Is there a "one-liner" way to get a list of keys from a dictionary in sorted order?

The list sort() method is a modifier function that returns None.

So if I want to iterate through all of the keys in a dictionary I cannot do:

for k in somedictionary.keys().sort():
    dosomething()

Instead, I must:

keys = somedictionary.keys()
keys.sort()
for k in keys:
    dosomething()

Is there a pretty way to iterate through these keys in sorted order without having to break it up in to multiple steps?

Asked by: Roman901 | Posted: 28-01-2022

Answer 1

for k in sorted(somedictionary.keys()):
    doSomething(k)

Note that you can also get all of the keys and values sorted by keys like this:

for k, v in sorted(somedictionary.iteritems()):
   doSomething(k, v)

Answered by: John391 | Posted: 01-03-2022

Answer 2

Can I answer my own question?

I have just discovered the handy function "sorted" which does exactly what I was looking for.

for k in sorted(somedictionary.keys()):
    dosomething()

It shows up in Python 2.5 dictionary 2 key sort

Answered by: Brad660 | Posted: 01-03-2022

Answer 3

Actually, .keys() is not necessary:

for k in sorted(somedictionary):
    doSomething(k)

[doSomethinc(k) for k in sorted(somedict)]

Answered by: Gianna398 | Posted: 01-03-2022

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