Sorting Python Dictionary based on Key?

Question

I have created a python dictionary which has keys in this form :

11, 10, 00, 01, 20, 21, 31, 30

The keys are string

I would like to maintain my dictionary in these sorted order:

00, 10, 20, 30, 01, 11, 21, 31

This is based on the second value of the key.

I tried this sorted(dict.items(), key = lambda s: s[1]) and got the keys like:

20, 30, 21, 31, 01, 11, 10, 00

Can somebody guide me?

Have you considered using an [OrderedDict](http://docs.python.org/2/library/collections.html#collections.OrderedDict)? — kojiro, Sep 22 '13 at 21:56
This question is most definitely a duplicate. :) I can't even count how many times I've seen it. — Shashank, Sep 22 '13 at 21:56
Hi. `items` returns list of tuples where first item is key and second is value. So if You would like to sort them by keys You should use sorted(dict.items(), key = lambda s: s[0]) but in this case You got keys ascending sorted If You would like to sort as in your example 11, 10, 00, 01, 20, 21, 31, 30 > I would like to maintain my dictionary in these sorted order: 00, 10, 20, 30, 01, 11, 21, 31 You should reverse your key in sorted function sorted(dict.items(), key = lambda s: ''.join(reversed(s[0]))) — oleg, Sep 22 '13 at 22:42

score 4 · Answer 1 · edited Sep 22 '13 at 22:12

4

You almost had it, but the key is the first item of the tuple:

sorted(dict.items(), key=lambda s: s[0])

edited Sep 22 '13 at 22:12

ThiefMaster

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answered Sep 22 '13 at 21:56

Ry-

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I'm reading that the "second value of the key" is probably: `s[0][1]` – Jon Clements Sep 22 '13 at 21:57
@JonClements: Ah, good point. Thanks. – Ry- Sep 22 '13 at 21:58
Also wouldn't hurt to use `key=lambda (key, val): key[1]` just to make it explicit – Jon Clements Sep 22 '13 at 21:59
@JonClements: I don’t think that’s valid lambda syntax. – Ry- Sep 22 '13 at 21:59
@minitech It's valid syntax. It's just the same as `def myfunc(key, val): return key[1]` – TerryA Sep 22 '13 at 22:01
@JonClements: Ah, not under Python 3. – Ry- Sep 22 '13 at 22:03
@Haidro: No, it’s actually tuple unpacking. – Ry- Sep 22 '13 at 22:03
@minitech Gah. Ignore me then. *flees* – TerryA Sep 22 '13 at 22:05
@Jon It's allowed but it assigns to key the tuple: (lambda function, val). It's not really valid lambda syntax unless you want key to be a tuple of a lambda and a val which is obviously not the case for a sort function keyword argument. – Shashank Sep 22 '13 at 22:06
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At this point isn't it about time someone suggested using an `operator.itemgetter` just to get out of the lambda mess? :P – kojiro Sep 22 '13 at 22:07
but it changes to list of tuple instead of keeping it dictionary object! – Anu Jul 15 '19 at 22:38
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@anu: dicts weren’t ordered at the time of writing. In Python 3.7, you can wrap this expression in `dict()` to get a dict again. – Ry- Jul 15 '19 at 22:40

score 0 · Answer 2 · answered Sep 22 '13 at 21:57

0

A dictionary uses something called hashing and it can't be sorted as such. What you can do is use dict.keys().sort(key = lambda s: s[1]) and iterate or go on from there.

answered Sep 22 '13 at 21:57

adarsh

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This won't do what you expect. The `.sort` method to a list operates on the list in-place, and that list is ephemeral. The next time you get dict.keys() it won't be sorted. It won't be the *same* list. – kojiro Sep 22 '13 at 22:05
yes you are right! `x = dict.keys()` and then `x.sort(...)` Thanks! – adarsh Sep 22 '13 at 22:11

Sorting Python Dictionary based on Key?

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