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 Countercover = 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' dictionariesfor 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_dicts['over'] = average_dicts(d['over'])s['under'] = average_dicts(d['under'])print(s)