CODEWITHSUNDEEP
pythonregex
python
python

Reputation: 1940

How to write a regex in python to match this?

The code is as follows:

#coding=utf-8

import re

str = "The output is\n"
str += "1) python\n"
str += "A dynamic language\n"
str += "easy to learn\n"
str += "2) C++\n"
str += "difficult to learn\n"
str += "3244) PHP\n"
str += "eay to learn\n"


pattern = r'^[1-9]+\) .*'
print re.findall(pattern,str,re.M)

The output is 

['1) python', '2) C++', '3244) PHP']

However, I want to split it like this:

['1) python\n'A dynamic language\n easy to learn\n'  2) C++\n difficult to learn\n', '3244) PHP\n easy to learn\n']

That is, ignore the first lines does not start with "number)",and when comes across a number, the following lines until next line start with a "number)" is consider to be the same group. How should I rewrite the pattern ?

Upvotes: 3

Views: 106

Answers (3)

Casimir et Hippolyte
Casimir et Hippolyte

Reputation: 89547

you can use this, that allow digits but not followed by a closing parenthesis:

re.findall(r'\d+\)\s(?:\D+|\d+(?!\d*\)))*',str)

Upvotes: 2

Ashwini Chaudhary
Ashwini Chaudhary

Reputation: 250881

>>> import re
>>> strs = 'The output is\n1) python\nA dynamic language\neasy to learn\n2) C++\ndifficult to learn\n3244) PHP\neay to learn\n'
>>> re.findall(r'\d+\)\s[^\d]+',strs)
['1) python\nA dynamic language\neasy to learn\n',
'2) C++\ndifficult to learn\n',
'3244) PHP\neay to learn\n']

Upvotes: 3

xgord
xgord

Reputation: 4776

You need to add the python regex for whitespace into your pattern to account for the newlines.

Try this:

regex = r"[1-9]+\) .*\s.*"

\s is the regex for any whitespace

Upvotes: 1

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