批处理文件重命名：使用正则表达式的填充时间为零？

Question

I have a whole set of files (10.000+) that include the date and time in the filename. The problem is that the date and time are not zero padded, causing problems with sorting.

The filenames are in the format: output 5-11-2018 9h0m.xml
What I would like is it to be in the format: output 05-11-2018 09h00m.xml

I've searched for different solutions, but most seem to use splitting strings and then recombining them. That seems pretty cumbersome, since in my case day, month, hour and minute then need to be seperate, padded and then recombined.

I thought regex might give me some better solution, but I can't quite figure it out.

I've edited my original code based on the suggestion of Wiktor Stribiżew that you can't use regex in the replacement and to use groups instead:

import os
import glob
import re

old_format = 'output [1-9]-11-2018 [1-2]?[1-9]h[0-9]m.xml'
dir = r'D:\Gebruikers\<user>\Documents\datatest\'   

old_pattern = re.compile(r'([1-9])-11-2018 ([1-2][1-9])h([0-9])m')

filelist = glob.glob(os.path.join(dir, old_format))
for file in filelist:
    print file
    newfile = re.sub(old_pattern, r'0\1-11-2018 \2h0\3m', file)
    os.rename(file, newfile)

But this still doesn't function completely as I would like, since it wouldn't change hours under 10. What else could I try?

Answer 1

You can pad the numbers in your file names with .zfill(2) using a lambda expression passed as the replacement argument to the re.sub method.

Also, fix the regex pattern to allow 1 or 2 digits: (3[01]|[12][0-9]|0?[1-9]) for a date, (2[0-3]|[10]?\\d) for an hour (24h), and ([0-5]?[0-9]) for minutes:

old_pattern = re.compile(r'\b(3[01]|[12][0-9]|0?[1-9])-11-2018 (2[0-3]|[10]?\d)h([0-5]?[0-9])m')

See the regex demo .

Then use:

for file in filelist:
    newfile = re.sub(old_pattern, lambda x: '{}-11-2018 {}h{}m'.format(x.group(1).zfill(2), x.group(2).zfill(2), x.group(3).zfill(2)), file)
    os.rename(file, newfile)

See Python re.sub docs:

If repl is a function, it is called for every non-overlapping occurrence of pattern . The function takes a single match object argument, and returns the replacement string.

Answer 2

I suggest going more generic with old_pattern for simplicity, assuming your filenames are only misbehaving with digits:

Because combinations of filenames matching a single-digit field that needs converting in any position but are double digits in other fields would need a long regex to list out more explicitly, I suggest this much simpler one to match the files to rename, which makes assumptions that there are only this matching type of file in the directory as it opens it up more widely in order to be simpler to write and read at a glance - find any single digit field in the filename (one or more of) - ie. non-digit, digit, non-digit:

old_format = r'output\\.*\\D\\d\\D.*\\.xml'

The fixing re.sub statement could then be:

newfile = re.sub(r'\\D(\\d)[hm-]', lambda x: x.group()[0]+x.group()[1].zfill(2)+x.group()[2], file)

This would also catch unicode non-ascii digits unless the appropriate re module flags are set.

If the year (2018 in example) might be given as just '18' then it would need special handling for that - could be separate case, and also adding a space into the re.sub regex pattern set (ie [-hm ] ).