我在試圖找出我正在編寫的程序的代碼時遇到了麻煩。我得到了一個格式如下的文件：student_firstnamestudent_lastnamestudent_numberassignment_markmidterm_markexam_gradefinal_mark這是一個示例：MarilynMalone 136238887270EstherMulcahy194563254591WilliamGray110031333862我試圖找到一種方法來返回所有學生的最高和最低 final_mark。到目前為止我所做的是：infile = open(fileName, "r")myList = []name = infile.readline().strip()passingGrades = 0failingGrades = 0avGrade = 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)當程序使用其中一個文本文件運行時，我得到：Number of passes: 9Number of fails: 1Average final grade: 64.55The Highest Grade: 79.5 Patty MarshallThe Lowest Grade: 79.5 Patty Marshall任何幫助深表感謝！
2 回答
慕妹3242003
TA貢獻1824條經驗 獲得超6個贊
每次您閱讀一個新學生時，您都將highestGrade和'重置'lowestGrade為其初始值（分別為0和99）。
while name != '':
# stuff that happens for each 'name'
highestGrade = [0, 'studentName']
lowestGrade = [99, 'studentName']
# ... do stuff ...
# read new name
name = infile.readline().strip()
通過在循環之外初始化這些值來輕松解決此問題。
highestGrade = [0, 'studentName']
lowestGrade = [99, 'studentName']
while name != '':
# stuff that happens for each 'name'
此外，您的lowGrade方法不必要地復雜和冗余。不要從列表中刪除項目然后添加新項目，只需創建一個新列表。事實上， atuple比 a 更合適list，盡管兩者都可以。
lowestGrade = (99, 'studentName')
def lowGrade(x):
if x < lowestGrade[0]:
lowestGrade = (x, name, lastName)
您不需要從此方法返回tuple/ list，因為您根本不使用返回的值。
互換的青春
TA貢獻1797條經驗 獲得超6個贊
每次找到最低等級時，都會從列表中刪除 2 個值并添加 3 個。這意味著永遠不會刪除第一個值。你能檢查一下這對你有用嗎？
def lowGrade(x):
if x < lowestGrade[0]:
lowestGrade.pop(0)
lowestGrade.pop(0)
lowestGrade.append(x)
lowestGrade.append(' '.join(name, lastName))
return lowestGrade
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