Split string before regex

Question

I'm trying to insert a tab (\t) before a regex, in a string. Before "x days ago", where x is a number between 0-999.

The text I have looks like this:

Great product, fast shipping! 22 days ago anon
Fast shipping. Got an extra free! Thanks! 42 days ago anon

Desired output:

Great product, fast shipping! \t 22 days ago anon
Fast shipping. Got an extra free! Thanks! \t 42 days ago anon

I am still new to this, and I'm struggling. I've looked around for answers, and found some that are close, but none that are identical.

This is what I have so far:

text = 'Great product, fast shipping! 22 days ago anon'
new_text = re.sub(r"\d+ days ago", "\t \d+", text)
print new_text

Output:

Great product, fast shipping!    \d+ anon

Again, what I need is (note the \t):

Great product, fast shipping!    22 days ago anon

halex · Accepted Answer · 2015-04-25 19:37:47Z

3

You can use backreferences in your replacement string. Put parantheses around the \d+ days ago to make it a captured group and use \\1 inside your replacement to refer to this group's text:

>>> text = 'Great product, fast shipping! 22 days ago anon'
>>> new_text = re.sub(r"(\d+ days ago)", "\t\\1", text)
>>> print new_text
Great product, fast shipping!    22 days ago anon

answered Apr 25, 2015 at 19:37

halex

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Wiktor Stribiżew · Accepted Answer · 2015-04-25 19:44:41Z

1

You were replacing with a regex pattern, and you needed just a \1 backreference.

In order to just insert a tab before n days ago, you can use a look-ahead, and replace the captured number with a \t\1:

import re
p = re.compile(ur'(\d+)(?=\s+days\s+ago)')
test_str = u"Great product, fast shipping! 22 days ago anon\nFast shipping. Got an extra free! Thanks! 42 days ago anon"
subst = u"\t\\1"
print re.sub(p, subst, test_str)

Result of a demo:

Great product, fast shipping!   22 days ago anon
Fast shipping. Got an extra free! Thanks!   42 days ago anon

And a sample program.

edited Apr 25, 2015 at 19:44

answered Apr 25, 2015 at 19:37

Wiktor Stribiżew

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Comments

dawg · Accepted Answer · 2015-04-25 20:06:07Z

You can use a lookahead to do zero width insertion and ' ' to find the leading literal space:

>>> import re
>>> txt='''\
... Great product, fast shipping! 22 days ago anon
... Fast shipping. Got an extra free! Thanks! 42 days ago anon'''
>>> repr(re.sub(r' (?=\d+)', ' \t', txt))
"'Great product, fast shipping! \\t22 days ago anon\\nFast shipping. Got an extra free! Thanks! \\t42 days ago anon'"

Note that all the patterns fitting ' \d+' become ' \t\d+' which is what I think you are after.

If you want to limit to ' \d+ days ago'' just add that to the lookahead:

>>> txt='''\
... Great product, fast shipping! 22 days ago anon
... Fast shipping. Got an extra free! Thanks! 42 weeks ago anon'''
>>> repr(re.sub(r' (?=\d+ days ago)', ' \t', txt))
"'Great product, fast shipping! \\t22 days ago anon\\nFast shipping. Got an extra free! Thanks! 42 weeks ago anon'"

kpie · Accepted Answer · 2015-04-25 19:48:52Z

0

you can use

Tabindex = re.search(r"\d days ago",text).start()
text = text[0:Tabindex]+'\t'+text[Tabindex:len(text)]

answered Apr 25, 2015 at 19:48

kpie

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