CODEWITHSUNDEEP
pythonpython-3.x
Gabriel Weich
Gabriel Weich

Reputation: 264

Convert list of dicts into a unique dict

I have:

[
  {'id': 1, 'name': 'foo'},
  {'id': 2, 'name': 'bar'},
  {'id': 1, 'name': 'gesiel'}
]

I want:

{
  1: [
    {'id': 1, 'name': 'foo'},
    {'id': 1, 'name': 'gesiel'}
  ],
  2: [
    {'id': 2, 'name': 'bar'}
  ]
}

This code does this:

organized = {d['id']:[] for d in data}
[organized[d['id']].append(d) for d in data]

Is there a more pythonic way to do this?

Upvotes: 4

Views: 225

Answers (4)

aydow
aydow

Reputation: 3801

Austin's answer is better, but here method just using dicts

In [175]: data = [{'id': 1, 'name': 'foo'}, {'id': 2, 'name': 'bar'}, {'id': 1, 'name': 'gesiel'}]

In [176]: organised = {}

In [177]: for d in data:
     ...:     if d['id'] in organised:
     ...:         organised[d['id']].append(d)
     ...:     else:
     ...:         organised[d['id']] = [d]
     ...:

In [178]: organised
Out[178]:
{1: [{'id': 1, 'name': 'foo'}, {'id': 1, 'name': 'gesiel'}],
 2: [{'id': 2, 'name': 'bar'}]}

Upvotes: 0

jpp
jpp

Reputation: 164693

There is nothing wrong with a 2-pass O(n) solution if you are, as here, working with an in-memory object. The main issue with your code is you are misusing the list comprehension.

A list comprehension should be used to construct a new list, not to process an in-place function or method in a loop. Taking your example, your logic will create a list which looks like:

[None, None, None, ..., None]

The side-effect of the comprehension means that organized values have items appended to them as required. Instead, you can rewrite using a simple for loop:

organized = {d['id']: [] for d in data}

for d in data:
    organized[d['id']].append(d)

Your logic can be made more efficient by not adding keys via an initial iteration. This common problem is resolved by collections.defaultdict, as in @Austin's solution. This solution gives an empty list for any key which does not exist:

from collections import defaultdict

res = defaultdict(list)

for d in data:
    res[d['i']].append(d)

print(res)

defaultdict(list,
            {1: [{'id': 1, 'name': 'foo'}, {'id': 1, 'name': 'gesiel'}],
             2: [{'id': 2, 'name': 'bar'}]})

Since defaultdict is a subclass of dict, there's usually no need to convert this back to a regular dictionary.

Upvotes: 0

vash_the_stampede
vash_the_stampede

Reputation: 4606

Using groupby.itertools we can create this dicitonary

from itertools import groupby
lista = [{'id': 1, 'name': 'foo'}, {'id': 2, 'name': 'bar'}, {'id': 1, 'name': 'gesiel'}]

d = {}
for k, g in groupby(sorted(lista, key=lambda x: x['id']), key=lambda x: x['id']):
    d[k] = list(g)
# {1: [{'id': 1, 'name': 'foo'}, {'id': 1, 'name': 'gesiel'}], 2: [{'id': 2, 'name': 'bar'}]}

or using dictionary comprehension

d = {k: list(g) for k, g in groupby(sorted(lista, key=lambda x: x['id']), key=lambda x: x['id'])}

Upvotes: 0

Austin
Austin

Reputation: 26039

Use collections.defaultdict:

from collections import defaultdict

data = [{'id': 1, 'name': 'foo'}, {'id': 2, 'name': 'bar'}, {'id': 1, 'name': 'gesiel'}]

d = defaultdict(list)

for x in data:
    d[x['id']].append(x)

print(d)
# defaultdict(<class 'list'>, {1: [{'id': 1, 'name': 'foo'}, {'id': 1, 'name': 'gesiel'}], 2: [{'id': 2, 'name': 'bar'}]})

Upvotes: 3

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