Pythonic way to increment value when building dictionary

Question

A snippet of code from a working script; I'm just curious if there's a "prettier" way to achieve the same result.

    if ctry in countries:
        countries[ ctry ] += 1
    else:
        countries[ ctry ] = 1

In awk I could have just used countries[ ctry ] += 1, but python threw a key error (understandably).

[Check if a given key already exists in a dictionary and increment it](https://stackoverflow.com/q/473099/2301450) — vaultah, Aug 19 '17 at 21:56
You might want [collections.Counter](https://docs.python.org/3/library/collections.html#collections.Counter) for that. — Izaak van Dongen, Aug 19 '17 at 22:33

score 6 · Accepted Answer · answered Aug 19 '17 at 22:01

Rather than using a normal dictionary, you could change countries to be a collections.defaultdict object instead. As the name implies, collections.defaultdict allows you to have a default value inserted in your dictionary if a key doesn't exist yet:

from collections import defaultdict
countries = defaultdict(int)

Then your snippit becomes one line:

countries[cntry] += 1

If you can't use collections.defaultdict, you can use the "ask forgiveness rather than permission" idiom instead:

try:
    countries[ ctry ] += 1
except KeyError:
    countries[ ctry ] = 1

Although the above will behave like your conditional statement, it is consider more "Pythonic" because try/except is used rather than if/else.

score 3 · Answer 2 · answered Aug 19 '17 at 21:55

3

The following is a bit more pythonic:

countries[ctry]  = 1 if ctry not in countries else countries[ctry] + 1

Or

countries[ctry] = countries.get(ctry, 0) + 1

answered Aug 19 '17 at 21:55

Dekel

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score 2 · Answer 3 · answered Aug 19 '17 at 22:02

Another option is to use defaultdict:

from collections import defaultdict
countries = defaultdict(int)
countries[ctry] += 1

Speed test:

%timeit countries['Westeros'] += 1
10000000 loops, best of 3: 79 ns per loop

countries = {}
%timeit countries['Westeros'] = countries.get('Westeros', 0) + 1
1000000 loops, best of 3: 164 ns per loop

Pythonic way to increment value when building dictionary

3 Answers3