Python Dictionary remove pairs when key not matching reference list

Question

This is what I have:

names = ['Bob', 'Mike', 'Nia', 'Tim', 'Holly', 'Liam', 'Dave']
temperatures = ['-0.3', '-0.6', '-0.8', '-0.2', '-1.3', '-2.1', '-0.4']

main_dictionary = {'Bob' : '-0.3', 'Mike': '-0.6', 'Nia' : '-0.8', 'Tim':'-0.2', 'Holly':'-1.3', 'Liam':'-2.1', 'Dave':'-0.4'}

I'm trying to produce a dictionary by comparing it to the list of selected_names against the main_dictionary to produce the updated_dictionary

selected_names = ['Nia', 'Holly', 'Liam', 'Dave']

updated_dictionary = {'Nia' : '-0.8',  'Holly':'-1.3', 'Liam':'-2.1', 'Dave':'-0.4'}

So far in my project Ive compared many lists using .intersection but can seem to work out how to apply it to a dictionary.

ypnos · Answer 1 · 2021-12-02T14:26:52.703

1

Probably a simple dict comprehension is enough:

updated_dictionary = {x: main_dictionary[x] for x in selected}

If selected may contain invalid keys, you could explicitely ignore them:

updated_dictionary = {x: main_dictionary[x] for x in selected if x in main_dictionary[x]}

See also Create a dictionary with list comprehension and Filter dict to contain only certain keys?

edited Dec 02 '21 at 14:26

answered Dec 02 '21 at 14:20

ypnos

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Not sure if this can happen with OP's data, but you'll get an error if the key is missing from `main_dictionary` – mozway Dec 02 '21 at 14:21
2

`{x: main_dictionary[x] for x in selected if x in main_dictionary}` – mama Dec 02 '21 at 14:25
Very good point, and I had already put that in, but still thanks mama! – ypnos Dec 02 '21 at 14:27
thank you this is awesome you all make it look too easy – Barry Marples Dec 02 '21 at 14:28

score 0 · Answer 2 · answered Dec 02 '21 at 14:20

0

Try this

main_dictionary = {'Bob' : '-0.3', 'Mike': '-0.6', 'Nia' : '-0.8', 'Tim':'-0.2', 'Holly':'-1.3', 'Liam':'-2.1', 'Dave':'-0.4'}
selected_names = ['Nia', 'Holly', 'Liam', 'Dave']

newd = {k:v for k,v in main_dictionary.items() if k in selected_names}

print(newd)

answered Dec 02 '21 at 14:20

mama

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Your solution is quite inefficient, as for each key in `main_dictionary` you have to go through the whole list of `selected_names`. – ypnos Dec 02 '21 at 14:21
python is an inefficient language – mama Dec 02 '21 at 14:22
@ypnos if a key is missing this approach won't throw an error.. so it has at least one advantage ;) – Hansanho Dec 02 '21 at 14:24
@ypnos it works great but by inefficient I guess you mean slow? If my project doesn't need to be quick does this matter? – Barry Marples Dec 02 '21 at 14:25
1

Yes if your `selected_names` is big or `main_dictionary` is big, then it becomes an issue. The question is, why start writing inefficient code in the first place? – ypnos Dec 02 '21 at 14:26

score -1 · Answer 3 · answered Dec 02 '21 at 14:34

Try this:

names = ['Bob', 'Mike', 'Nia', 'Tim', 'Holly', 'Liam', 'Dave']
temperatures = ['-0.3', '-0.6', '-0.8', '-0.2', '-1.3', '-2.1', '-0.4']

main_dictionary = {
    'Bob': '-0.3',
    'Mike': '-0.6',
    'Nia': '-0.8',
    'Tim': '-0.2',
    'Holly': '-1.3',
    'Liam': '-2.1',
    'Dave': '-0.4'
}

selected_names = ['Nia', 'Holly', 'Liam', 'Dave']
output = {
    'Nia': '-0.8',
    'Holly': '-1.3',
    'Liam': '-2.1',
    'Dave': '-0.4'
}
updated_dictionary = {k: v for k, v in main_dictionary.items() if k in selected_names}

if __name__ == '__main__':
    print("main", main_dictionary)
    print("updated", updated_dictionary)
    print(updated_dictionary == output)

# It produces:
# {'Nia': '-0.8', 'Holly': '-1.3', 'Liam': '-2.1', 'Dave': '-0.4'}
# updated {'Nia': '-0.8', 'Holly': '-1.3', 'Liam': '-2.1', 'Dave': '-0.4'}
# True

Python Dictionary remove pairs when key not matching reference list

3 Answers3