How to merge a list of multiple dictionaries into a dictionary of lists?

Question

I have the following list of dictionaries in Python3.x:

list_of_dictionaries = [{0:3523, 1:3524, 2:3540, 4:3541, 5:3542}, 
                        {0:7245, 1:7246, 2:7247, 3:7248, 5:7249, 6:7250},
                        {1:20898, 2:20899, 3:20900, 4:20901, 5:20902}]

In this case, it's a single list with three dictionaries.

I would like to efficiently merge this into a single dictionary with lists as values; here is the correct answer:

correct = {0:[3523, 7245], 1:[3524, 7246, 20898], 2:[3540, 7247, 20899], 
               3:[7248, 20900], 4:[3541, 20901], 5:[3542, 7249, 20902], 6:[7250]}

My first thought was a list comprehension like this:

dict(pair for dictionary in list_of_dictionaries for pair in dictionary.items())

But this is wrong, as it doesn't include lists of values:

{0: 7245, 1: 20898, 2: 20899, 4: 20901, 5: 20902, 3: 20900, 6: 7250}

I'm also worried about how to efficiently as possible create value lists. It may not scale to large lists/large dictionaries either.

How could I accomplish this?

Answer 1

defaultdict

You can use collections.defaultdict . Your dictionary comprehension will never work as you are not defining any lists. This is likely to be more efficient than using a dictionary comprehension, which would involve iterating each dictionary for each unique key.

from collections import defaultdict

dd = defaultdict(list)

for d in list_of_dictionaries:
    for k, v in d.items():
        dd[k].append(v)

Result:

print(dd)

defaultdict(list,
            {0: [3523, 7245],
             1: [3524, 7246, 20898],
             2: [3540, 7247, 20899],
             4: [3541, 20901],
             5: [3542, 7249, 20902],
             3: [7248, 20900],
             6: [7250]})

Dictionary comprehension

A dictionary comprehension is possible but this requires calculating the union of keys and iterating the list of dictionaries for each of these keys:

allkeys = set().union(*list_of_dictionaries)

res = {k: [d[k] for d in list_of_dictionaries if k in d] for k in allkeys}

{0: [3523, 7245],
 1: [3524, 7246, 20898],
 2: [3540, 7247, 20899],
 3: [7248, 20900],
 4: [3541, 20901],
 5: [3542, 7249, 20902],
 6: [7250]}

Time complexity

Consider these terms:

n = sum(map(len, list_of_dictionaries))
m = len(set().union(*list_of_dictionaries))
k = len(list_of_dictionaries)

In this context, the defaultdict solution will have complexity O( n ), while the dictionary comprehension will have complexity O( mk ), where mk >= n .

Answer 2

why not just use for loops? for example:

final = {}

for i in list_of_dictionaries:
    for k in i:
        if not k in final:
            final[k] = []
        final[k].append(i[k])


print(final)

Outputs final as:

{0: [3523, 7245], 1: [3524, 7246, 20898], 2: [3540, 7247, 20899], 4: [3541, 20901], 5: [3542, 7249, 20902], 3: [7248, 20900], 6: [7250]}

Answer 3

Using groupby and itemgetter we could first create a flat list of tuples that represent the keys and values of each subdict. Then we can use groupby on our sorted new list. From there we can create our new dictionary using k and the items in index[1] of list(g)

from itertools import groupby
from operator import itemgetter

d = {}
new_lod = sorted([(j, i[j]) for i in lod for j in i], key=itemgetter(0))
for k, g in groupby(new_lod, key=itemgetter(0)):
    d[k] = [i[1] for i in list(g)]

# {0: [3523, 7245], 1: [3524, 7246, 20898], 2: [3540, 7247, 20899], 3: [7248, 20900], 4: [3541, 20901], 5: [3542, 7249, 20902], 6: [7250]}

Answer 4

You first need to flatten the dictionaries:

flattened_pairs = (
    pair for dictionary in list_of_dictionaries for pair in dictionary.items()
)

Then you can use itertools.groupby to group the values. It expects the values to be sorted by key.

key_fn = lambda pair: pair[0]

merged = {
    k: [pair[1] for pair in g]
    for k, g in groupby(
        sorted(flattened_pairs, key=key_fn),
        key=key_fn
    )
}

print(merged)

Output:

{0: [3523, 7245], 1: [3524, 7246, 20898], 2: [3540, 7247, 20899], 3: [7248, 20900], 4: [3541, 20901], 5: [3542, 7249, 20902], 6: [7250]}