Create or append to a list in a dictionary - can this be shortened?

Question

Can this Python code be shortened and still be readable using itertools and sets?

result = {}
for widget_type, app in widgets:
    if widget_type not in result:
        result[widget_type] = []
    result[widget_type].append(app)

I can think of this only:

widget_types = zip(*widgets)[0]
dict([k, [v for w, v in widgets if w == k]) for k in set(widget_types)])

score 130 · Answer 1 · answered Nov 10 '10 at 11:12

130

An alternative to defaultdict is to use the setdefault method of standard dictionaries:

 result = {}
 for widget_type, app in widgets:
     result.setdefault(widget_type, []).append(app)

This relies on the fact that lists are mutable, so what is returned from setdefault is the same list as the one in the dictionary, therefore you can append to it.

answered Nov 10 '10 at 11:12

Daniel Roseman

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And that's my Python nugget to learn for the day. Thanks, Daniel. :) – Walter Nov 10 '10 at 12:39
1

My personal opinion: I am opting for `defaultdict` because `setdefault` is, at least in my experience, the method with the most unintelligible name I have ever seen across any language. Also [python docs](https://docs.python.org/3/library/collections.html#defaultdict-examples) suggest to use `defaultdict` _This technique is simpler and faster than an equivalent technique using `dict.setdefault()`_ – Antonio Aug 17 '22 at 12:06

score 94 · Accepted Answer · answered Nov 10 '10 at 10:57

94

You can use a defaultdict(list).

from collections import defaultdict

result = defaultdict(list)
for widget_type, app in widgets:
    result[widget_type].append(app)

answered Nov 10 '10 at 10:57

Mark Byers

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+1 As stated in https://docs.python.org/3.7/library/collections.html#defaultdict-objects : *This technique is simpler and faster than an equivalent technique using `dict.setdefault()`* – Marco Nov 10 '19 at 11:57
1

Related: [How to convert a `default_dict` to `dict`](https://stackoverflow.com/a/20428703/2436175). – Antonio Aug 17 '22 at 12:20

score 6 · Answer 3 · answered Mar 25 '13 at 10:55

6

may be a bit slow but works

result = {}
for widget_type, app in widgets:
    result[widget_type] = result.get(widget_type, []) + [app]

answered Mar 25 '13 at 10:55

user1139002

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why would this be slower than the other solutions? – David Schumann Jul 18 '17 at 14:35
3

It would be slower because the concat operator (+) creates a new list every time, while append() adds the element to the existing list. – Diego Aragão Sep 14 '18 at 17:45

Create or append to a list in a dictionary - can this be shortened?

3 Answers3

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