计算列表中字典值的中值

Question

我想计算列表中"score"（字典值）的中值。

my_dict = {"John": [{"class": "math", "score": 100, "year": 2014}, {"class": "english", "score": 85, "year": 2015}, {"class": "science", "score": 90, "year": 2015}], 
"Timmy": [{"class": "math", "score": 87, "year": 2014}, {"class": "english", "score": 91, "year": 2015}], 
"Sally":[{"class": "math", "score": 95, "year": 2014}]}

输出如下：

new_dict = {"John": 90, "Timmy": 89, "Sally": 95}

我想我需要根据得分对 my_dict 进行排序，然后计算中值。如果不使用外部包装，就无法完全弄清楚任何一步。

如有任何帮助，我们将不胜感激！ Python.

的新手

Answer 1

my_dict = {"John": [{"class": "math", "score": 100, "year": 2014}, {"class": "english", "score": 85, "year": 2015}, {"class": "science", "score": 90, "year": 2015}], 
"Timmy": [{"class": "math", "score": 87, "year": 2014}, {"class": "english", "score": 91, "year": 2015}], 
"Sally":[{"class": "math", "score": 95, "year": 2014}]}

import numpy as np
medians = {}
for k in my_dict.keys():
    ls = []
    for d in my_dict[k]:
        ls.append(d['score'])
    medians[k] = np.median(ls)

print(medians)

输出：

{'Sally': 95.0, 'Timmy': 89.0, 'John': 90.0}

您可以只使用 numpy.median 来计算中位数。

Answer 2

您可以使用 statistics 的中位数。

from statistics import median

my_dict = {"John": [{"class": "math", "score": 100, "year": 2014}, {"class": "english", "score": 85, "year": 2015}, {"class": "science", "score": 90, "year": 2015}], 
"Timmy": [{"class": "math", "score": 87, "year": 2014}, {"class": "english", "score": 91, "year": 2015}], 
"Sally":[{"class": "math", "score": 95, "year": 2014}]}

new_dict = {}

for k, v in my_dict.items():
  m = []
  for l in v:
    m.append(l["score"])
  new_dict[k] = median(m)

print(new_dict)

如果你不想使用包而自己写函数，你可以这样调用：

def median(lst):
    n = len(lst)
    s = sorted(lst)
    return (sum(s[n//2-1:n//2+1])/2.0, s[n//2])[n % 2] if n else None

Answer 3

使用 median 和听写理解

的一行

from statistics import median 
{k:median([v.get('score',0) for v in my_dict[k]]) for k in my_dict.keys()}

输出：

{'John': 90, 'Timmy': 89.0, 'Sally': 95}

计算列表中字典值的中值

Calculate median of dictionary values inside list

python

dictionary

list-comprehension

list

dictionary-comprehension