Python Order Alphabetically without preceding zeros for numbers less than 10

Question

I would like to list a directory with Python. My directory only has files with name :

A1, A2, A3,..., A10, A11,..., B1,B2, ..., B10, B11 ...

Problem is that when I try with alphabetically order it with Python :

listQuery = os.listdir('C:\\query\\')
listQuery.sort()

I got the following order :

A1, A10, A11, ... ,A2 ... 

So my question is how can I first alphabetically order those and then order it with numbers

Answer 1

Make a sorting key function, like:

def my_order(value):
    return (value[0], int(value[1:]))

Then use it to sort your list:

listQuery.sort(key=my_order)

This calls the my_order function on every value in the list, then sorts the list based on those newly computed values. This is also known as a "decorate-sort-undecorate" ("DSU") or "Schwartzian transform".

In this case, it creates a list of tuples like ('A', 2) , ('A', 11) , etc. Python sorts tuples based on their individual values. If two tuples have the same first value (like 'A' ), it moves on to the next pair of values ( 2 and 11 here). Since both of those are integers, it would sort them numerically. It uses that ordering to sort the original list.