2

I am trying to remove all digits from a string that are not attached to a word. Examples:

 "python 3" => "python"
 "python3" => "python3"
 "1something" => "1something"
 "2" => ""
 "434" => ""
 "python 35" => "python"
 "1 " => ""
 " 232" => ""

Till now I am using the following regular expression:

((?<=[ ])[0-9]+(?=[ ])|(?<=[ ])[0-9]+|^[0-9]$)

which can correctly do some of the examples above, but not all. Any help and some explanation?

Mpizos Dimitris
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3 Answers3

5

Why not just use word boundaries?

\b\d+\b

Here is an example:

>>> import re
>>> words = ['python 3', 'python3', '1something', '2', '434', 'python 35', '1 ', ' 232']
>>> for word in words:
...     print("'{}' => '{}'".format(word, re.sub(r'\b\d+\b', '', word)))
...
'python 3' => 'python '
'python3' => 'python3'
'1something' => '1something'
'2' => ''
'434' => ''
'python 35' => 'python '
'1 ' => ' '
' 232' => ' '

Note that this will not remove spaces before and after. I would advise using strip(), but if not you can probably do \b\d+\b\s* (for space after) or something similar.

brianpck
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3

You could just split the words and remove any words that are digits which is a lot easier to read:

new = " ".join([w for w in s.split() if not w.isdigit()])

And also seems faster:

In [27]: p = re.compile(r'\b\d+\b')

In [28]: s =  " ".join(['python 3', 'python3', '1something', '2', '434', 'python
    ...:  35', '1 ', ' 232'])

In [29]: timeit " ".join([w for w in s.split() if not w.isdigit()])

100000 loops, best of 3: 1.54 µs per loop

In [30]: timeit p.sub('', s)

100000 loops, best of 3: 3.34 µs per loop

It also removes the space like your expected output:

In [39]:  re.sub(r'\b\d+\b', '', " 2")
Out[39]: ' '

In [40]:  " ".join([w for w in " 2".split() if not w.isdigit()])
Out[40]: ''

In [41]:  re.sub(r'\b\d+\b', '', s)
Out[41]: 'python  python3 1something   python     '

In [42]:  " ".join([w for w in s.split() if not w.isdigit()])
Out[42]: 'python python3 1something python'

So both approaches are significantly different.

Padraic Cunningham
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  • The OP mentioned in a comment that he has a solution without regexp already. (But yes, this is definitely the best way) – brianpck Oct 21 '16 at 14:17
  • @brianpck, the OP also wants `"1 "` to become `""` which this does and also more efficiently so I will leave the answer as it is a better overall approach for future readers and the fact it does the correct thing that the OP seems to want. – Padraic Cunningham Oct 21 '16 at 14:36
  • actually, you could just pass the generator into `join` instead of creating a list: `" ".join(w for w in s.split() if not w.isdigit())` – Bahrom Oct 21 '16 at 16:47
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    @Bahrom, that is slower as python will build a list internally if you pass a generator, – Padraic Cunningham Oct 21 '16 at 16:47
  • @Bahrom, some comparisons here http://stackoverflow.com/a/9061024/2141635, join has to do two passes over the data, one to check the size needed and one to create the new string. Because you can only iterate a generator once, it first has to be converted a list. – Padraic Cunningham Oct 21 '16 at 16:53
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    Thanks, will check that out! – Bahrom Oct 21 '16 at 16:54
0

This regex, (\s|^)\d+(\s|$), could work as shown below in javascript

var value = "1 3@bar @foo2 * 112";
var matches = value.replace(/(\s|^)\d+(\s|$)/g,"");
console.log(matches)

It works in 3 parts:

  1. It first matches a space or begging of string using (\s|^) with \s matching a white-space | meaning or and ^ meaning beginning of string.
  2. next matching digits from 1 to times using \d for a digit and + to match 1 to N times but as many as possible.
  3. Finally (\s|$) matches a space or end of sting with \s matching space, | meaning or, and $ matching end of string.

You can replace $ with end of line or \n if you have several lines or just add it in next to it like this (\s|$|\n). Hope this is what your're looking for.

milo.farrell
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