Python regular expression example code

The re module makes the Python language have all the regular expression functions.

Syntax that will be used

note:

1. Escape character

 s
'( abc)def'
 m = re.search("(\(.*\)).*", s)
 print m.group(1)(abc)

re.match function

re.match tries to match a pattern from the beginning of the string. If the match is not successful at the beginning, match() returns none.

Example 1:

#! /usr/bin/python
# - *- coding: UTF-8-*-import re
print(re.match('www','www.zalou.cn').span()) #Match at the start
print(re.match('net','www.zalou.cn'))     #Does not match at the beginning

Output result:

(0, 3)
None

Example 2:

#! /usr/bin/python
import re
 
line ="Cats are smarter than dogs"
 
matchObj = re.match( r'(.*) are (.*?) .*', line, re.M|re.I)if matchObj:
 print "matchObj.group() : ", matchObj.group()
 print "matchObj.group(1) : ", matchObj.group(1)
 print "matchObj.group(2) : ", matchObj.group(2)else:
 print "No match!!"

Output result:

matchObj.group() : Cats are smarter than dogs
matchObj.group(1) : Cats
matchObj.group(2) : smarter

The above is the print output of python2. Just remember to add () in python, and the output of python is similar to the \n in other languages to match the obtained content.

python group()

In regular expressions, group() is used to propose the string intercepted by grouping, and () is used to group

Repeat the previous string multiple times

 a ="kdlal123dk345"
 b ="kdlal123345"
 m = re.search("([0-9]+(dk){0,1})[0-9]+", a)
 m.group(1), m.group(2)('123dk','dk')
 m = re.search("([0-9]+(dk){0,1})[0-9]+", b)
 m.group(1)'12334'
 m.group(2)

The reason

1. The three sets of brackets in the regular expression divide the matching results into three groups

group() is the same as group(0) to match the overall result of the regular expression
group(1) lists the first bracket matching part, group(2) lists the second bracket matching part, group(3) lists the third bracket matching part.
2. If there is no successful match, re.search() returns None

3. Of course, there are no parentheses in the regular expression, and group(1) is definitely wrong.

Example

1. Determine whether the string is all lowercase

# - *- coding: cp936 -*-import re 
s1 ='adkkdk'
s2 ='abc123efg'

an = re.search('^[a-z]+$', s1)if an:
 print 's1:', an.group(),'All lowercase'else:
 print s1,"Not all lowercase!"

an = re.match('[a-z]+$', s2)if an:
 print 's2:', an.group(),'All lowercase'else:
 print s2,"Not all lowercase!"

result

The reason

1. Regular expressions are not part of python, you need to quote the re module when using

2. The matching form is: re.search (regular expression, with matching string) or re.match (regular expression, with matching string). The difference between the two is that the latter starts with a start character (^) by default. therefore,

re.search('^[1]+′,s1) is equivalent to re.match('[a−z]+', s2)

3. If the match fails, an = re.search('^[2]+$', s1) returns None

group is used to group matching results

E.g

import re
a ="123abc456"
print re.search("([0-9]*)([a-z]*)([0-9]*)",a).group(0)  #123abc456,Back to the whole
print re.search("([0-9]*)([a-z]*)([0-9]*)",a).group(1)  #123
print re.search("([0-9]*)([a-z]*)([0-9]*)",a).group(2)  #abc
print re.search("([0-9]*)([a-z]*)([0-9]*)",a).group(3)  #456

Output result

123 abc456
123
abc
456

1 ) The three sets of parentheses in the regular expression divide the matching results into three groups

group() is the same as group(0) to match the overall result of the regular expression

group(1) lists the first bracket matching part, group(2) lists the second bracket matching part, group(3) lists the third bracket matching part.

2 ) If there is no match, re.search() returns None

3 ) Of course, there are no parentheses in the regular expression, and group(1) is definitely wrong.

2. Acronym expansion

Concrete example

FEMA Federal Emergency Management Agency
IRA Irish Republican Army
DUP Democratic Unionist Party

FDA Food and Drug Administration
OLC Office of Legal Counsel
analysis

Abbreviation FEMA
Decomposed into F*** E*** M*** A***
Regular uppercase letters + lowercase (more than or equal to 1) + spaces

Reference Code

import re
def expand_abbr(sen, abbr):
 lenabbr =len(abbr)
 ma =''for i inrange(0, lenabbr):
 ma += abbr[i]+"[a-z]+"+' '
 print 'ma:', ma
 ma = ma.strip(' ')
 p = re.search(ma, sen)if p:return p.group()else:return''

print expand_abbr("Welcome to Algriculture Bank China",'ABC')

result

problem

The above code is correct for the first 3 in the example, but the latter two are wrong, because the words beginning with capital letters are also mixed with lowercase letters.

law

Uppercase letter + lowercase (more than or equal to 1) + space + [lower case + space] (0次或1次)

Reference Code

import re
def expand_abbr(sen, abbr):
 lenabbr =len(abbr)
 ma =''for i inrange(0, lenabbr-1):
 ma += abbr[i]+"[a-z]+"+' '+'([a-z]+ )?'
 ma += abbr[lenabbr-1]+"[a-z]+"
 print 'ma:', ma
 ma = ma.strip(' ')
 p = re.search(ma, sen)if p:return p.group()else:return''

print expand_abbr("Welcome to Algriculture Bank of China",'ABC')

skill

The set of lowercase letters in the middle + a space, as a whole, add parentheses. Either at the same time or not at the same time, so need to use it? , To match the whole ahead.

3. Remove the comma in the number

Concrete example

When dealing with natural language, if 123,000,000 is divided by punctuation, there will be a problem. A good number is dismembered by a comma, so you can deal with the number first (remove the comma).

analysis

The number is often a group of 3 numbers, followed by a comma, so the rule is: ***, ***, ***

Regular expression

[ a-z]+,[a-z]?

Reference code 3-1

import re

sen ="abc,123,456,789,mnp"
p = re.compile("\d+,\d+?")for com in p.finditer(sen):
 mm = com.group()
 print "hi:", mm
 print "sen_before:", sen
 sen = sen.replace(mm, mm.replace(",",""))
 print "sen_back:", sen,'\n'

result

skill

Use the function finditer(string[, pos[, endpos]]) | re.finditer(pattern, string[, flags]):

Search for string and return an iterator that sequentially accesses each match result (Match object).

Reference code 3-2

sen ="abc,123,456,789,mnp"while1:
 mm = re.search("\d,\d", sen)if mm:
 mm = mm.group()
 sen = sen.replace(mm, mm.replace(",",""))
 print sen
 else:break

result

extend

Such a program is aimed at a specific problem, that is, a group of 3 digits. If the numbers are mixed with letters, eliminate the commas between the numbers, that is, convert "abc,123,4,789,mnp" into "abc,1234789,mnp"

Ideas

More specifically, find the regular expression "number, number" and replace it with the comma removed

Reference code 3-3

sen ="abc,123,4,789,mnp"while1:
 mm = re.search("\d,\d", sen)if mm:
 mm = mm.group()
 sen = sen.replace(mm, mm.replace(",",""))
 print sen
 else:break
print sen

result

4. Year conversion for Chinese processing (for example: 1949-1949)

Chinese processing involves coding issues. For example, when the program below recognizes the year (****year)

# - *- coding: cp936 -*-import re
m0 ="New China was founded in 1949"
m1 ="5.2% lower than in 1990"
m2 ='Man defeated the Russian army in 1996,Achieved substantial independence'

def fuc(m):
 a = re.findall("[zero|One|two|three|four|Fives|six|Seven|Eight|nine]+year", m)if a:for key in a:
  print key
 else:
 print "NULL"fuc(m0)fuc(m1)fuc(m2)

operation result

It can be seen that there are errors in the second and third.

Improvement-standardization into unicode recognition

# - *- coding: cp936 -*-import re
m0 ="New China was founded in 1949"
m1 ="5.2% lower than in 1990"
m2 ='Man defeated the Russian army in 1996,Achieved substantial independence'

def fuc(m):
 m = m.decode('cp936')
 a = re.findall(u"[\u96f6|\u4e00|\u4e8c|\u4e09|\u56db|\u4e94|\u516d|\u4e03|\u516b|\u4e5d]+\u5e74", m)if a:for key in a:
  print key
 else:
 print "NULL"fuc(m0)fuc(m1)fuc(m2)

result

It can be recognized by replacing Chinese characters with numbers.

reference

numHash ={}
numHash['zero'.decode('utf-8')]='0'
numHash['One'.decode('utf-8')]='1'
numHash['two'.decode('utf-8')]='2'
numHash['three'.decode('utf-8')]='3'
numHash['four'.decode('utf-8')]='4'
numHash['Fives'.decode('utf-8')]='5'
numHash['six'.decode('utf-8')]='6'
numHash['Seven'.decode('utf-8')]='7'
numHash['Eight'.decode('utf-8')]='8'
numHash['nine'.decode('utf-8')]='9'

def change2num(words):
 print "words:",words
 newword =''for key in words:
 print key
 if key in numHash:
  newword += numHash[key]else:
  newword += key
 return newword

def Chi2Num(line):
 a = re.findall(u"[\u96f6|\u4e00|\u4e8c|\u4e09|\u56db|\u4e94|\u516d|\u4e03|\u516b|\u4e5d]+\u5e74", line)if a:
 print "------"
 print line
 for words in a:
  newwords =change2num(words)
  print words
  print newwords
  line = line.replace(words, newwords)return line

5. Multiple mobile phone numbers, separated by |

For example:

Null value
12222222222
12222222222|12222222222
12222222222|12222222222|12222222444

expression

s = “[\d]{11}(\|[\d]{11})*|”

4. Recommended

Python regular expression guide

So far this article about python regular expression example code is introduced. For more relevant python regular expression example content, please search for previous articles on ZaLou.Cn or continue to browse related articles below. Hope you will support ZaLou.Cn more in the future!