CODEWITHSUNDEEP
pythonregex
aeupinhere
aeupinhere

Reputation: 2983

regex to remove floating point values

I want to be able to remove numbers with decimal places from within a string of text using regex. See here

import re
obj = '''This is my #1 [email protected] <body/> 2 3 4 5 2345! 23542 312453 76666374 56s34534 
        1. [email protected] 
        1978-12-01 12:00:00 1.23 21.243
        <script>function stripScripts(s) {
            var div = document.createElement('div');
            div.innerHTML = s;
            var scripts = div.getElementsByTagName('script');
            var i = scripts.length;
            while (i--) {
              scripts[i].parentNode.removeChild(scripts[i]);
            }
            return div.innerHTML;
          }</script> 99.258 245.643.3456!'''
regex1 = re.compile('(?is)(<script[^>]*>)(.*?)(</script>)|(<.*?>)|(?<!\S)\d+(?!\S)')
out1 = re.sub(regex1, ' ', obj)
print out1

data = ' '.join(out1.split()).strip()
print data

This regex removes most of what I need it to but leaves 1.23, 21.243 and 99.258. I would like to append this current regex to remove those values as well...

regex = (?is)(<script[^>]*>)(.*?)(</script>)|(<.*?>)|(?<!\S)\d+(?!\S)

Upvotes: 1

Views: 2392

Answers (2)

aeupinhere
aeupinhere

Reputation: 2983

Thanks @Joran Beasley! I tried this and it worked.

(?is)(<script[^>]*>)(.*?)(</script>)|(<.*?>)|(?<!\S)\d+(?!\S)|([0-9]+\.[0-9]+ )

What is the advantage of adding the first "d" here?

(\d+\.[0-9 ]+)

Upvotes: 0

Joran Beasley
Joran Beasley

Reputation: 114038

re.sub("\d*\.\d+","",the_text)

wouldnt work? or maybe 

re.sub("(\d*\.\d+)|(\d+\.[0-9 ]+)","",the_text)

Upvotes: 2

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