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python - 检索各个列表中元素的排名,计算其排名分数的加权平均值 Python

转载 作者:太空宇宙 更新时间:2023-11-03 17:48:30 25 4
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我有两个排序的字典,即它们现在表示为列表。我想检索每个列表中每个元素的排名位置并将其存储在一个变量中,以便最终我可以计算两个列表中每个元素的排名分数的加权平均值。这是一个例子。

dict1 = {'class1': 15.17, 'class2': 15.95, 'class3': 15.95}

sorted_dict1 = [('class1', 15.17), ('class2', 15.95), ('class3', 15.95)]

sorted_dict2 = [('class2', 9.10), ('class3', 9.22), ('class1', 10.60)]

到目前为止,我可以检索列表中每个元素的排名位置并打印排名,但是当我尝试计算排名分数的加权平均值时,即 [(w1*a + w2*b)/(w1+w2) )],其中“a”是sorted_dict1中的排名位置,“b”是sorted_dict2中的排名位置,我得到的数字不是正确的加权平均数。

尝试了各种事情,这里是一个:

for idx, val in list(enumerate(sorted_dict1, 1)):
for idx1, val1 in list(enumerate(sorted_dict2, 1)):
position_dict1 = idx
position_dict2 = idx1
weighted_average = float((0.50*position_dict1 + 0.25*position_dict2))/0.75
print weighted_average

我也没有考虑如果两个类在列表中排名相同会发生什么。我也将很感激得到任何有关这方面的提示/帮助。

我认为我可能需要创建一个函数来解决这个问题,但我也没有走得太远。

任何帮助以及解释代码的注释都会很棒。

所以我想计算列表中元素排名位置的加权平均值。例如的加权平均值:

1 类:加权平均值 = ((0.50 * 1) + (0.25 * 3))/0.75 = 1.5

2级:那么加权平均值 = ((0.50 *2)+(0.25*1))/0.75 = 1.6666..7

谢谢!

最佳答案

我采取了简单的路线,并给同等分数的类(class)下一个整数排名,所以 class3class2两者均在 sorted_dict1 中排名第 2

#!/usr/bin/env python

#Get the ranks for a list of (class, score) tuples sorted by score
#and return them in a dict
def get_ranks(sd):
#The first class in the list has rank 1
k, val = sd[0]
r = 1
rank = {k: r}

for k, v in sd[1:]:
#Only update the rank number if this value is
#greater than the previous
if v > val:
val = v
r += 1
rank[k] = r
return rank

def weighted_mean(a, b):
return (0.50*a + 0.25*b) / 0.75

sorted_dict1 = [('class1', 15.17), ('class2', 15.95), ('class3', 15.95)]
sorted_dict2 = [('class2', 9.10), ('class3', 9.22), ('class1', 10.60)]

print sorted_dict1
print sorted_dict2

ranks1 = get_ranks(sorted_dict1)
ranks2 = get_ranks(sorted_dict2)

print ranks1
print ranks2

keys = sorted(k for k,v in sorted_dict1)

print [(k, weighted_mean(ranks1[k], ranks2[k])) for k in keys]

输出

[('class1', 15.17), ('class2', 15.949999999999999), ('class3', 15.949999999999999)]
[('class2', 9.0999999999999996), ('class3', 9.2200000000000006), ('class1', 10.6)]
{'class2': 2, 'class3': 2, 'class1': 1}
{'class2': 1, 'class3': 2, 'class1': 3}
[('class1', 1.6666666666666667), ('class2', 1.6666666666666667), ('class3', 2.0)]
<小时/>

在评论中我提到有一个很好的方法来创建 weighted_mean()具有自定义权重的函数。当然,我们可以将权重作为附加参数传递给 weighted_mean() ,但这会调用 weighted_mean()比需要的更加困惑,使程序更难以阅读。

技巧是使用一个函数,该函数将自定义权重作为参数并返回所需的函数。从技术上讲,这样的函数生成函数称为 closure .

这里有一个关于如何做到这一点的简短演示。

#!/usr/bin/env python

#Create a weighted mean function with weights w1 & w2
def make_weighted_mean(w1, w2):
wt = float(w1 + w2)
def wm(a, b):
return (w1 * a + w2 * b) / wt
return wm

#Make the weighted mean function
weighted_mean = make_weighted_mean(1, 2)

#Test
print weighted_mean(6, 3)
print weighted_mean(3, 9)

输出

4.0
7.0
<小时/>

这是上面第一个程序的更新版本,它处理任意数量的sorted_dict列表。它使用原来的get_ranks()函数,但它使用比上面的示例稍微复杂的闭包来对数据列表(或元组)进行加权平均值。

#!/usr/bin/env python

''' Weighted means of ranks

From https://stackoverflow.com/q/29413531/4014959

Written by PM 2Ring 2015.04.03
'''

from pprint import pprint

#Create a weighted mean function with weights from list/tuple weights
def make_weighted_mean(weights):
wt = float(sum(weights))
#A function that calculates the weighted mean of values in seq
#weighted by the weights passed to make_weighted_mean()
def wm(seq):
return sum(w * v for w, v in zip(weights, seq)) / wt
return wm


#Get the ranks for a list of (class, score) tuples sorted by score
#and return them in a dict
def get_ranks(sd):
#The first class in the list has rank 1
k, val = sd[0]
r = 1
rank = {k: r}

for k, v in sd[1:]:
#Only update the rank number if this value is
#greater than the previous
if v > val:
val = v
r += 1
rank[k] = r
return rank


#Make the weighted mean function
weights = [0.50, 0.25]
weighted_mean = make_weighted_mean(weights)

#Some test data
sorted_dicts = [
[('class1', 15.17), ('class2', 15.95), ('class3', 15.95), ('class4', 16.0)],
[('class2', 9.10), ('class3', 9.22), ('class1', 10.60), ('class4', 11.0)]
]
print 'Sorted dicts:'
pprint(sorted_dicts, indent=4)

all_ranks = [get_ranks(sd) for sd in sorted_dicts]
print '\nAll ranks:'
pprint(all_ranks, indent=4)

#Get a sorted list of the keys
keys = sorted(k for k,v in sorted_dicts[0])
#print '\nKeys:', keys

means = [(k, weighted_mean([ranks[k] for ranks in all_ranks])) for k in keys]
print '\nWeighted means:'
pprint(means, indent=4)

输出

Sorted dicts:
[ [ ('class1', 15.17),
('class2', 15.949999999999999),
('class3', 15.949999999999999),
('class4', 16.0)],
[ ('class2', 9.0999999999999996),
('class3', 9.2200000000000006),
('class1', 10.6),
('class4', 11.0)]]

All ranks:
[ { 'class1': 1, 'class2': 2, 'class3': 2, 'class4': 3},
{ 'class1': 3, 'class2': 1, 'class3': 2, 'class4': 4}]

Weighted means:
[ ('class1', 1.6666666666666667),
('class2', 1.6666666666666667),
('class3', 2.0),
('class4', 3.3333333333333335)]

这是 get_ranks() 的替代版本如果两个或多个类在列表中排名相同,则跳过排名数字

def get_ranks(sd):
#The first class in the list has rank 1
k, val = sd[0]
r = 1
rank = {k: r}
#The step size from one rank to the next. Normally
#delta is 1, but it's increased if there are ties.
delta = 1

for k, v in sd[1:]:
#Update the rank number if this value is
#greater than the previous.
if v > val:
val = v
r += delta
delta = 1
#Otherwise, update delta
else:
delta += 1
rank[k] = r
return rank

这是使用 get_ranks() 的替代版本的程序的输出:

Sorted dicts:
[ [ ('class1', 15.17),
('class2', 15.949999999999999),
('class3', 15.949999999999999),
('class4', 16.0)],
[ ('class2', 9.0999999999999996),
('class3', 9.2200000000000006),
('class1', 10.6),
('class4', 11.0)]]

All ranks:
[ { 'class1': 1, 'class2': 2, 'class3': 2, 'class4': 4},
{ 'class1': 3, 'class2': 1, 'class3': 2, 'class4': 4}]

Weighted means:
[ ('class1', 1.6666666666666667),
('class2', 1.6666666666666667),
('class3', 2.0),
('class4', 4.0)]

关于python - 检索各个列表中元素的排名,计算其排名分数的加权平均值 Python,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29413531/

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