Is there a “one-liner” way to get a list of keys from a dictionary in sorted order?
-
11-07-2019 - |
Question
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?
Solution
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)
OTHER TIPS
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
Actually, .keys() is not necessary:
for k in sorted(somedictionary):
doSomething(k)
or
[doSomethinc(k) for k in sorted(somedict)]
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