Sort list of strings in natural fashion

user_607846 0
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I have the following:

sorted( ("A","a","b","B","CC","c"), key=lambda x: x.lower() )

Which gives:

['A', 'a', 'b', 'B', 'c', 'CC']

How do I sort it like this:

['a', 'A', 'b', 'B', 'c', 'CC']

where if there are two values the same, the lower form comes first - a before A.

I'm guessing I do something like this:

sorted( ("A","a","b","B","CC","c"), key=lambda x: (x.lower(),x) )

but this gives A before a:

['A', 'a', 'B', 'b', 'c', 'CC']

How do I do the opposite?

UPDATE

To clarify things further, the following strings: ["A","a","aA","aa"] should be sorted as:

["a","A","aa","aA"]

So "aa" before "aA" and "aaA" before "aAa" etc.

So:

lst = ["A","aA","aa","aaa","aAa","aaA","b","B","CC","c"]

should sort to:

['A', 'aa', 'aA', 'aaa', 'aaA', 'aAa', 'b', 'B', 'c', 'CC']
python-3.x python
4 Answers

You could do:

lst = ["A","a","b","B","CC","c"]

result = sorted(lst, key= lambda x: (x.lower(), not x.islower()))
print(result)

Output

['a', 'A', 'b', 'B', 'c', 'CC']

UPDATE

Given the new you examples you could use the following key:

lst = ["A", "aA", "aa", "aaa", "aAa", "aaA", "b", "B", "CC", "c"]
result = sorted(lst, key=lambda x: (x.lower(), ''.join(str(c.isupper()) for c in x)))
print(result)

Output

['A', 'aa', 'aA', 'aaa', 'aaA', 'aAa', 'b', 'B', 'c', 'CC']

For the other example (["A","a","aA","aa"]) it gives:

['a', 'A', 'aa', 'aA']
user_40015920

Here is one way:

sorted( lst, key=lambda x: (x.lower(), *map(str.isupper, x)) )
#['A', 'aa', 'aA', 'aaa', 'aaA', 'aAa', 'b', 'B', 'c', 'CC']

First sort by case insensitive letter. Then call str.upper on all the characters of each string as the second sort condition. This will rank lower case letters first for the same length strings.

Python 2 Version

The above syntax only works in python3, but you can equivalently do the following in python2:

sorted( lst, key=lambda x: ((x.lower(),) + tuple(map(str.isupper, x))) )
user_58588510

You can do it this way:

>>> sorted( ("A","a","b","B","CC","c"), key=lambda x: (x.lower() + str(x != x.lower()) ,x) )
['a', 'A', 'b', 'B', 'c', 'CC']

It works by comparing the lowercase version of the letter to the actual letter. If there is a match on lower case then str(x != x.lower() will be 0, otherwise 1.

user_20843840

Here is a solution that works based on previous answers:

sorted(lst, key= lambda x: (x.lower(), [c.isupper() for c in x]))

and this also seems to work:

sorted(lst, key= lambda x: (x.lower(), x.swapcase()))
user_6078460
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