Question

我在尝试找出正在编写的程序的代码时遇到麻烦。我得到的文件格式为：

student_firstname
student_lastname
student_number
assignment_mark
midterm_mark
exam_grade
final_mark

以下是示例：

Marilyn
Malone    
136238
88
72
70
Esther
Mulcahy
194563
25
45
91
William
Gray
110031
33
38
62

我正在尝试寻找一种方法来从所有学生中退还最高和最低的final_mark。到目前为止，我所做的是：

infile = open(fileName, "r")
myList = []
name = infile.readline().strip()

passingGrades = 0
failingGrades = 0
avGrade = 0    

while name != '':
    highestGrade = [0, 'studentName']
    lowestGrade = [99, 'studentName']
    lastName = infile.readline().strip()
    studentNum = infile.readline().strip()
    assignGrade = infile.readline().strip()
    midGrade = infile.readline().strip()
    examGrade = infile.readline().strip()
    averageGrade = ((int(assignGrade) * 0.25) + (int(midGrade) * 0.25) +
                    (int(examGrade) * 0.50)) 

    def lowGrade(x):
        if x < lowestGrade[0]:
            lowestGrade.pop(0)
            lowestGrade.pop(0)
            lowestGrade.append(x)
            lowestGrade.append(name)
            lowestGrade.append(lastName)
        return lowestGrade   

    if averageGrade >= 50 and int(examGrade) >= 50:
        #print(name)
        passingGrades += 1
        avGrade += averageGrade

        if averageGrade > highestGrade[0]:
            highestGrade.pop(0)
            highestGrade.pop(0)
            highestGrade.append(averageGrade)
            highestGrade.append(name)
            highestGrade.append(lastName)
        else:
            pass

    else:
        failingGrades += 1
        avGrade += averageGrade

    lowGrade(averageGrade)

    name = infile.readline().strip()

finalAverage = avGrade / (passingGrades + failingGrades)

highFinal = ' '.join(str(x) for x in highestGrade)
lowFinal = ' '.join(str(x) for x in lowestGrade)

使用其中一个文本文件运行程序时，我得到：

Number of passes: 9
Number of fails: 1
Average final grade: 64.55
The Highest Grade: 79.5 Patty Marshall
The Lowest Grade: 79.5 Patty Marshall

非常感谢您的帮助！

Answer 1

每次找到最低成绩时，您将从列表中删除2个值并加3。这意味着从不删除第一个值。您能检查一下是否适合您吗？

def lowGrade(x):
    if x < lowestGrade[0]:
        lowestGrade.pop(0)
        lowestGrade.pop(0)
        lowestGrade.append(x)
        lowestGrade.append(' '.join(name, lastName))
    return lowestGrade

Answer 2

每次阅读新学生时，您都将{ event: 'subscribe', user: <USER_ID>, subscribe: { transactionId: <CUSTOMER>, transactionAffiliation: <AFFILIATION>, transactionTotal: <PRICE>, transactionProducts: [ { id: <CUSTOMER>, sku: <SUBSCRIPTION>, name: <PLAN>, price: <PRICEPER>, quantity: <QUANTITY> } ] } };和highestGrade重置为其初始值（分别为lowestGrade和0）。

通过初始化循环中外部的值来轻松解决此问题。

while name != '':
     # stuff that happens for each 'name'
    highestGrade = [0, 'studentName']
    lowestGrade = [99, 'studentName']

    # ... do stuff ...

    # read new name
    name = infile.readline().strip()

此外，您的highestGrade = [0, 'studentName'] lowestGrade = [99, 'studentName'] while name != '': # stuff that happens for each 'name'方法不必要地复杂且多余。无需从列表中删除项目然后添加新项目，只需创建一个新列表即可。而且实际上，tuple比list更合适，尽管两者都可以。

lowGrade

您完全不需要使用返回的值，因此无需通过此方法返回lowestGrade = (99, 'studentName') def lowGrade(x): if x < lowestGrade[0]: lowestGrade = (x, name, lastName) / tuple。

如何将文本文件中的最高值附加到列表中？

2 个答案: