下面的腳本有效，但我想知道是否有更快的解決方案？對于非常大的字典串列，我注意到一個小的延遲。
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]]})
uj5u.com熱心網友回復：
字典中的鍵似乎是相同的，因此您可以使用字典理解：
out = {k:[d[k] for d in input] for k in input[0]}
另一個非常快速的替代方法是使用該cytoolz模塊。
# pip install cytoolz
from cytoolz.dicttoolz import merge_with
out = merge_with(list, *input)
輸出：
{'first': [1.56, 7.786, 4.4],
'second': [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]}
時間：
>>> 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)
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