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Home/ Questions/Q 8833847
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T08:52:50+00:00 2026-06-14T08:52:50+00:00

Let’s say I have a Counter object: Counter({‘test’: 2, ‘this’: 1, ‘is’: 1}) I

  • 0

Let’s say I have a Counter object: Counter({'test': 2, 'this': 1, 'is': 1})

I would like to iterate over this object in the following way:

c = Counter({'test': 2, 'this': 1, 'is': 1})

for i,s in my_counter_iterator(c):
    print i, ":", s

>> 1 : ['this', 'is']
>> 2 : ['test']

How do I do that efficiently (this code should run on a web server for each request…)?

EDIT

I have tried this, but I have the feeling there are more efficient ways. Are there?

from itertools import groupby

for k,g in groupby(sorted(c.keys(), key=lambda x: c[x]),key=lambda x: c[x]):
    print k, list(g)


1 ['this', 'is']
2 ['test']
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  1. Editorial Team
    Editorial Team
    2026-06-14T08:52:51+00:00Added an answer on June 14, 2026 at 8:52 am

    If you want to do this with large Counters, you really have no choice but to invert the mapping.

    inv_c = defaultdict(list)
    for k, v in c.iteritems():
        inv_c[v].append(k)
    

    Then inv_c.iteritems() is what you want.

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