nested list to nested dict python3

Question

I have a list as below:

L = [[0,[1,1.0]],
     [0,[2,0.5]],
     [1,[3,3.0]],
     [2,[1,0.33],
     [2,[4,1.5]]]

I would like to convert it into a nested dict as below:

D = {0:{1: 1.0,
        2: 0.5},
     1:{3: 3.0},
     2:{1: 0.33,
        4: 1.5}
     }

I'm not sure how to convert it. Any suggestion? Thank you!

I tried this https://docs.python.org/2/library/collections.html#collections.defaultdict — MLam, Feb 22 '18 at 20:52
I asked a similar question a while ago and got some good answers. https://stackoverflow.com/questions/45019709/how-to-process-dicts-within-nested-lists-in-an-ordered-manner-with-python — Calvin Ellington, Feb 22 '18 at 20:56
If it's given that the input and output are depth-2, the question and answers are not very general. — smci, Dec 10 '19 at 04:31

Reck · Answer 1 · 2018-02-22T21:06:53.350

7

Beginners friendly,

D = {}
for i, _list in L:
    if i not in D:
        D[i] = {_list[0] : _list[1]}
    else:
        D[i][_list[0]] = _list[1]})

Result:

{0: {1: 1.0, 2: 0.5}, 1: {3: 3.0}, 2: {1: 0.33, 4: 1.5}}

edited Feb 22 '18 at 21:06

answered Feb 22 '18 at 21:02

Reck

1,388
11
20

This isn't bad, but instead of `D[i].update({_list[0] : _list[1]})` use `D[i][_list[0]] = _list[1]`. and better yet, use `for k, (k2, v) in L: ... `D[k][k2] = v` – juanpa.arrivillaga Feb 22 '18 at 21:03
Yeah sure. Correct. @juanpa.arrivillaga – Reck Feb 22 '18 at 21:04
1

Use `.upate` to update the dict with *another dynamic dict* with many possible entries. You want to update a single key-value pair, don't creat an unecessary intermediate `dict` to pass to the `update` method. – juanpa.arrivillaga Feb 22 '18 at 21:05
1

Yeah makes sense. – Reck Feb 22 '18 at 21:06

RomanPerekhrest · Accepted Answer · 2018-02-22T21:07:18.413

6

With collections.defaultdict([default_factory[, ...]]) class:

import collections

L = [[0,[1,1.0]],
     [0,[2,0.5]],
     [1,[3,3.0]],
     [2,[1,0.33]],
     [2,[4,1.5]]]

d = collections.defaultdict(dict)
for k, (sub_k, v) in L:
    d[k][sub_k] = v

print(dict(d))

The output:

{0: {1: 1.0, 2: 0.5}, 1: {3: 3.0}, 2: {1: 0.33, 4: 1.5}}

collections.defaultdict(dict) - the first argument provides the initial value for the default_factory attribute; it defaults to None. Setting the default_factory to dict makes the defaultdict useful for building a dictionary of dictionaries.

edited Feb 22 '18 at 21:07

answered Feb 22 '18 at 20:56

RomanPerekhrest

88,541
4
65
105

3

use `for k, (k2, v) in L: ` – juanpa.arrivillaga Feb 22 '18 at 20:58
my mistake I see now @juanpa.arrivillaga – Calvin Ellington Feb 22 '18 at 20:59
1

@juanpa.arrivillaga, let's do it – RomanPerekhrest Feb 22 '18 at 21:00

score 1 · Answer 3 · answered Feb 22 '18 at 20:53

1

You can use itertools.groupby:

import itertools
L = [[0,[1,1.0]],
 [0,[2,0.5]],
 [1,[3,3.0]],
 [2,[1,0.33]],
 [2,[4,1.5]]
 ]

new_l = {a:dict([i[-1] for i in b]) for a, b in itertools.groupby(L, key=lambda x:x[0])}

Output:

{0: {1: 1.0, 2: 0.5}, 1: {3: 3.0}, 2: {1: 0.33, 4: 1.5}}

answered Feb 22 '18 at 20:53

Ajax1234

69,937
8
61
102

4

@MLam the other answer to this is better – Calvin Ellington Feb 22 '18 at 21:01

nested list to nested dict python3

3 Answers3