我如何找出字典中分配给键的值的平均值？

Question

我编写了一些代码，将文本文件中的考试成绩结果添加到 python 字典中，格式如下：

{'Ruan': '22', 'hello': [22, 1], 'kurun': '29'}

我想计算出每个人的平均分数，这是我迄今为止尝试过的方法：

while choice == 'av':
    if schClass == '1':
     schClass = open("scores1.txt", 'r')
     li = open("scores1.txt", 'r')
     data = li.read().splitlines()
     for li in data:
        name = li.split(":")[0]
        score = li.split(":")[1]
        if name not in diction1:
            diction1[name] = score
        if name in diction1:
                    diction1[name] = [int(diction1[name]),int(score)]       
        print(diction1)
        averages_dct = {}
        for name in diction1:
            student_average = sum((diction1[name])) / len((diction1[name]))
            averages_dct.update({name: student_average})
        reversed_dct = {averages_dct[k]: [] for k in averages_dct}
        for average in reversed_dct:
            for name in averages_dct:
                if average == averages_dct[name]:
                           reversed_dct[average].append(name)
                           for av in sorted(reversed_dct, reverse=True):
                               print('average: %s, students: %s' % (av, reversed_dct[av]))

这是错误：

    student_average = sum((diction1[name])) / len((diction1[name]))
TypeError: unsupported operand type(s) for +: 'int' and 'str'

我完全理解这意味着什么，但不知道如何解决它？

Answer 1

在数据结构中混用字符串和整数列表是相当不明智的。你应该尝试类似的东西。这将有助于进一步计算：

while choice == 'av':
    if schClass == '1':
     schClass = open("scores1.txt", 'r')
     li = open("scores1.txt", 'r')
     data = li.read().splitlines()
     for li in data:
        name = li.split(":")[0]
        score = li.split(":")[1]

        diction1.setdefault(name,[]).append(int(score))

     # The following loop should work,
     # even if it can be optimized (see Padraic's answer)
     for name in diction1:
        student_average = sum(diction1[name]) / len(diction1[name])
        averages_dct[name] = student_average
     ...

详情设置setdefault文档

因为我没有你的输入数据文件，我无法真正测试它，但这应该会产生类似的东西：

{'Ruan': [22], 'hello': [22, 1], 'kurun': [29]}

在那之后，您的其余代码应该可以正常工作，因为您现在统一拥有整数列表。无论同一玩家的"scores"个数是多少。

Answer 2

不确定所有代码在做什么，但使用 defaultdict 并将所有分数存储在列表中将更容易求和和平均，默认字典将添加名称并在该键不存在时追加或者只是附加每个分数，它比使用 dict.setdefault:

更有效

from collections import defaultdict


diction1 = defaultdict(list)
averages_dct = {}
student_average = {}
while choice == 'av': # 
    if schClass == '1': # not sure what this is supposed to do
        schClass = open("scores1.txt", 'r')
    with open("scores1.txt") as f:
        for li in f: # just iterate over the file object
            name, score = li.split(":") # split once and unpack
            # append score cast as int to the list
            diction1[name].append(int(score))
    # now average scores for each using calling sum on lists of ints
    for name,scores in diction1.items():
        student_average = sum(scores) / len(scores)
        averages_dct[name] = student_average

我假设您的下一个循环是查找具有相同平均分数的名称，因此我们可以再次使用 defaultdict，使用平均值作为键并附加具有相同平均值的名称：

common_dct = defaultdict(list)
# use items to get the key and value
for name, average in averages_dct.items():
    common_dct[averages_dct].append(name)

如果您不想实际使用 common_dict，您可以在之前的循环中对名称进行分组，使用分数作为键并附加名称来反转逻辑。

您还可以让 statistics 模块处理平均值，将代码替换为：

from statistics import mean
for name,scores in diction1.items():
    student_average = mean(scores)
    averages_dct[name] = student_average

我如何找出字典中分配给键的值的平均值？

How do i find out the average of values in a dictionary which are assigned to a key?

python

dictionary