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Home/ Questions/Q 815305
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Editorial Team
  • 0
Editorial Team
Asked: May 15, 20262026-05-15T01:39:47+00:00 2026-05-15T01:39:47+00:00

There are many ways to write a Python program that computes a histogram. By

  • 0

There are many ways to write a Python program that computes a histogram.

By histogram, I mean a function that counts the occurrence of objects in an iterable and outputs the counts in a dictionary. For example:

>>> L = 'abracadabra'
>>> histogram(L)
{'a': 5, 'b': 2, 'c': 1, 'd': 1, 'r': 2}

One way to write this function is:

def histogram(L):
    d = {}
    for x in L:
        if x in d:
            d[x] += 1
        else:
            d[x] = 1
    return d

Are there more concise ways of writing this function?

If we had dictionary comprehensions in Python, we could write:

>>> { x: L.count(x) for x in set(L) }

but since Python 2.6 doesn’t have them, we have to write:

>>> dict([(x, L.count(x)) for x in set(L)])

Although this approach may be readable, it is not efficient: L is walked-through multiple times. Furthermore, this won’t work for single-life generators; the function should work equally well for iterator generators such as:

def gen(L):
    for x in L:
        yield x

We might try to use the reduce function (R.I.P.):

>>> reduce(lambda d,x: dict(d, x=d.get(x,0)+1), L, {}) # wrong!

Oops, this does not work: the key name is 'x', not x. 🙁

I ended with:

>>> reduce(lambda d,x: dict(d.items() + [(x, d.get(x, 0)+1)]), L, {})

(In Python 3, we would have to write list(d.items()) instead of d.items(), but it’s hypothethical, since there is no reduce there.)

Please beat me with a better, more readable one-liner! 😉

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  1. Editorial Team
    Editorial Team
    2026-05-15T01:39:48+00:00Added an answer on May 15, 2026 at 1:39 am

    Python 3.x does have reduce, you just have to do a from functools import reduce. It also has “dict comprehensions”, which have exactly the syntax in your example.

    Python 2.7 and 3.x also have a Counter class which does exactly what you want:

    from collections import Counter
    cnt = Counter("abracadabra")
    

    In Python 2.6 or earlier, I’d personally use a defaultdict and do it in 2 lines:

    d = defaultdict(int)
    for x in xs: d[x] += 1
    

    That’s clean, efficient, Pythonic, and much easier for most people to understand than anything involving reduce.

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