How to sort numeric strings with 2 decimal points in python
Question:
I have some directories in linux having version as directory name :
1.1.0 1.10.0 1.5.0 1.7.0 1.8.0 1.8.1 1.9.1 1.9.2
I want to sort the above directories from lowest to highest version
when i try to use .sort in python i end up getting below
['1.1.0', '1.10.0', '1.5.0', '1.7.0', '1.8.0', '1.8.1', '1.9.1']
which is actually incorrect , the 1.10.0 version is the gretest among all which should lie in the last index , is there a way to handle these things using python..
Thanks in advance
Answers:
Since your versions are of string datatype. We would have to split after each dot.
v_list = ['1.1.0','1.10.0','1.5.0','1.7.0','1.8.0','1.8.1','1.9.1','1.9.2']
v_list.sort(key=lambda x: list(map(int, x.split('.'))))
or you can also try this:
v_list = ['1.1.0','1.10.0','1.5.0','1.7.0','1.8.0','1.8.1','1.9.1','1.9.2']
v_list.sort(key=lambda x: [int(y) for y in x.split('.')])
I have some directories in linux having version as directory name :
1.1.0 1.10.0 1.5.0 1.7.0 1.8.0 1.8.1 1.9.1 1.9.2
I want to sort the above directories from lowest to highest version
when i try to use .sort in python i end up getting below
['1.1.0', '1.10.0', '1.5.0', '1.7.0', '1.8.0', '1.8.1', '1.9.1']
which is actually incorrect , the 1.10.0 version is the gretest among all which should lie in the last index , is there a way to handle these things using python..
Thanks in advance
Since your versions are of string datatype. We would have to split after each dot.
v_list = ['1.1.0','1.10.0','1.5.0','1.7.0','1.8.0','1.8.1','1.9.1','1.9.2']
v_list.sort(key=lambda x: list(map(int, x.split('.'))))
or you can also try this:
v_list = ['1.1.0','1.10.0','1.5.0','1.7.0','1.8.0','1.8.1','1.9.1','1.9.2']
v_list.sort(key=lambda x: [int(y) for y in x.split('.')])