I am trying to sort list of strings containing numbers:
a = ["1099.0","9049.0"]
a.sort()
a
['1099.0', '9049.0']
b = ["949.0","1099.0"]
b.sort()
b
['1099.0', '949.0']
a
['1099.0', '9049.0']
But list b is sorting and not list a.
4 Answers
You want to sort based on the float values (not string values), so try:
>>> b = ["949.0","1099.0"]
>>> b.sort(key=float)
>>> b
['949.0', '1099.0']
Comments
Use a lambda inside 'sort' to convert them to float and then sort properly:
a = sorted(a, key=lambda x: float(x))
So you will maintain them as strings, but sorted by value and not lexicographically.
2 Comments
arshajii
@Keyser My point is you don't need a lambda, you can just use
key=float.
Samuele Mattiuzzo
to make my answer cool? actually, I forgot about the key functions. @Keyser using
sort(key=float) achieves the sameIn case anybody is dealing with numbers and extensions such as 0.png, 1.png, 10.png, 2.png... We need to retrieve and sort the characters before the extension since this extension does not let us to convert the names to floats:
myList = sorted(myList, key=lambda x: int(x[:-4]))
PD: After many years I have edited this answer changing float to int because I guess it is more efficient and more precise. It is a rare case to need floats for this purpose and it might bring errors.
Convert them to int or float or even decimal (since it has trailing numbers)
>>> b = [float(x) for x in b]
>>> b.sort()
>>> b
[949.0, 1099.0]
1 Comment
arshajii
This would mean changing the types of the elements in
b, which may or may not be ideal.
ais already sorted.1is smaller that9.0always and smaller number at index0