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]]})
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Answer
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)