我怎么能算每个年级的每个类别中的金额为一组标记

Posted 2023-02-03

我在给定一组商标

midtermMarks = [48.0, 64.25, 71.80, 82.0, 53.45, 59.75, 62.80, 26.5,
55.0, 67.5, 70.25, 52.45, 66.25, 94.0, 65.5, 34.5, 69.25, 52.0]

我必须找到多少个A的。 B的，和C的有在该组中，假定A是100-80 B是79-70，C是69-60

答案

首先，你需要创建具有密钥'60字典-70' ，'70 -80' 和'80 -100' 为0的每个值并使用循环获取每个标记和使用if语句检查每一个条件，如果条件满足再加入1到相应的键。

我做给你。希望它的工作原理。

[码]

import pprint     # This module prints dictionary in a pretty way

marks = [48.0, 64.25, 71.80, 82.0, 53.45, 59.75, 62.80, 26.5,
     55.0, 67.5, 70.25, 52.45, 66.25, 94.0, 65.5, 34.5, 69.25, 52.0]

counting = 'Number of A\'s': 0, 'Number of B\'s': 0, 'Number of C\'s': 0  # Dictionary to count the number in given range

for mark in marks:  # Getting each marks from marks list
    if mark >= 80:  # Checking if mark is more than 80
        counting['Number of A\'s'] += 1  # Adding 1 if there is repetition to dictionary

    elif mark >= 70 and mark < 80:  # Checking if mark is more than and equals to 70 but less than 80
        counting['Number of B\'s'] += 1  # Adding 1 if there is repetition to dictionary

    elif mark >= 60 and mark < 70:  # Checking if mark is more than and equals to 60 but less than 70
        counting['Number of C\'s'] += 1  # Adding 1 if there is repetition to dictionary

pprint.pprint(counting)    # Printing the output

[OUTPUT]

"Number of A's": 2, "Number of B's": 2, "Number of C's": 6