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

问题

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?

回答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)

回答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

回答3:

Actually, .keys() is not necessary:

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

or

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

来源：https://stackoverflow.com/questions/327191/is-there-a-one-liner-way-to-get-a-list-of-keys-from-a-dictionary-in-sorted-ord