The Python Oracle

How to create key or append an element to key?

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Chapters
00:00 How To Create Key Or Append An Element To Key?
00:59 Accepted Answer Score 394
01:15 Answer 2 Score 102
01:54 Answer 3 Score 39
02:16 Answer 4 Score 16
02:27 Answer 5 Score 13
02:31 Thank you

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Full question
https://stackoverflow.com/questions/1290...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #dictionary

#avk47



ACCEPTED ANSWER

Score 394


Use dict.setdefault():

dict.setdefault(key,[]).append(value)

help(dict.setdefault):

    setdefault(...)
        D.setdefault(k[,d]) -> D.get(k,d), also set D[k]=d if k not in D



ANSWER 2

Score 102


Here are the various ways to do this so you can compare how it looks and choose what you like. I've ordered them in a way that I think is most "pythonic", and commented the pros and cons that might not be obvious at first glance:

Using collections.defaultdict:

import collections
dict_x = collections.defaultdict(list)

...

dict_x[key].append(value)

Pros: Probably best performance. Cons: Not available in Python 2.4.x.

Using dict().setdefault():

dict_x = {}

...

dict_x.setdefault(key, []).append(value)

Cons: Inefficient creation of unused list()s.

Using try ... except:

dict_x = {}

...

try:
    values = dict_x[key]
except KeyError:
    values = dict_x[key] = []
values.append(value)

Or:

try:
    dict_x[key].append(value)
except KeyError:
    dict_x[key] = [value]



ANSWER 3

Score 39


You can use a defaultdict for this.

from collections import defaultdict
d = defaultdict(list)
d['key'].append('mykey')

This is slightly more efficient than setdefault since you don't end up creating new lists that you don't end up using. Every call to setdefault is going to create a new list, even if the item already exists in the dictionary.




ANSWER 4

Score 16


You can use defaultdict in collections.

An example from doc:

s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
d = defaultdict(list)
for k, v in s:
    d[k].append(v)