I'm using Python 3. I'm trying to remove from each line of a file a specific part of it. Each line is a string and they have the same similar format, like so:

/* 7 */
margin-top:1.5rem!important/* 114 */
}/* 115 *//* 118 *//* 121 */
.mb-2{/* 122 */
margin-bottom:.5rem!important/* 123 */
}/* 124 *//* 127 *//* 130 *//* 133 *//* 137 */

I want to remove in each line that "has multiple quoted numbers" like this for example (with 3 quoted numbers):

}/* 115 *//* 118 *//* 121 */

or this for example (with 5 quoted numbers)

}/* 124 *//* 127 *//* 130 *//* 133 *//* 137 */

I want to remove all quoted numbers except the "first one", so for the first example, output should be like this:

}/* 115 */

and for the second example, output should be like this:

}/* 124 */

Here is my code:

def clean_file(file_name, new_output_file_name):
    with open(file_name, 'r') as src:
        with open(new_output_file_name, 'w') as output_file:
            for line in src:
                if line.strip().startswith('/*'):
                    output_file.write('%s' % '')
                elif line # How can I isolate and remove the final part I don't want?
                    output_file.write('%s\n' % line.rstrip('\n')) # How can I isolate and remove the final part I don't want?
                else:
                    output_file.write('%s\n' % line.rstrip('\n'))


clean_file('new.css', 'clean.css')

How can I isolate and remove the final part of the string that I don't want with Python?

CodePudding user response：

You can use re.sub for this. Use this regex to search:

(/\* \d  \*/)/\*. 

And replace it with r"\1"

RegEx Demo

Code:

import re
src = '}/* 124 *//* 127 *//* 130 *//* 133 *//* 137 */'
print (re.sub(r'(/\* \d  \*/)/\*. ', r'\1', src))
## }/* 124 */

RegEx Breakup:

• (/\* \d \*/): Match a string that has /* <number> */ and capture in group #1
• /\*: Match /*
• . : Followed by 1 of any char till end
• `\1': Is replacement that puts capture value of group #1 back