Python列表中两个字典值的最大平均值

Question

这是我的字典列表：

array_of_dictionaries = [{

    "name": "Budi",

    "age": 23,

    "test_scores": [100.0, 98.0, 89.0]

},

{

    "name": "Charlie",

    "age": 24,

    "test_scores": [90.0, 100.0]

}]

这是我的代码：

def get_all_time_max_avg_score(dictionary_list):

  for element in dictionary_list:

    lensum = len(element['test_scores'])

    elesum = sum(element['test_scores'])

    meansum = elesum / lensum

    for attribute in element.keys():

        print("Mr. " + str(element['name']) + " with age(years) " + str(element['age']) + " get the highest average scores, " + str(round(meansum, 2)))

        break

get_all_time_max_avg_score(array_of_dictionaries)

所以我想从这两个词典中获得最高的平均分。我想要的输出是：

Mr. Budi with age(years) 23 get the highest average scores, 95.67

但我得到的是：

Mr. Budi with age(years) 23 get the highest average scores, 95.67

Mr. Charlie with age(years) 24 get the highest average scores, 95.0

我真的很感激我能得到的每一个帮助。

Answer 1

此代码迭代并打印列表中的每个元素。相反，通过与平均值max比较来执行列表。最大（可迭代，*，默认=无，键=功能）这是一个列表理解的解决方案。def get_all_time_max_avg_score(dictionary_list):    text = "Mr. {name} with age(years) {age} get the highest average scores, {average:.2f}"    def average(element):        return sum(element['test_scores']) / len(element['test_scores'])    e = max((element for element in dictionary_list), key=average, default=0)    return text.format(name=e['name'], age=e['age'], average=average(e))print(get_all_time_max_avg_score(array_of_dictionaries))输出：Mr. Budi with age(years) 23 get the highest average scores, 95.67

Answer 2

在打印任何内容之前，您需要计算每一项的平均值def get_all_time_max_avg_score(dictionary_list):    for element in dictionary_list:        lensum = len(element['test_scores'])        elesum = sum(element['test_scores'])        element['mean'] = elesum / lensum    first = sorted(dictionary_list, key=lambda x: x['mean'], reverse=True)[0]    print("Mr.", first['name'], "with age(years)", first['age'], "get the highest average scores,",          round(first['mean'], 2))或者使用Dataframedef get_all_time_max_avg_score(dictionary_list):    df = pd.DataFrame(dictionary_list)    df['mean'] = df['test_scores'].apply(np.mean)    first = df.loc[df['mean'].idxmax()]    print("Mr.", first['name'], "with age(years)", first['age'],          "get the highest average scores,", round(first['mean'], 2))

Answer 3

使用带有自定义键的内置max函数来找出平均分数最高的条目。array_of_dictionaries = [{    "name": "Budi",    "age": 23,    "test_scores": [100.0, 98.0, 89.0]},{    "name": "Charlie",    "age": 24,    "test_scores": [90.0, 100.0]}]all_time_max_avg_score = max(    array_of_dictionaries,    key=lambda d: sum(d['test_scores']) / len(d['test_scores']))meansum = sum(all_time_max_avg_score['test_scores']) / len(all_time_max_avg_score['test_scores'])print("Mr. " + str(all_time_max_avg_score['name']) + " with age(years) " + str(all_time_max_avg_score['age']) + " get the highest average scores, " + str(round(meansum, 2)))

Answer 4

收集所有名称的平均值并选择最大值def get_all_time_max_avg_score(dictionary_list):    means = []    for d in dictionary_list:        means.append(sum(d['test_scores']) / len(d['test_scores']))    maxmean = max(means)    winner = dictionary_list[means.index(maxmean)]    print(        f'Mr. {winner["name"]}, with {winner["age"]} years of age, has the ' +        f'highest average scores: {maxmean:.2f}'    )get_all_time_max_avg_score(array_of_dictionaries)输出Mr. Budi, with 23 years of age, has the highest average scores: 95.67

Answer 5

通过一些额外的循环，您可以确定具有最大平均值的人def get_all_time_max_avg_score(dictionary_list):     maxPersonIndex=0    maxAver=0    index=0    for element in dictionary_list:        lensum = len(element['test_scores'])        elesum = sum(element['test_scores'])        meansum = elesum / lensum        if(meansum>maxAver):            maxAver=meansum            maxPersonIndex=index        index+=1                for attribute in dictionary_list[maxPersonIndex].keys():        print("Mr. " + str(element['name']) + " with age(years) " + str(element['age']) + " get the highest average scores, " + str(round(meansum, 2)))        breakget_all_time_max_avg_score(array_of_dictionaries)