Python等级分布函数，计算和打印等级出现的次数

我正在使用以下代码创建成绩分布函数：

def distribution(grades):
available_grades = [ 'A+','A','A-','B+','B','B-','C+','C','C-','F']     
fin = open(grades,'r')
gradesList = fin.readline().split(' ')
for c_grade in available_grades:
    if c_grade in gradesList:
        print('students got '+c_grade)

我正在使用一个grades.txt文件，其中包含所有的成绩，我需要一种方法来计算该文本文件中出现成绩的次数，并打印该数字，如下所示：

 distribution('grades.txt')
6 students got A
2 students got A-
3 students got B+
2 students got B
2 students got B-
4 students got C
1 student got C-
2 students got F

但目前它只打印以下内容：

students got A
students got A-
students got B+
students got B
students got B-
students got C
students got C-
students got F

My grades.txt文件包括：

A A- C F C C B- B A A A- B B+ B+ B+ C C- B- A A A F

使用：

---

您可以为此使用python计数器类

from collections import counter

# creates a new Counter object
c = Counter()

# your file contents
lst_grades = "A A- C F C C B- B A A A- B B+ B+ B+ C C- B- A A A F"

# turn lst_grades to a list containing the grades
lst_grades = lst_grades.split(" ")

# printing lst_grades for a better idea what lst_grades looks like now
# lst_grades
# > ['A', 'A-', 'C', 'F', 'C', 'C', 'B-', 'B', 'A', 'A', 'A-', 'B', 'B+',
#    'B+', 'B+', 'C', 'C-', 'B-', 'A', 'A', 'A', 'F']

# call the update method of the counter c
# update takes an iterable (e.g. a list)
# and counts the values inside this iterable
c.update(lst_grades)

# now our counter contains a dictionary with the counted grads
# c
# > Counter({'A': 6, 'C': 4, 'B+': 3, 'B': 2, 'F': 2,
#            'B-': 2, 'A-': 2, 'C-': 1})

# you can simply access every counted entry
# c["A"]
# > 6

---

如果您不想使用集合，则以下操作应起作用：

def distribution(grades):
    available_grades = [ 'A+','A','A-','B+','B','B-','C+','C','C-','F']
    grade_dict = {}
    for grade in available_grades:
        grade_dict[grade] = 0

    fin = open(grades,'r')
    gradesList = fin.readline().split()
    fin.close()

    for grade in gradesList:
        grade_dict[grade] += 1

    for grade in available_grades:
        print(str(grade_dict[grade]) + ' students got ' + grade)

请注意，我删除了要拆分的参数，因此它将删除所有空白，而不仅仅是空格。

或者，解决方案：

fin = open(grades,'r')
grades_given = fin.readline().split()
fin.close()

pd.Series(grades_given).value_counts().reset_index(name='cnt').\
apply(lambda x: "{cnt} students got {grade}".format(cnt=x['cnt'],
      grade=x['index']),axis=1)

0     6 students got A
1     4 students got C
2    3 students got B+
3    2 students got A-
4    2 students got B-
5     2 students got F
6     2 students got B
7    1 students got C-

---

你需要提供你的成绩档案的样本。很抱歉，我会在我的问题中添加这个。你为什么不想使用这些集合？这是更快，更简单，更少的代码。太好了，我喜欢你的答案，因为我更愿意使用这个方法而不是集合方法，非常感谢。你为什么不想使用集合呢？这是更快、更简单和更少的代码。啊，对不起，我想在问题下发布，而不是你的答案^^^没关系。

fin = open(grades,'r')
grades_given = fin.readline().split()
fin.close()

pd.Series(grades_given).value_counts().reset_index(name='cnt').\
apply(lambda x: "{cnt} students got {grade}".format(cnt=x['cnt'],
      grade=x['index']),axis=1)

0     6 students got A
1     4 students got C
2    3 students got B+
3    2 students got A-
4    2 students got B-
5     2 students got F
6     2 students got B
7    1 students got C-