Update item in dict comprehension python

Question

words = ['rocky','mahesh','surendra','mahesh','rocky','mahesh','deepak','mahesh','mahesh','mahesh','surendra']

words_count = {}
for word in words:
    words_count[word] = words_count.get(word, 0) + 1

print(words_count)
# Expected Output
# {'rocky': 2, 'mahesh': 6, 'surendra': 2, 'deepak': 1}

In this example, I just want to modify value of dict key while dict comprehension

Note: not looking other way to find occurrence/count of each key in dict.

kederrac · Answer 1 · 2020-04-17T09:56:58.853

3

You can use collections.Counter:

from collections import Counter

words_count = Counter(words)

edited Apr 17 '20 at 09:56

answered Apr 17 '20 at 09:23

kederrac

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score 1 · Answer 2 · answered Apr 17 '20 at 09:20

1

A short, simple, one-liner code:

{i:words.count(i) for i in words}

Here, we create a dictionary based on the count of the word.
Gives:

{'rocky': 2, 'mahesh': 6, 'surendra': 2, 'deepak': 1}

answered Apr 17 '20 at 09:20

Joshua Varghese

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score 1 · Accepted Answer · answered Apr 17 '20 at 09:52

1

You could count that without using any import and using as few .counts as possible following way:

words = ['rocky','mahesh','surendra','mahesh','rocky','mahesh','deepak','mahesh','mahesh','mahesh','surendra']
words_count = {i:words.count(i) for i in set(words)}
print(words_count)  # {'surendra': 2, 'mahesh': 6, 'rocky': 2, 'deepak': 1}

Converting list to set will result in unique values.

answered Apr 17 '20 at 09:52

Daweo

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This is slow!! It runs in quadratic time!! That's why I am downvoting – Riccardo Bucco Apr 17 '20 at 10:00
yes, i like it.... even this might execute same inside counter module. – Surendra Tarai Apr 19 '20 at 07:24

Update item in dict comprehension python

3 Answers3