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

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

Answer 2

For a while there, anything using itertools was by definition Pythonic. Still, this is a bit on the opaque side:

>>> from itertools import groupby
>>> grouplen = lambda grp : sum(1 for i in grp)
>>> hist = dict((a[0], grouplen(a[1])) for a in groupby(sorted("ABRACADABRA")))
>>> print hist
{'A': 5, 'R': 2, 'C': 1, 'B': 2, 'D': 1}

I'm currently running Python 2.5.4.

Answer 3

import pandas as pd

pd.Series(list(L)).value_counts()

Answer 4

I needed a histogram implementation to work in python 2.2 up to 2.7, and came up with this:

>>> L = 'abracadabra'
>>> hist = {}
>>> for x in L: hist[x] = hist.setdefault(x,0)+1
>>> print hist
{'a': 5, 'r': 2, 'b': 2, 'c': 1, 'd': 1}

I was inspired by Eli Courtwright's post of a defaultdict. These were introduced in python 2.5 so can't be used. But they can be emulated with the dict.setdefault(key,default).

This is basically the same thing gnibbler is doing, but I had to write this first before I could completely understand his lambda function.

Answer 5

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 6

One that works back to 2.3 (slightly shorter than Timmerman's, I think more readable) :

L = 'abracadabra'
hist = {}
for x in L: hist[x] = hist.pop(x,0) + 1
print hist
{'a': 5, 'r': 2, 'b': 2, 'c': 1, 'd': 1}