CODEWITHSUNDEEP
pythonregex
historystamp
historystamp

Reputation: 1478

Python regular expression for r.findall

I am using findall to separate text.

I started with this expression re.findall(r'(.?)(\$.?\$)' but it doesn't give me the data after the last piece of text found. I missed the '6\n\n'

How do I get the last piece of text?

Here is my python code:

#!/usr/bin/env python

import re

allData = '''
1
2
3 here Some text in here 
$file1.txt$
4 Some text in here and more  $file2.txt$
5 Some text $file3.txt$ here  
$file3.txt$
6

'''

for record in re.findall(r'(.*?)(\$.*?\$)|(.*?$)',allData,flags=re.DOTALL) :
print repr(record)

The output I get for this is:

('\n1\n2\n3 here Some text in here \n', '$file1.txt$', '')
('\n4 Some text in here and more  ', '$file2.txt$', '')
('\n5 Some text ', '$file3.txt$', '')
(' here  \n', '$file3.txt$', '')
('', '', '\n6\n')
('', '', '')
('', '', '')

I really would like this output:

('\n1\n2\n3 here Some text in here \n', '$file1.txt$')
('\n4 Some text in here and more  ', '$file2.txt$')
('\n5 Some text ', '$file3.txt$')
(' here  \n', '$file3.txt$')
('\n6\n', '', )

Background info in case you need to see the larger picture.

I case your are interested, I'm re-writing this in python. I have the rest of the code under control. I am just getting too much stuff out of findall.

https://discussions.apple.com/message/21202021#21202021

Upvotes: 0

Views: 474

Answers (4)

Janne Karila
Janne Karila

Reputation: 25207

Here's one way to solve your substitution problem with findall. 

def readfile(name):
    with open(name) as f:
        return f.read()

r = re.compile(r"\$(.+?)\$|(\$|[^$]+)")

print "".join(readfile(filename) if filename else text 
    for filename, text in r.findall(allData))

Upvotes: 1

sotapme
sotapme

Reputation: 4903

If I understand correctly from that Apple link you want to do something like:

import re


allData = '''
1
2
3 here Some text in here
$file1.txt$
4 Some text in here and more  $file2.txt$
5 Some text $file3.txt$ here
$file3.txt$
6

'''


def read_file(m):
    return open(m.group(1)).read()

# Sloppy matching :D
# print re.sub("\$(.*?)\$",  read_file, allData)
# More precise.
print re.sub("\$(file\d+?\.txt)\$",  read_file, allData)

EDIT As Oscar suggests make match more precise.

ie. take the filename between $s and read the file for the data and that's what the above would do.

Example output:

1
2
3 here Some text in here

I'am file1.txt

4 Some text in here and more  
I'am file2.txt

5 Some text 
I'am file3.txt
 here

I'am file3.txt

6

Files:

==> file1.txt <==

I'am file1.txt

==> file2.txt <==

I'am file2.txt

==> file3.txt <==

I'am file3.txt

Upvotes: 2

ThomasH
ThomasH

Reputation: 23536

To achieve the output you want you need to restrict your pattern to 2 capture groups. (If you use 3 capture groups, you will have 3 elements in every "record").

You could make the second group optional, that should do the job:

r'([^$]*)(\$.*?\$)?'

Upvotes: 1

kiriloff
kiriloff

Reputation: 26333

This one is partly solving your problem

import re

allData = '''
1
2
3 here Some text in here 
$file1.txt$
4 Some text in here and more  $file2.txt$
5 Some text $file3.txt$ here  
$file3.txt$
6

'''

for record in re.findall(r'(.*?)(\$.*?\$)|(.*?$)',allData.strip(),flags=re.DOTALL) :
    print  [ x for x in record if x]

producing output

['1\n2\n3 here Some text in here \n', '$file1.txt$']
['\n4 Some text in here and more  ', '$file2.txt$']
['\n5 Some text ', '$file3.txt$']
[' here  \n', '$file3.txt$']
['\n6']
[]

Avoid last empty list with

for record in re.findall(r'(.*?)(\$.*?\$)|(.*?$)',allData.strip(),flags=re.DOTALL) :
    if ([ x for x in record if x] != []):
        print  [ x for x in record if x]

Upvotes: 0

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