Python update a key in dict if it doesn't exist
--
Music by Eric Matyas
https://www.soundimage.org
Track title: Hypnotic Puzzle3
--
Chapters
00:00 Question
00:30 Accepted answer (Score 272)
01:04 Answer 2 (Score 163)
01:20 Answer 3 (Score 52)
01:46 Answer 4 (Score 14)
02:32 Thank you
--
Full question
https://stackoverflow.com/questions/4231...
Answer 1 links:
[dict.setdefault()]: https://docs.python.org/3/library/stdtyp...
Answer 3 links:
[image]: https://i.stack.imgur.com/lxC54.png
--
Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
--
Tags
#python #dictionary
#avk47
ACCEPTED ANSWER
Score 315
You do not need to call d.keys(), so
if key not in d:
d[key] = value
is enough. There is no clearer, more readable method.
You could update again with dict.get(), which would return an existing value if the key is already present:
d[key] = d.get(key, value)
but I strongly recommend against this; this is code golfing, hindering maintenance and readability.
ANSWER 2
Score 197
Use dict.setdefault():
>>> d = {'key1': 'one'}
>>> d.setdefault('key1', 'some-unused-value')
'one'
>>> d # d has not changed because the key already existed
{'key1': 'one'}
>>> d.setdefault('key2', 'two')
'two'
>>> d
{'key1': 'one', 'key2': 'two'}
ANSWER 3
Score 65
Since Python 3.9 you can use the merge operator | to merge two dictionaries. The dict on the right takes precedence:
new_dict = old_dict | { key: val }
For example:
new_dict = { 'a': 1, 'b': 2 } | { 'b': 42 }
print(new_dict) # {'a': 1, 'b': 42}
Note: this creates a new dictionary with the updated values.
ANSWER 4
Score 15
With the following you can insert multiple values and also have default values but you're creating a new dictionary.
d = {**{ key: value }, **default_values}
I've tested it with the most voted answer and on average this is faster as it can be seen in the following example, .
Speed test comparing a for loop based method with a dict comprehension with unpack operator method.
if no copy (d = default_vals.copy()) is made on the first case then the most voted answer would be faster once we reach orders of magnitude of 10**5 and greater. Memory footprint of both methods are the same.