Split list of names into alphabetic dictionary, in Python.

Question

List.

['Chrome', 'Chromium', 'Google', 'Python']

Result.

{'C': ['Chrome', 'Chromium'], 'G': ['Google'], 'P': ['Python']}

I can make it work like this.

alphabet = dict()
for name in ['Chrome', 'Chromium', 'Google', 'Python']:
  character = name[:1].upper()
  if not character in alphabet:
    alphabet[character] = list()
  alphabet[character].append(name)

It is probably a bit faster to pre-populate the dictionary with A-Z, to save the key check on every name, and then afterwards delete keys with empty lists. I'm not sure either is the best solution though.

Is there a pythonic way to do this?

Answer 1

I don't know if it's Pythonic, but it's more succinct:

import itertools
def keyfunc(x):
   return x[:1].upper()
l = ['Chrome', 'Chromium', 'Google', 'Python']
l.sort(key=keyfunc)
dict((key, list(value)) for (key,value) in itertools.groupby(l, keyfunc))

EDIT 2 made it less succinct than previous version, more readable and more correct (groupby works as intended only on sorted lists)

Answer 2

Anything wrong with this? I agree with Antoine, the oneliner solution is rather cryptic.

import collections

alphabet = collections.defaultdict(list)
for word in words:
    alphabet[word[0].upper()].append(word)

Answer 3

I would do it like this-

import string

#init thedict
group_d = dict((letter, []) for letter in string.ascii_uppercase)

#populate the dict
[group_d[x[0].upper()].append(x) for x in ['Chrome','Chromium','Google','Python']]

This updates your dict.

UPDATE: Some of them seem bent-out-of shape since I used list-comprehension for "side-effect". So one can do it like this also -

import string

group_d = dict((letter, []) for letter in string.ascii_uppercase)
for x in ['Chrome','Chromium','Google','Python']:
    group_d[x[0].upper()].append(x)