Sorting os.listdir()'s arbitrary order for desired output-CodePudding

Image of Txt Files in Folder
I currently possess a folder with .txt files that are named "Month day.txt"

Running os.listdir('.') outputs in this order:

["April 30.txt", "April 4.txt", "May 1.txt", "May 10.txt", "May 11.txt", "May 2.txt"]

The order I would like to output:

["April 4.txt", "April 30.txt", "May 1.txt", "May 2.txt", "May 10.txt", "May 11.txt"]

To get it, I have a feeling to somehow sort the names by numbers first and then alphabetical.
I spent the last hour researching similar problems, and the best I was able to find was this.

files = os.listdir('.')
re_pattern = re.compile('. ?(\d )\.([a-zA-Z0-9 ])')
files_ordered = sorted(files, key=lambda x: int(re_pattern.match(x).groups()[0]))

Which I understood used a regex to capture the digits and used it as a key for sorting, which sensibly arranged the files based on the numbers only. (May 12, April 13, etc)
Furthermore, I tried to mess around with it using capture groups on the month, but to no avail.

CodePudding user response：

You can actually sort this rather easily with the datetime module.

For example:

from datetime import datetime
a = ["April 30.txt", "April 4.txt", "May 1.txt", "May 10.txt", "May 11.txt", "May 2.txt"]

b = sorted(a,key=lambda x: datetime.strptime(x[:-4], "%B %d"))
print(b)

output:

['April 4.txt', 'April 30.txt', 'May 1.txt', 'May 2.txt', 'May 10.txt', 'May 11.txt']

What this does is it takes the sequence of filenames, and for each filename it removes the extension .txt and then converts the remaining date string into a date object with datetime.strptime function. The builtin sorted function automatically knows how to sot date objects.

If you were in a situation where you had multiple file extensions with different lengths, e.g. .txt , .json instead of using x[:-4] you could instead use os.path.splitext(x)[0] or use regex to isolate and remove the file extension.