Is it possible to access specific item of Counter without using for loop?

Question

Is it possible to access specific item of Counter?

from collections import Counter
s=['a','b','r','e', 'a', 'r','r']
print(Counter(s))

Output

Counter({'r': 3, 'a': 2, 'b': 1, 'e': 1})

I want to access ('a':2) or for example ('a': 2, 'b': 1) items, how to do it without using for loop?

I couldn't access ('a':2) using print(Counter(s).items([1])).

My answer:

s=sorted(['a','e','r','b', 'a', 'r','r'])

x=Counter(s)
y={k: v for k, v in sorted(x.items(), key=lambda item: item[1], reverse=True)}

i=0
for key,value in y.items():
    if i<3:
        print(str(key) + " " + str(value))
        i+=1

I sorted the list in the beginning so that if the values of the items are equal, then the alphabetic order will be preserved.

The Counter returns in dictionary format and an element can be accessed by it key... — Deven Ramani, Aug 03 '22 at 12:11
Does this answer your question? [How do I sort a dictionary by value?](https://stackoverflow.com/questions/613183/how-do-i-sort-a-dictionary-by-value) — OrenIshShalom, Aug 03 '22 at 12:22
step 1: get the dict: `counter.items()`. step 2: use the link to sort dict by values ... — OrenIshShalom, Aug 03 '22 at 12:24
`Counter` is a subclass of `dict`; `Counter(s)` *is* a `dict`, just with extra methods. — chepner, Aug 03 '22 at 12:30

score 1 · Answer 1 · answered Aug 03 '22 at 12:35

1

Counter has a most_common method that returns a list of key-value pairs in decreasing order of their count.

>>> s = ['a','b','r','e', 'a', 'r','r']
>>> Counter(s).most_common(3)
[('r', 3), ('a', 2), ('b', 1)]

answered Aug 03 '22 at 12:35

chepner

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if you change 'b' and 'e' inside list s, then your code prints [('r', 3), ('a', 2), ('e', 1)], but since both b and e have same value, I want to print b, since it is first in alphabetic order – Lee Aug 03 '22 at 12:45
Would sorting the input first be sufficient? `Counter(sorted(s)).most_common(3)`. `Counter` is breaking ties based on which key was added first, not necessarily which values are lexicographically smaller. – chepner Aug 03 '22 at 13:19
I did exactly the same thing and yes it works, I have added my new answer to the question. thanks! – Lee Aug 03 '22 at 13:31

score 0 · Answer 2 · answered Aug 03 '22 at 12:09

0

Try this?

>>> c = Counter(s)
>>> find = 'a'
>>> (find, c[find])
('a', 2)

answered Aug 03 '22 at 12:09

The Thonnu

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this is more of a comment rather than an answer ... – OrenIshShalom Aug 03 '22 at 12:12
problem is that I don't know exactly what is the second item. For example some problems want me to print first 3 items, 1 item per line. OR print second and third item – Lee Aug 03 '22 at 12:13
1

maybe this will help you: `for letter, num_occurrences in counter.items(): ...` – OrenIshShalom Aug 03 '22 at 12:15

score 0 · Answer 3 · answered Aug 03 '22 at 12:28

0

This should do the trick:

import json
from collections import Counter
counter = Counter(['a','b','r','e', 'a', 'r','r'])
sorted_data = {
    letter: occurrences
    for letter, occurrences in sorted(
        counter.items(),
        key=lambda item: item[1],
        reverse=True
    )
}
print(json.dumps(sorted_data, indent=4))

answered Aug 03 '22 at 12:28

OrenIshShalom

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Or, just sort the input to `Counter`, then dump `dict(counter.most_common())`. – chepner Aug 03 '22 at 12:33
if you change 'b' and 'e' inside list s, then your code prints [('r', 3), ('a', 2), ('e', 1), ('b',1)], but since both b and e have same value, I want to print b first, since it is first in alphabetic order – Lee Aug 03 '22 at 12:47

Is it possible to access specific item of Counter without using for loop?

3 Answers3