在字典列表中获取平均值的最佳方法

Question

我有一个字典列表如下：rois = [{'player': 'kraftvk', 'over': {1.5: 67.97, 1.75: 51.005}, 'under': {1.5: -77.97, 1.75: -59.12}}, {'player': 'meltosik', 'over': {1.5: 61.635, 1.75: 37.455}, 'under': {1.5: -71.635, 1.75: -44.765}}]我想得到每本字典的平均值。预期输出：d = {'over': {1.5: 64.80, 1.75: 44.23}, 'under': {1.5: -74.80, 1.75: -51.9425}}因此，映射每个字典中每个键的值并将它们平均并放入新字典中。不确定这样做的最佳/最pythonic方式。第一个列表的结构与第二个列表的结构相同，所以可以这样做：d = {'over': {}, 'under': {}}for k,v in rois[0].items():    if k != 'player':        for quote, roi in rois[0][k].items():            if k == 'over':                d[k][quote] = (roi + rois[1]['over'][quote])/2            if k == 'under':                d[k][quote] = (roi + rois[1]['under'][quote])/2这似乎有点矫枉过正，可能有比这更优雅的解决方案。帮助表示赞赏。

Answer 1

可以争论它们是否更“Pythonic”，但字典理解是一种选择。zip()这是一种使用字典理解和内置map()函数的方法

>>> rois = [{'player': 'kraftvk', 'over': {1.5: 67.97, 1.75: 51.005}, 'under': {1.5: -77.97, 1.75: -59.12}}, {'player': 'meltosik', 'over': {1.5: 61.635, 1.75: 37.455}, 'under': {1.5: -71.635, 1.75: -44.765}}]

>>> d1 = { key : zip(*[ elem[key].values() for elem in rois]) for key in rois[0].keys() if key != "player" }

>>> d2 = { key : value.keys() for key,value in rois[0].items() if key != "player" }

>>> avg = lambda l : round(sum(l)/float(len(l)), 2)

>>> d3 = { key : map(avg, value) for key, value in d1.items() }

>>> d = { key : dict(zip(value, d3[key])) for key, value in d2.items() }

>>> d

{'under': {1.5: -74.8, 1.75: -51.94}, 'over': {1.5: 64.8, 1.75: 44.23}}

Answer 2

您可以使用Counter查找所有 dict 值的总和，然后最后将其除以列表的长度以获得平均值

from collections import Counter

cover = Counter(); cunder = Counter()

for d in rois:

    cover.update(d['over'])

    cunder.update(d['under'])

>>> l = len(rois)

>>> res = {}

>>> res['over'] = {k:round(v/l,2) for k,v in cover.items()}

>>> res['under'] = {k:round(v/l,2) for k,v in cunder.items()}

>>> 

>>> res

{'over': {1.5: 64.8, 1.75: 44.23}, 'under': {1.5: -74.8, 1.75: -51.94}}

Answer 3

d = {'over': [], 'under': []}

rois = [{'player': 'kraftvk', 'over': {1.5: 67.97, 1.75: 51.005}, 'under': {1.5: -77.97, 1.75: -59.12}}, {'player': 'meltosik', 'over': {1.5: 61.635, 1.75: 37.455}, 'under': {1.5: -71.635, 1.75: -44.765}}]

# Collecting players 'over' and 'under' dictionaries

for player_dict in rois:

    d['over'].append(player_dict['over'])

    d['under'].append(player_dict['under'])

print(d)

s = {'over': {}, 'under': {}}

def average_dicts(dicts):

    sum_dict = {}

    # Going over the ['over' and 'under'] dictionaries collection

    for d in dicts:

        # Going over the [{1.5 : value_1, 1.75 : value_2}] dictionary, that's under either ['over' and 'under'] dictionaries

        for k, v in d.items():

            # Summing all values of either [ 1.5, 1.75]

            if k not in sum_dict.keys():

                sum_dict[k] = 0

            sum_dict[k] += d.get(k,0)

    # Dividing sum by number of dictionaries to get average

    for k, v in sum_dict.items():

        sum_dict[k] = v / len(dicts)

    return sum_dict

s['over'] = average_dicts(d['over'])

s['under'] = average_dicts(d['under'])

print(s)