获取字典列表平均值的最佳方法

Question

我有一个dicts列表，如下所示：

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数。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关键字的值，并对其进行平均，然后放入一个新的字典中。

不确定最好/最像蟒蛇的方式。

第一个列表的结构与第二个列表的架构相同，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

这似乎有点过头了，可能还有比这更优雅的解决方案。

感谢您的帮助。

Answer 1

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)

Answer 2

您可以使用Counter找到所有dict值的总和，然后最后将其除以列表的长度，得到平均

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

可以争论它们是否更"Pythonic"；但字典理解是一种选择。这里有一种dict理解和内置zip()和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}}