Sort Python dictionary by it's keys using another Python list

Question

I have a list of words:

['apple', 'zoo', 'chicken', 'needle', 'car', 'computer']

I also have a dictionary with keys and values:

{'zoo': 42, 'needle': 32, 'computer': 18, 'apple': 39, 'car': 11, 'chicken': 12}

The keys of my dictionary are all from the list of words. How can I sort the dictionary so that the order of its keys is the same as the order of the words in the list? So once sorted, my dictionary should look like this:

{'apple': 39, 'zoo': 42, 'chicken': 12, 'needle': 32, 'car': 11, 'computer': 18}

Thanks so much!

Dictionaries don't preserve order. An available ordered version of a dictionary is collections.OrderedDict — Vinícius Figueiredo
– Vinícius Figueiredo, Commented Jul 16, 2017 at 18:07
First of all you should be aware that dictionaries are unordered in Python 3.5 and below. For ordered dictionaries you can use collections.OrderedDict. — jmd_dk
– jmd_dk, Commented Jul 16, 2017 at 18:08
@jmd_dk pardon me for hijacking the comment's section for this question, but are dictionaries really ordered in Python 3.6? Looks like I missed that part of the docs then. Nvm found the docs! :) — Vinícius Figueiredo
– Vinícius Figueiredo, Commented Jul 16, 2017 at 18:11
@ViníciusAguiar They are indeed. I recommend this nice talk for an overview of the recent (and historic) changes to dicts: youtube.com/watch?v=npw4s1QTmPg — jmd_dk
– jmd_dk, Commented Jul 16, 2017 at 18:18

cs95 · Accepted Answer · 2017-07-16 18:14:15Z

3

For python versions < 3.6, dictionaries do not maintain order, and sorting a dictionary is consequently not possible.

You may use the collections.OrderedDict to build a new dictionary with the order you want:

In [269]: from collections import OrderedDict

In [270]: keys = ['apple', 'zoo', 'chicken', 'needle', 'car', 'computer']
     ...: dict_1 = {'zoo': 42, 'needle': 32, 'computer': 18, 'apple': 39, 'car': 11, 'chicken': 12}
     ...: 

In [271]: dict_2 = OrderedDict()

In [272]: for k in keys:
     ...:     dict_2[k] = dict_1[k]
     ...:     

In [273]: dict_2
Out[273]: 
OrderedDict([('apple', 39),
             ('zoo', 42),
             ('chicken', 12),
             ('needle', 32),
             ('car', 11),
             ('computer', 18)])

In Python3.6, a simple dict comprehension suffices:

>>> {x : dict_1[x] for x in keys}
{'apple': 39, 'zoo': 42, 'chicken': 12, 'needle': 32, 'car': 11, 'computer': 18}

edited Jul 16, 2017 at 18:14

answered Jul 16, 2017 at 18:09

cs95

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2 Comments

user4889732 Over a year ago

Great, that works! (I'm using 3.6) What if my list has values that my dictionary doesn't? How do I bypass the KeyError without stopping the loop?

cs95 Over a year ago

@QuantumJuker You'll need to add an if condition: if k in dict_1: dict_2[k] = dict_1[k]

user2399453 · Accepted Answer · 2017-07-16 18:12:50Z

0

You can used OrderedDict since regular dictionaries are unordered. For your case you could do this:

from collections import OrderedDict
od = OrderedDict()

ll = ['apple', 'zoo', 'chicken', 'needle', 'car', 'computer']
d = {'zoo': 42, 'needle': 32, 'computer': 18, 'apple': 39, 'car': 11, 'chicken': 12} 

for f in ll:
  od[f] = d[f]
#Outputs: OrderedDict([('apple', 39), ('zoo', 42), ('chicken', 12), ('needle', 32), ('car', 11), ('computer', 18)])

answered Jul 16, 2017 at 18:12

user2399453

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Comments

Arman Ordookhani · Accepted Answer · 2017-07-16 18:15:16Z

-1

Python dict doesn't preserve order by default, you should use collections.OrderedDict. The first item you put into OrderedDict is the first item you will get when you enumerate it (e.g. using for).

from collections import OrderedDict

order_list = ['apple', 'zoo', 'chicken', 'needle', 'car', 'computer']
unordered_dict = {'zoo': 42, 'needle': 32, 'computer': 18, 'apple': 39, 'car': 11, 'chicken': 12}
ordered_dict = OrderedDict()
for item in order_list:
    ordered_dict[item] = unordered_dict[item]

for k, v in unordered_dict.items():
    print(k, v)

for k, v in ordered_dict.items():
    print(k, v)

answered Jul 16, 2017 at 18:15

Arman Ordookhani

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