remove entry in counter object if value meets condition

Question

Is there a way to remove entries from a counter object if the value matches a certain condition. For example:

Counter({'a': 1142,'b':1004,'c':100,'d':5})

I want to drop all indexes where it is less than 1000, so I just have 'a' and 'b' left. I know I can loop through each and then delete if it doesnt match the condition as shown in this solution. Just looking for a more efficient way.

Are you asking about an arbitrary condition, or specifically about a comparison with a threshold value? — mkrieger1, Nov 13 '22 at 21:32
@mkrieger1 I currently need a comparison for specifically a threshold value, but would be interested in seeing both if they defer. — Nikita Belooussov, Nov 13 '22 at 21:37

score 1 · Accepted Answer · answered Nov 13 '22 at 21:23

1

I think it can be useful for you:

from collections import Counter
counter = Counter({'a': 1142, 'b': 1004, 'c': 100, 'd':5})
Counter({k: c for k, c in counter.items() if c >= 1000})

Output:

Counter({'a':1142 , 'b': 1004})

This way is more effective as you mentioned.

answered Nov 13 '22 at 21:23

Amirhossein Sefati

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5
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How do you know that it is more effective? – mkrieger1 Nov 13 '22 at 21:24
Because `"For loop" is around 50% slower than a list comprehension` according to https://switowski.com/blog/for-loop-vs-list-comprehension/ – Amirhossein Sefati Nov 13 '22 at 21:27

score 0 · Answer 2 · answered Nov 13 '22 at 21:21

0

You can use a simple loop to delete the keys in place:

from collections import Counter

c = Counter({'a': 1142,'b':1004,'c':100,'d':5})

for k in list(c):
    if c[k] < 1000:
        del c[k]

print(c)

Output:

Counter({'a': 1142, 'b': 1004})

If you want a copy:

c2 = Counter({k:v for k,v in c.items() if v>= 1000})

answered Nov 13 '22 at 21:21

mozway

194,879
13
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75

I know you can use a loop, I was wondering if there was a better solution to it. – Nikita Belooussov Nov 13 '22 at 21:25
How do you define "*better*" what limitations have you encountered? – mozway Nov 13 '22 at 21:27
mainly one that has a faster run time. – Nikita Belooussov Nov 13 '22 at 21:38
If you're fine with a copy, the dictionary comprehension (see my second example) will be faster. If you want the original object, you have to use a classical loop – mozway Nov 13 '22 at 21:40

remove entry in counter object if value meets condition

2 Answers2