Find a number between 2 regular expressions

[jvdb]

Hello,

I'm pretty new to Python.

Can someone please show me how to find a number between 2 regular expressions?
Something like this:

Text:
' MyText (mymessage): 0 (Time 0 hours)'
' MyText (mymessage): 344 (Time 34 hours)'
' MyText (mymessage): 1000 (Time 2000 hours)'

I'd like to "grep" for:
0
344
1000

And then later grep for:
0
34
2000

But am sure I will manage the second one when I can figure out how to do the first one.

I've played with the "re" function, but the output looks like this:
[' 0 ']

And I can't figure out how to get rid of the white space and quotes. The plan is to get the integer and compare that with another integer.

Thank you

[jvdb]

That output from my "re" did not post right, there are white spaces before the first quote and the 0.

[Laharl]

You can use the string function strip() to remove all whitespace from it (or lstrip()/rstrip() for just one end). As to the regex, what's your actual regex pattern?

[jvdb]

Hello,

Thanks for the reply.

This is my regexp:

s=' MyText (mymessage): 1000 (Time 2000 hours)'
pattern = re.compile(r':( .*?)\(')

pattern.findall (s)

I'm still playing with the pattern to try and figure out how the whole thing works.

Cheers

[jlm699]

pattern = re.compile(r':( .*?)\(')

Seeing that there is a space in your text on either side of the number, you should add spaces to your pattern so that it is like this:
[code=python]
pattern = re.compile(r': (.*?) \(')
[/code]

[jvdb]

Thanks for that.

There won't always be only 1 space on either side though.

My output looks something like this:

>>> s=' MyText (mymessage): 100 (Time 2000 hours)'
>>> pattern = re.compile(r':( .*?)\(')
>>> pattern.findall (s)
[' 100 ']


How do I get rid of the annoying [' and '] ?

I want to be able to say, x=100, compare that to y.

[bvdet]

I would probably do something like this:
[CODE=Python]import re

s1 = 'MyText (mymessage): 0 (Time 0 hours)'
s2 = 'MyText (mymessage): 344(Time 34 hours)'
s3 = 'MyText (mymessage):100 0 (Time 2000 hours)'

patt = re.compile(r'\) : *([0-9]+) *\(')


for s in [s1,s2,s3]:
m = patt.search(s)
if m:
print int(m.group(1))[/CODE]The pattern in parentheses (one or more digits, matching as many as possible) is saved as matchObj.group( 1). You can have no spaces or multiple spaces surrounding the digits.

[ripat]

Hi,

This is my first post on this forum and my very first attempt in Python. So forgive me if my answer is not to the point.

import re

s="""
MyText (mymessage):   0   (Time 0 hours)
MyText (mymessage): 344(Time 34 hours)
MyText (mymessage):1000 (Time 2000 hours)
"""

patt = re.compile(r':[^0-9]*([0-9]+)[^0-9]+([0-9]+)')

print patt.findall(s)

This returns:

[('0', '0'), ('344', '34'), ('1000', '2000')]