result = Regex.Split (s) print (2,result) Regex = Re.compile (' ^[\s\d]+ ', re. M) result = Regex.Split (s) print (3,result) regex = Re.compile (' \s\d+\s+ ') result = Regex.Split (' +s ') print (4,result)

Grouping

The data captured using the parentheses pattern is placed in the group. The match, search function returns a Match object, FindAll returns a list of strings, Finditer returns a match object if the pattern uses a grouping, if the match has a result, it will be in the Match object 1, using Group (N) The method returns the corresponding grouping, 1 to n is the corresponding grouping, 0 returns the entire match character 2, if a named grouping is used, you can use group (' name ') to take the group 3, use Groupdict () to return all named groupings import re s = "' Bottle\nbag \nbig\napple "for I,c in Enumerate (s,1): Print ((i-1,c), end = ' \ n ' if i%8==0 Else ') print () #分组regex = Re.compile ( ' (b\w+) ') result = Regex.match (s) print (type (result)) print (1, ' Match ', result.groups ()) result = Regex.search (s,8) print (2, ' Match ', Result.groups ()) #命名分组regex = Re.compile (' (b\w+) \ n (? p<name2>b\w+) \ n (? p<name3>b\w+) ' result = Regex.match (s) print (3, ' match ', result) print (4,result.group (3), Result.group (2), Result.group (1)) print (5,result.group (0). Encode ()) # 0 returns the entire matching string, which is Matchprint (6,result.group (' name2 '), Result.group ( ' Name3 ') print (7,result.groups ()) print (8,result.groupdict ()) result = Regex.findall (s) for x in result: # string list Prin T (type (x), x) regex = Re.compile (? p

Python Regular expression