Get list items that don't match most frequent value

Asked Nov 16 '21 at 17:54

Active Nov 16 '21 at 18:10

Viewed 48 times

I have a list of dictionaries looking like this:

[{'customer': 'Charles', 'city': 'Paris'}, {'customer': 'John', 'city': 'New York'}, {'customer': 'Jean', 'city': 'Paris'}]

I tried something using collections which will return me the name of the most common city in this list of dictionaries:

city_counts = Counter(c['city'] for c in customers)
return city_counts .most_common(1)[0][0]

From this, I would like to return a list of all customers who are not in this city. So, if I take the list I gave above, ['John'] should be the output.

I there a best way to do it ?

edited Nov 16 '21 at 18:10

Barmar

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asked Nov 16 '21 at 17:54

Synops

Can a customer be in multiple cities? – Homer512 Nov 16 '21 at 17:58
Don't limit the output of `most_common`, take the _last_ value? – jonrsharpe Nov 16 '21 at 17:59
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`[c['customer'] for c in customers if c['city'] != most_common_city]` – Barmar Nov 16 '21 at 17:59
Your title isn't the same as your question. Do you want the least frequent city, or do you want all the customers who aren't in the most frequent city? – Barmar Nov 16 '21 at 18:00
@barmar, I want all the customers who aren't in the most frequent city – Synops Nov 16 '21 at 18:03
Please fix the title so you don't get wrong answers. – Barmar Nov 16 '21 at 18:04
@barmar, I can't find any good one, if you have any, you are welcome to share – Synops Nov 16 '21 at 18:09
1

There you go, is that a good description? – Barmar Nov 16 '21 at 18:11

Get list items that don't match most frequent value

0 Answers0