如何计算平均分计算器中A、B、C、D、F的数量？

Question

我正在尝试编写一个 Python 脚本，该脚本根据用户输入的考试次数和考试分数计算平均分数以及 A、B、C、D 和 F 的数量。

我已经为函数中当前的脚本部分尝试了多种方法： “def letter_score（分数）：” 但未能成功地为输入的考试成绩分配相应的字母。

    sum_of_scores = 0
    number_of_exams = int(input("What is the size of the class? "))
    print("Now enter the scores below.")
    for i in range(1, number_of_exams + 1):
     scores = int(input("Student %d : " %(i)))
     sum_of_scores += scores

    def letter_score(scores):
      if scores >= 90:
        scores = "A"
      elif scores < 90 and scores >= 80:
        scores = "B"
      elif scores < 80 and scores >= 70:
        scores = "C"
      elif scores < 70 and scores >= 60:
        scores = "D"
      else:
        scores = "F"

    average_score = sum_of_scores/number_of_exams

    print("The average is " + str(average_score))

    print("There are " + str(scores) + " A's.")
    print("There are " + str(scores) + " B's.")
    print("There are " + str(scores) + " C's.")
    print("There are " + str(scores) + " D's.")
    print("There are " + str(scores) + " F's.")

我成功地找到了平均分数，但字母分数给我带来了麻烦。

Answer 1

考虑使用pythondictionary来存储letter_score的计数

您可以通过以下方式将字典初始化为全0：

letter_score_count = { "A": 0, "B": 0, "C": 0, "D": 0, "F": 0}

在函数letter_score中使用return关键字给出返回值

    def letter_score(scores):
        if scores >= 90:
            return "A"
        elif scores < 90 and scores >= 80:
            return "B"
        elif scores < 80 and scores >= 70:
            return "C"
        elif scores < 70 and scores >= 60:
            return "D"
        else:
            return"F"

此时，调用letter_score方法，将结果作为字典的key，在输入分数时增加每个字母分数的计数器

letter_score_count[letter_score(scores)] += 1

最后就可以打印结果了

print("There are " + str(letter_score_count["A"]) + " A's.")

您的代码中的编辑版本。希望对您有所帮助

    sum_of_scores = 0
    number_of_exams = int(input("What is the size of the class? "))
    print("Now enter the scores below.")
    letter_score_count = { "A": 0, "B": 0, "C": 0, "D": 0, "F": 0}
    def letter_score(scores):
        if scores >= 90:
            return "A"
        elif scores < 90 and scores >= 80:
            return "B"
        elif scores < 80 and scores >= 70:
            return "C"
        elif scores < 70 and scores >= 60:
            return "D"
        else:
            return"F"

    for i in range(1, number_of_exams + 1):
        scores = int(input("Student %d : " % (i)))
        sum_of_scores += scores
        letter_score_count[letter_score(scores)] += 1

    average_score = sum_of_scores / number_of_exams

    print("The average is " + str(average_score))

    print("There are " + str(letter_score_count["A"]) + " A's.")
    print("There are " + str(letter_score_count["B"]) + " B's.")
    print("There are " + str(letter_score_count["C"]) + " C's.")
    print("There are " + str(letter_score_count["D"]) + " D's.")
    print("There are " + str(letter_score_count["F"]) + " F's.")

Answer 2

如果这有帮助，请告诉我。您需要在字母评分函数中包含计数逻辑。

from collections import defaultdict

def letter_score(scores,counter_store):
      if scores >= 90:
           scores = "A"
           counter_store["A"]+=1
      elif scores < 90 and scores >= 80:
           scores = "B"
           counter_store["B"]+=1
      elif scores < 80 and scores >= 70:
           scores = "C"
           counter_store["C"]+=1
      elif scores < 70 and scores >= 60:
           scores = "D"
           counter_store["D"]+=1
      else:
           scores = "F"
           counter_store["F"]+=1
      return(counter_store)
sum_of_scores = 0
counter_store=defaultdict(int)
number_of_exams = int(input("What is the size of the class? "))
print("Now enter the scores below.")
for i in range(1, number_of_exams + 1):
     scores = int(input("Student %d : " %(i)))
     score_count=letter_score(scores,counter_store)
     sum_of_scores += scores


average_score = sum_of_scores/number_of_exams
print("The average is " + str(average_score))

print("There are " + str(counter_store["A"]) + " A's.")
print("There are " + str(counter_store["B"]) + " B's.")
print("There are " + str(counter_store["C"]) + " C's.")
print("There are " + str(counter_store["D"]) + " D's.")
print("There are " + str(counter_store["F"]) + " F's.")