减少字典列表的优雅方法？ [英] elegant way to reduce a list of dictionaries?

问题描述

我有一个词典列表，每个词典包含完全相同的键。我想找到每个键的平均值，我想知道如何使用reduce进行操作（或者，如果不可能，使用比嵌套 for s更为优雅的方法） ）。

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 reduce (or if not possible with another more elegant way than using nested fors).

以下是列表：

[
  {
    "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
  }, ...
]

我想找回我这样的字典：

I would like to get back I dictionary like this:

{
  "accuracy": 0.81,
  "f_measure": 0.83,
  "precision": 0.84,
  "recall": 0.83
}

这是我到目前为止的内容，但我不喜欢它：

Here is what I had so far, but I don't like it:

folds = [ ... ]

keys = folds[0].keys()
results = dict.fromkeys(keys, 0)

for fold in folds:
    for k in keys:
        results[k] += fold[k] / len(folds)

print(results)

推荐答案

作为替代方案，如果要对数据进行此类计算，那么您不妨使用 pandas （这对于一次过来说是过大的，但是会大大简化此类任务...）

As an alternative, if you're going to be doing such calculations on data, then you may wish to use pandas (which will be overkill for a one off, but will greatly simplify such tasks...)

import pandas as pd

data = [
  {
    "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
  }, # ...
]

result = pd.DataFrame.from_records(data).mean().to_dict()

哪一个给您：

{'accuracy': 0.79000000000000004,
 'f_measure': 0.8184953755563118,
 'precision': 0.82016728940624295,
 'recall': 0.81870370370370382}

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