Sort list of string based on number in string [duplicate]

Answer 1

list.sort accepts optional key function. Each item is passed to the function, and the return value of the function is used to compare items instead of the original values.

>>> my_list= ['image101.jpg', 'image2.jpg', 'image1.jpg']
>>> my_list.sort(key=lambda x: int(''.join(filter(str.isdigit, x))))
>>> my_list
['image1.jpg', 'image2.jpg', 'image101.jpg']

---

filter, str.isdigit were used to extract numbers:

>>> ''.join(filter(str.isdigit, 'image101.jpg'))
'101'
>>> int(''.join(filter(str.isdigit, 'image101.jpg')))
101

• ''.join(..) is not required in Python 2.x

Answer 2

Actually you don't need any regex patern. You can parse easily like that.

>>> 'image101.jpg'[5:-4]
'101'

Solution:

>>> sorted(my_list, key=lambda x: int(x[5:-4]))
['image1.jpg', 'image2.jpg', 'image101.jpg']

Answer 3

If you want to do this in the general case, I would try a natural sorting package like natsort.

from natsort import natsorted
my_list = ['image101.jpg', 'image2.jpg', 'image1.jpg']
natsorted(my_list)

Returns:

['image1.jpg', 'image2.jpg', 'image101.jpg']

You can install it using pip i.e. pip install natsort

Answer 4

Use a regex to pull the number from the string and cast to int:

import  re
r = re.compile("\d+")
l = my_list= ['image101.jpg', 'image2.jpg', 'image1.jpg']
l.sort(key=lambda x: int(r.search(x).group()))

Or maybe use a more specific regex including the .:

import  re

r = re.compile("(\d+)\.")
l = my_list= ['image101.jpg', 'image2.jpg', 'image1.jpg']
l.sort(key=lambda x: int(r.search(x).group()))

Both give the same output for you example input:

['image1.jpg', 'image2.jpg', 'image101.jpg']

If you are sure of the extension you can use a very specific regex:

 r = re.compile("(\d+)\.jpg$")
 l.sort(key=lambda x: int(r.search(x).group(1)))