elegant way to reduce a list of dictionaries?

Question

I have a list of dictionaries and each dictionary contains exactly the same keys. I want to find the average value for each key and I would like to know how to do it using r

Answer 1

Here you go, a solution using reduce():

from functools import reduce  # Python 3 compatibility

summed = reduce(
    lambda a, b: {k: a[k] + b[k] for k in a},
    list_of_dicts,
    dict.fromkeys(list_of_dicts[0], 0.0))
result = {k: v / len(list_of_dicts) for k, v in summed.items()}

This produces a starting point with 0.0 values from the keys of the first dictionary, then sums all values (per key) into a final dictionary. The sums are then divided to produce an average.

Demo:

>>> from functools import reduce
>>> list_of_dicts = [
...   {
...     "accuracy": 0.78,
...     "f_measure": 0.8169374016795885,
...     "precision": 0.8192088044235794,
...     "recall": 0.8172222222222223
...   },
...   {
...     "accuracy": 0.77,
...     "f_measure": 0.8159133315763016,
...     "precision": 0.8174754717495807,
...     "recall": 0.8161111111111111
...   },
...   {
...     "accuracy": 0.82,
...     "f_measure": 0.8226353934130455,
...     "precision": 0.8238175920455686,
...     "recall": 0.8227777777777778
...   }, # ...
... ]
>>> summed = reduce(
...     lambda a, b: {k: a[k] + b[k] for k in a},
...     list_of_dicts,
...     dict.fromkeys(list_of_dicts[0], 0.0))
>>> summed
{'recall': 2.4561111111111114, 'precision': 2.4605018682187287, 'f_measure': 2.4554861266689354, 'accuracy': 2.37}
>>> {k: v / len(list_of_dicts) for k, v in summed.items()}
{'recall': 0.8187037037037038, 'precision': 0.820167289406243, 'f_measure': 0.8184953755563118, 'accuracy': 0.79}
>>> from pprint import pprint
>>> pprint(_)
{'accuracy': 0.79,
 'f_measure': 0.8184953755563118,
 'precision': 0.820167289406243,
 'recall': 0.8187037037037038}