In Python 2.4, how can I strip out characters after ';'?

Question:

Let’s say I’m parsing a file, which uses ; as the comment character. I don’t want to parse comments. So if I a line looks like this:

example.com.              600     IN      MX      8 s1b9.example.net ; hello!

Is there an easier/more-elegant way to strip chars out other than this:

rtr = ''
for line in file:
    trig = False
    for char in line:
        if not trig and char != ';':
            rtr += char
        else:
            trig = True
    if rtr[max(rtr)] != 'n':
        rtr += 'n'
Asked By: lfaraone

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Answers:

just do a split on the line by comment then get the first element
eg

line.split(";")[0]
Answered By: ghostdog74

I’d recommend saying

line.split(";")[0]

which will give you a string of all characters up to but not including the first “;” character. If no “;” character is present, then it will give you the entire line.

Answered By: Eli Courtwright

For Python 2.5 or greater, I would use the partition method:

rtr = line.partition(';')[0].rstrip() + 'n'
Answered By: Sinan Ünür
file = open(r'c:temptest.txt', 'r')
for line in file:   print
   line.split(";")[0].strip()
Answered By: Gern Blanston

Reading, splitting, stripping, and joining lines with newline all in one line of python:

rtr = 'n'.join(line.split(';')[0].strip() for line in open(r'c:temptest.txt', 'r'))
Answered By: hughdbrown

I have not tested this with python but I use similar code else where.

import re
content = open(r'c:temptest.txt', 'r').read()
content = re.sub(";.+", "n")
Answered By: Matthew FL

So you’ll want to split the line on the first semicolon, take everything before it, strip off any lingering whitespace, and append a newline character.

rtr = line.split(";", 1)[0].rstrip() + 'n'

Links to Documentation:

Answered By: tgray

Here is another way :

In [6]: line = "foo;bar"
In [7]: line[:line.find(";")] + "n"
Out[7]: 'foon'
Answered By: Chmouel Boudjnah
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