Converting a list of tuples into a dict

Dan_18909 0
Viewed 23773

I have a list of tuples like this:

[
('a', 1),
('a', 2),
('a', 3),
('b', 1),
('b', 2),
('c', 1),
]

I want to iterate through this keying by the first item, so, for example, I could print something like this:

a 1 2 3
b 1 2
c 1

How would I go about doing this without keeping an item to track whether the first item is the same as I loop around the tuples? This feels rather messy (plus I have to sort the list to start with)...

tuples iteration list dictionary python
5 Answers
l = [
('a', 1),
('a', 2),
('a', 3),
('b', 1),
('b', 2),
('c', 1),
]

d = {}
for x, y in l:
    d.setdefault(x, []).append(y)
print d

produces:

{'a': [1, 2, 3], 'c': [1], 'b': [1, 2]}
user_143430

Slightly simpler...

from collections import defaultdict

fq = defaultdict(list)
for n, v in myList:
    fq[n].append(v)
    
print(fq) # defaultdict(<type 'list'>, {'a': [1, 2, 3], 'c': [1], 'b': [1, 2]})
SLott_106610

A solution using groupby

from itertools import groupby
l = [('a',1), ('a', 2),('a', 3),('b', 1),('b', 2),('c', 1),]
[(label, [v for l,v in value]) for (label, value) in groupby(l, lambda x:x[0])]

Output:

[('a', [1, 2, 3]), ('b', [1, 2]), ('c', [1])]

groupby(l, lambda x:x[0]) gives you an iterator that contains

['a', [('a', 1), ...], c, [('c', 1)], ...]
user_74610

I would just do the basic

answer = {}
for key, value in list_of_tuples:
  if key in answer:
    answer[key].append(value)
  else:
    answer[key] = [value]

If it's this short, why use anything complicated. Of course if you don't mind using setdefault that's okay too.

lacker_26520
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