Follow

Keep Up to Date with the Most Important News

By pressing the Subscribe button, you confirm that you have read and are agreeing to our Privacy Policy and Terms of Use
Contact

How to optimise the grouping of values in lists by key in a dictionary list?

The script below works but I was wondering if there is a faster solution? With very large dictionary lists I noticed a small delay.

from collections import defaultdict

input = [{"first": 1.56,
          "second": [1, 2, 3, 4]}, {"first": 7.786,
                                    "second": [5, 6, 7, 8]}, {"first": 4.4,
                                                              "second": [9, 10, 11, 12]}]
output = [{"first": [1.56, 7.786, 4.4],
          "second":[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]}]

my_dictionary = defaultdict(list)
for item in input:
    for key, value in item.items():
        my_dictionary[key].append(value)
print(my_dictionary)

#defaultdict(<class 'list'>, {'first': [1.56, 7.786, 4.4], 'second': [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]})

>Solution :

MEDevel.com: Open-source for Healthcare and Education

Collecting and validating open-source software for healthcare, education, enterprise, development, medical imaging, medical records, and digital pathology.

Visit Medevel

It seems keys in the dictionaries are the same, so you could use a dict comprehension:

out = {k:[d[k] for d in input] for k in input[0]}

Another pretty fast alternative is to use the cytoolz module.

# pip install cytoolz
from cytoolz.dicttoolz import merge_with
out = merge_with(list, *input)

Output:

{'first': [1.56, 7.786, 4.4],
 'second': [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]}

Timings:

>>> my_input = input * 10000
>>> %%timeit
... my_dictionary = defaultdict(list)
... for item in my_input:
...     for key, value in item.items():
...         my_dictionary[key].append(value)
20.3 ms ± 2.49 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

>>> %timeit out = {k:[d[k] for d in my_input] for k in my_input[0]}
4.65 ms ± 541 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

>>> %timeit out = merge_with(list, *my_input)
5.58 ms ± 2.09 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)
Add a comment

Leave a Reply

Keep Up to Date with the Most Important News

By pressing the Subscribe button, you confirm that you have read and are agreeing to our Privacy Policy and Terms of Use

Discover more from Dev solutions

Subscribe now to keep reading and get access to the full archive.

Continue reading