在python中查找列表中唯一数据的平均值答案

【问题标题】：find average of unique data in list in python在python中查找列表中唯一数据的平均值
【发布时间】：2021-04-18 06:20:54
【问题描述】：

数据列如下：城市、名字、姓氏、性别、年龄、体重。客户可以多次输入，并以不同的权重记录。

我通过执行以下操作唯一标识了客户和性别：

现在我想找出唯一客户的平均年龄。不知道该怎么做？另外，我想报告有多少客户被多次称重。客户将被列出的次数与记录的体重一样多。这是我的课

class clients:
    def __init__(self, city, name, gender, age, weight):
        self.City = city
        self.Name = name
        self.Gender = gender
        self.Age = age
        self.Weight = weight

【问题讨论】：

1.您能否指定应该用于决定客户端“唯一性”的属性集，因为您写道，同一个客户端可能会被多次列出？ 2. 同一个客户有可能有历史数据（不同年龄）吗？ 3. 客户数据如何存储？是列表，还是....
客户端名称是否唯一？你的班级名字很糟糕；这是一个体重记录，而不是“客户”

标签： python filter average

【解决方案1】：

对该主题的部分回答：

对类声明进行了一些更正，以便说明唯一性

class Clients:
    def __init__(self, city, first_name, last_name, gender, age, weight, uuid):
        self.City = city
        self.FirstName = first_name
        self.LastName = last_name
        self.Gender = gender
        self.Age = age
        self.Weight = weight
        self.UUID = uuid
        
    def __eq__(self, other):
        if self.City == other.City and \
            self.FirstName == other.FirstName and \
            self.LastName == other.LastName and \
            self.Gender == other.Gender and \
            self.Age == other.Age and \
            self.Weight == other.Weight and \
            self.UUID == other.UUID:
            return True
        return False

    def __hash__(self):
        return hash((self.City, self.FirstName, self.LastName, self.Gender, self.Age, self.Weight, self.UUID))

假设您有一个包含不同条目的简单列表，例如：

import uuid

list_of_measurements = []
list_of_measurements.append(Clients("Spamburg", "Foo", "Bar", "Apache", 31, 80, uuid.uuid4()))
list_of_measurements.append(Clients("Spamburg", "Foo", "Bar", "Apache", 31, 82, list_of_measurements[0].UUID))  # new weight measurement
list_of_measurements.append(Clients("Spamburg", "Foo", "Bar", "Apache", 32, 83, list_of_measurements[0].UUID))  # new weight measurement, age changed
list_of_measurements.append(Clients("Spamville", "Foo", "Bar", "Apache", 33, 83, list_of_measurements[0].UUID))  # new weight measurement, client moved
list_of_measurements.append(Clients("Spamville", "Foo", "Bar", "Apache", 33, 83, list_of_measurements[0].UUID))  # duplicate measurement
list_of_measurements.append(Clients("Spamville", "Foo", "Bar", "Hind", 33, 83, list_of_measurements[0].UUID))  # client underwent gender change
list_of_measurements.append(Clients("Spamville", "Foo", "ZanziBar", "Hind", 33, 83, list_of_measurements[0].UUID))  # client got married and changed their last name
list_of_measurements.append(Clients("Spamville", "Foo Too", "Bar None", "Hind", 23, 63, uuid.uuid4()))  # second unique client
list_of_measurements.append(Clients("Spamville", "Foo Three", "Bar Some", "Apache", 18, 60, uuid.uuid4()))  # third unique client

请注意，除了 UUID 之外的所有内容可能会因每个客户而实际发生变化。

查找唯一客户数量 - 可以通过使用集合获得：

set_of_clients = set([c.UUID for c in list_of_measurements]) # will only contain unique UUIDs

报告有多个体重测量值的客户 - 可以使用 dict 和 sum 完成：

measurements_dict = {uid: sum(1 for c in list_of_measurements if c.UUID == uid) for uid in set_of_clients}
list_of_customers_with_more_than_one_measurement = [uid for uid, m in measurements_dict.items() if m > 1]
print(list_of_customers_with_more_than_one_measurement)

删除可能重复的测量值 - 可以通过强制转换为列表和设置来完成，这需要之前定义的类的 eq 和 hash 属性

list_of_deduplicated_measurements = list(set(list_of_measurements))

平均年龄，假设您只想对客户最后记录的年龄进行平均

from statistics import mean

ages_dict = {uid: max(c.Age for c in list_of_measurements if c.UUID == uid) for uid in set_of_clients}
average_age = mean(ages_dict.values())

【讨论】：