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

Answer 1

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 2

Your one-liner using reduce was almost ok, you only needed to tweak it a little bit:

>>> reduce(lambda d, x: dict(d, **{x: d.get(x, 0) + 1}), L, {})
{'a': 5, 'b': 2, 'c': 1, 'd': 1, 'r': 2}

Of course, this won't beat in-place solutions (nor in speed, nor in pythonicity), but in exchange you've got yourself a nice purely functional snippet. BTW, this would be somewhat prettier if Python had a method dict.merge().

Answer 3

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}

Answer 4

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 5

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 6

For python 2.7, you can use this small list comprehension:

v = list('abracadabra')
print {x: v.count(x) for x in set(v)}