python histogram one-liner

Question

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! ;)

Answer 1

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.

Answer 2

It's kinda cheaty to import modules for oneliners, so here's a oneliner that is O(n) and works at least as far back as Python2.4

>>> f=lambda s,d={}:([d.__setitem__(i,d.get(i,0)+1) for i in s],d)[-1]
>>> f("ABRACADABRA")
{'A': 5, 'R': 2, 'B': 2, 'C': 1, 'D': 1}

And if you think __ methods are hacky, you can always do this

>>> f=lambda s,d=lambda:0:vars(([setattr(d,i,getattr(d,i,0)+1) for i in s],d)[-1])
>>> f("ABRACADABRA")
{'A': 5, 'R': 2, 'B': 2, 'C': 1, 'D': 1}

:)

Answer 3

$d{$_} += 1 for split //, 'abracadabra';