CODEWITHSUNDEEP
pythonlistdictionarytuplesiteration
Dan
Dan

Reputation: 35453

Converting a list of tuples into a dict

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)...

Upvotes: 44

Views: 25262

Answers (5)

user7461
user7461

Reputation: 119

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)], ...]

Upvotes: 11

S.Lott
S.Lott

Reputation: 391852

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]})

Upvotes: 39

jfs
jfs

Reputation: 414275

Print list of tuples grouping by the first item

This answer is based on the @gommen one.

#!/usr/bin/env python

from itertools import groupby
from operator  import itemgetter

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

key = itemgetter(0)
L.sort(key=key) #NOTE: use `L.sort()` if you'd like second items to be sorted too
for k, group in groupby(L, key=key):
    print k, ' '.join(str(item[1]) for item in group)

Output:

a 1 2 3
b 1 2
c 1

Upvotes: 3

lacker
lacker

Reputation: 5550

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.

Upvotes: 3

Ned Batchelder
Ned Batchelder

Reputation: 375594

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]}

Upvotes: 49

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