我有一個包含 5 個鍵的字典，每個鍵都有 10 個以串列形式分配給它的值：
{'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]}
我還設定了一個空字典，我想在原始字典中存盤每個串列的均值和標準差：
{'a_analysis': [],
'b_analysis': [],
'c_analysis': [],
'd_analysis': [],
'e_analysis': []}
到目前為止我有這個代碼：
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
任何幫助將不勝感激。
uj5u.com熱心網友回復：
您也可以pandas為此使用：
import pandas as pd
out = pd.DataFrame(your_dict).agg(['mean', 'std']).to_dict()
輸出：
{'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}}
uj5u.com熱心網友回復：
這是一個解決方案，希望您喜歡：
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]),
}
這是輸出：
{'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}}
uj5u.com熱心網友回復：
這應該有效。希望我能幫到你 :)
import statistics
direc = {'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 direc.items():
sum = 0
lenght = len(key[1])
for i in key[1]:
sum = sum int(i)
mean = sum / lenght
string = str(key[0]) '_analysis'
sd = statistics.stdev(key[1])
analyse[string] = [mean, sd]
print(analyse)
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