Question

我有一份上学的任务，其中一项任务是显示学生将获得的成绩。成绩是：

答：90％+

B：80％ - 89％

C：70％ - 79％

D：60％ - 69％

E：50％ - 59％

这是一些文件，它是逗号分隔的csv文件：

StudentName分数

Harrison 64

杰克68

杰克61

Hayley 86

我想知道/得到一些指导，以便我更好地了解如何创建成绩计算器。我花了很多年时间试图解决这个问题，但却没有希望。我的代码：

def determine_grade(scores):
    if scores >= 90 and <= 100:
        return 'A'
    elif scores >= 80 and <= 89:
        return 'B'
    elif scores >= 70 and <= 79:
        return 'C'
    elif scores >= 60 and <= 69:
        return 'D'
    elif scores >= 50 and <= 59:
        return 'E'
    else:
        return 'F'

Answer 1

import bisect 

def determine_grade(scores, breakpoints=[50, 60, 70, 80, 90], grades='FEDCBA'):
    i = bisect.bisect(breakpoints, scores)
    return grades[i]

Answer 2

你可以使用 pandas 和 numpy 来做到这一点。类似这样：

import pandas as pd
import numpy as np

#Create a DataFrame

d = {                                                # Creating a dict for dataframe
    'StudentName':['Harrison','Jake','Jake','Hayley'],    
    'Score':[64,68,61,86]}
 
df = pd.DataFrame(d)   # converting dict to dataframe
                       # Keys get converted to column names and values to column values

#get grade by adding a column to the dataframe and apply np.where(), similar to a nested if


df['Grade'] = np.where((df.Score < 60 ), 
                  'F', np.where((df.Score >= 60) & (df.Score <= 69), 
                  'D', np.where((df.Score >= 70) & (df.Score <= 79), 
                  'C', np.where((df.Score >= 80) & (df.Score <= 89), 
                  'B', np.where((df.Score >= 90) & (df.Score <= 100), 
                  'A', 'No Marks')))))
print(df)

result:

    StudentName Score   Grade
0   Harrison    64      D
1   Jake        68      D
2   Jake        61      D
3   Hayley      86      B

Answer 3

试试这个：

def determine_grade(scores):
    if scores >= 90 and scores <= 100:
        return 'A'
    elif scores >= 80 and scores <= 89:
        return 'B'
    elif scores >= 70 and scores <= 79:
        return 'C'
    elif scores >= 60 and scores <= 69:
        return 'D'
    elif scores >= 50 and scores <= 59:
        return 'E'
    else:
        return 'F'

如果你必须将得分与数值进行比较，这就是if scores >= 90 and <= 100:不正确的原因，但经过简短的编辑就可以了。

Answer 4

对于scores >= 90 and <= 100，您可以撰写90 <= scores <= 100

我不知道，如果得分是浮点数或整数。如果得分是浮点数，那么你的比较是不够的。

if scores >= 90 and <= 100:
        return 'A'
elif scores >= 80 and <= 89:
        return 'B'

如果得分为89.99会怎样？

这是我的解决方案。有一个GRADES_PATTERN。如果有什么变化，你不能改变你的功能。

GRADES_PATTERN = {'A':[90, float('inf')], 'B':[80, 90], 'C':[70, 80], 'D':[60, 70], 'E':[50, 60], 'F':[0, 50]}

def check_grade(score, pattern):
   for grade, score_range in pattern.iteritems():
      if score_range[0] <= score < score_range[1]:
          return grade
   raise Exception("score is out of pattern range")

print check_grade(89.99, GRADES_PATTERN)

students = {'Harrison':64, 'Jake': 68, 'Hayley':86}

for name, score in students.iteritems():
   print("Student {} hat score {} and grade {}".format(name, score, check_grade(score, GRADES_PATTERN)))

Answer 5

另一种选择......

如果你不想要一个整数，你可以为浮点数改变它。

grade = int(input("What was your score?"))

if grade >=90 and grade <=100:
    print("A*")
elif grade >=80 and grade <=89:
    print("A")
elif grade >=70 and grade <=79:
    print("B")
else:
    print("Unknown grade")

Answer 6

您可以这样做，注意if比较的顺序。

def determine_grade(scores):
    if scores >= 90:
        return 'A'
    elif scores >= 80:
        return 'B'
    elif scores >= 70:
        return 'C'
    elif scores >= 60:
        return 'D'
    elif scores >= 50:
        return 'E'
    else:
        return 'F'

Answer 7

# Score   Grade
# if > 1.0 error Bad score
# >= 0.9     A
# >= 0.8     B
# >= 0.7     C
# >= 0.6     D
# < 0.6      F

# ss (string score)
ss = input ("Enter your score: ")
# fs (float score)
try:
    fs = float (ss)
except:
    print("Error Bad score")
    quit()

#print(fs)
if fs >= 1.0:
    print("Error Bad Score")

if fs >= 0.9 :
    print ("A")
elif fs >= 0.8 :
    print ("B")
elif fs >= 0.7 :
    print ("C")
elif fs >= 0.6 :
    print ("D")
else:
    print ("F")

我试图解决这个很棒的网站所引用的问题：https://www.py4e.com/html3/03-conditional

Answer 8

def determine_grade(scores):
    if scores >= 0 and <= 39:
        return 'U'
    elif scores >= 40 and <= 49:
        return 'D'
    elif scores >= 50 and <= 59:
        return 'C'
    elif scores >= 60 and <= 69:
        return 'B'
    elif scores >= 70 and <= 79:
        return 'A'
    else:
        return 'No Marks'

如何在python中创建一个评分系统？

8 个答案: