CODEWITHSUNDEEP
pythonlistdirectory
YosiFZ
YosiFZ

Reputation: 7900

Python list folders list by numeric order

I'm getting list of all the folders in specific directory with this code:

TWITTER_FOLDER = os.path.join('TwitterPhotos')
dirs = [d for d in os.listdir(TWITTER_FOLDER) if os.path.isdir(os.path.join(TWITTER_FOLDER, d))]

This is the array: ['1','2','3','4','5','6','7','8','9','10','11'].

And I want to get the array in this order:['11','10','9','8','7','6','5','4','3','2','1']

So I use this code for that:

dirs.sort(key=lambda f: int(filter(str.isdigit, f)))

and when I use it I get this error:

int() argument must be a string, a bytes-like object or a number, not 'filter'

Any idea what is the problem? Or how I can sort it in another way? It's important that the array will be sort by numeric order like:

12 11 10 9 8 7 6 5 4 3 2 1

And not:

9 8 7 6 5 4 3 2 12 11 1

Thanks!

Upvotes: 0

Views: 295

Answers (2)

Sayse
Sayse

Reputation: 43300

Filter returns an iterator, you need to join them back into a string before you can convert it to an integer

dirs.sort(key=lambda f: int(''.join(filter(str.isdigit, f))))

Upvotes: 2

Osman Mamun
Osman Mamun

Reputation: 2882

Use sorted with key:

In [4]: sorted(f, key=lambda x: int(x), reverse=True)
Out[4]: ['11', '10', '9', '8', '7', '6', '5', '4', '3', '2', '1']

Or you can do f.sort(key=lambda x:int(x), reverse=True) for inplace sort.

Upvotes: 2

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