如何从字典中的多个列表中计算平均值和 SD 并将它们排序到新字典中？答案

【问题标题】：How to calculate the mean and SD from multiple lists in a dictionary and sort them into a new dictionary?如何从字典中的多个列表中计算平均值和 SD 并将它们排序到新字典中？
【发布时间】：2022-01-21 10:19:00
【问题描述】：

我有一个包含 5 个键的字典，每个键都有 10 个以列表形式分配给它的值：

{'a': [5, 6, 3, 1, 3, 2, 5, 8, 6, 7], 
 'b': [3, 5, 2, 7, 0, 2, 10, 4, 3, 4], 
 'c': [9, 7, 11, 10, 8, 9, 7, 10, 7, 9], 
 'd': [6, 4, 5, 7, 6, 8, 5, 6, 5, 7], 
 'e': [2, 5, 1, 4, 2, 3, 4, 2, 5, 1]}

我还设置了一个空字典，我想将每个列表的平均值和标准差存储在原始字典中：

{'a_analysis': [],
 'b_analysis': [],
 'c_analysis': [],
 'd_analysis': [],
 'e_analysis': []}

到目前为止我有这个代码：

for key in original_dictionary:   #for each key in the dictionary
   for value in original_dictionary:   #iterate through the values in each key
      sum =+ value   #add the value to a sum variable
   mean = sum / len(orignal_dictionary[key])   #get the mean by dividing the sum by the len of each key
   #here I want to return the mean value to the respective key in the new dictionary 
   sd =    #then I need to get a value for the standard deviation here 
   #and also return it to the respective key in the new dictionary

任何帮助将不胜感激。

【问题讨论】：

标签： python list dictionary for-loop

【解决方案1】：

只是我的解决方案又小了一点，并添加了一些 cmets。希望现在得到解释。

import statistics

input = {'a': [5, 6, 3, 1, 3, 2, 5, 8, 6, 7], 
 'b': [3, 5, 2, 7, 0, 2, 10, 4, 3, 4], 
 'c': [9, 7, 11, 10, 8, 9, 7, 10, 7, 9], 
 'd': [6, 4, 5, 7, 6, 8, 5, 6, 5, 7], 
 'e': [2, 5, 1, 4, 2, 3, 4, 2, 5, 1]}

analyse = {'a_analysis': [],
 'b_analysis': [],
 'c_analysis': [],
 'd_analysis': [],
 'e_analysis': []}

for key in input.keys(): #We look up every key we have in the directory "input"
    mean = sum(input[key]) / len(input[key]) #We calculate the mean by summing up all the values for the current key and dividing them through the length of our array 
    sd = statistics.stdev(input[key]) #We simply calculate the sd with a function from the libarary "statistics" 
    analyse[key+'_analysis'] = ["mean: "+str(mean), "sd: "+str(sd)] #We put the mean and sd into the directory "analyse" at the key our_current_key+"_analyse"

for item in analyse.items(): print(item)

【讨论】：

您的答案可以通过添加有关代码的作用以及它如何帮助 OP 的更多信息来改进。

【解决方案2】：

这里有一个解决方案，希望你喜欢：

def standard_deviation(numbers):
    mean = sum(numbers) / len(numbers)
    result = (sum( [((x - mean) ** 2) for x in numbers] ) / len(numbers))**0.5
    return result

input ={
'a': [5, 6, 3, 1, 3, 2, 5, 8, 6, 7],
'b': [3, 5, 2, 7, 0, 2, 10, 4, 3, 4],
'c': [9, 7, 11, 10, 8, 9, 7, 10, 7, 9],
'd': [6, 4, 5, 7, 6, 8, 5, 6, 5, 7],
'e': [2, 5, 1, 4, 2, 3, 4, 2, 5, 1]
}
output = {}
for key in input.keys():
    output[key+"_analysis"] = {"mean": sum(input[key])/len(input[key]),
                               "standard deviation":standard_deviation(input[key]),
                               }

这是输出：

{'a_analysis': {'mean': 4.6, 'standard deviation': 2.1540659228538015},
 'b_analysis': {'mean': 4.0, 'standard deviation': 2.6832815729997477},
 'c_analysis': {'mean': 8.7, 'standard deviation': 1.345362404707371},
 'd_analysis': {'mean': 5.9, 'standard deviation': 1.1357816691600546},
 'e_analysis': {'mean': 2.9, 'standard deviation': 1.445683229480096}}

【讨论】：

【解决方案3】：

您也可以为此使用pandas：

import pandas as pd
out = pd.DataFrame(your_dict).agg(['mean', 'std']).to_dict()

输出：

{'a': {'mean': 4.6, 'std': 2.2705848487901865},
 'b': {'mean': 4.0, 'std': 2.8284271247461903},
 'c': {'mean': 8.7, 'std': 1.4181364924121764},
 'd': {'mean': 5.9, 'std': 1.1972189997378646},
 'e': {'mean': 2.9, 'std': 1.5238839267549948}}

【讨论】：