Question

我写了这个代码，取5个等级取平均值并根据平均值显示消息，这里是代码：

counter=0
grd=0
while counter<5:
    a=int(input('Please enter a grade.'))
    grd=grd+a
    counter=counter+1
avg=int(grd/5)
if avg>=90&avg<=100:
    print('Your average is', avg, 'Excellent!')
if avg>=80&avg<=89:
    print('Your average is', avg, 'Very Good!')
if avg>=70&avg<=79:
    print('Your average is', avg, 'Good')
if avg>=60&avg<=69:
    print('Your average is', avg, 'Satisfactory')
if avg<=50:`enter code here`
    print('Your average is', avg, 'Go Home

我遇到的问题是，无论平均值是多少，它们都会显示前4个if语句。

Answer 1

你不能使用＆amp;在python中，除非你正在处理按位操作。否则，您使用and

您正在寻找：

counter=0
grd=0
while counter<5:
    a=int(input('Please enter a grade.'))
    grd=grd+a
    counter=counter+1
avg=int(grd/5)
if avg>=90 and avg<=100:
    print('Your avarage is', avg, 'Excellent!')
if avg>=80 and avg<=89:
    print('Your avarage is', avg, 'Very Good!')
if avg>=70 and avg<=79:
    print('Your avarage is', avg, 'Good')
if avg>=60 and avg<=69:
    print('Your avarage is', avg, 'Satisfactory')
if avg<=50:
    print('Your avarage is', avg, 'Go Home

Answer 2

这是一个范围测试，所以你也可以像这样编码：

import bisect as bisect
counter,grd = 0,0

while counter < 5:
    a = float(input('Please enter a grade.'))
    grd += a
    counter += 1
avg = grd/5

breakPoint = [50,60,70,80,90]
comment = [' Go Home','',' Satisfactory',' Good',' Very Good!',' Excellent!']
index = bisect.bisect(breakPoint,avg)
remark = 'Your avarage is {}' + comment[index]
print (remark.format(avg) )

Answer 3

如之前的回复中所述，请勿使用＆amp;而不是和。也处理非数字值，不要让他们破坏你的程序将是有益的。

counter=0
grd=0
while counter<5:
    a=input('Please enter a grade.')
    try: 
        a = int(a) # check if input is convertible to integer    
        grd=grd+a
        counter=counter+1
    except: #if not, print a message
        print ("enter numeric value ") 
        a = 0 # assign zero
        counter = counter # reset counter increment
avg=int(grd/5)
if avg>=90 and avg<=100:
    print('Your avarage is', avg, 'Excellent!')
elif avg>=80 and avg<=89:
    print('Your avarage is', avg, 'Very Good!')
elif avg>=70 and avg<=79:
    print('Your avarage is', avg, 'Good')
elif avg>=60 and avg<=69:
    print('Your avarage is', avg, 'Satisfactory')
elif avg<=50:
    print('Your avarage is', avg, 'Go Home')
else:
    print ("some other message for further development")

Answer 4

m1=int(input("Enter 1st Sub marks"))
m2=int(input("Enter 2nd Sub marks"))
m3=int(input("Enter 3rd Sub marks"))
m4=int(input("Enter 4th Sub marks"))
m5=int(input("Enter 5th Sub marks"))

avg=(m1+m2+m3+m4+m5)/5
print (int(avg))

if (avg>=90 and avg<=100):
  print ("Grade A")
elif (avg>=80 and avg<90):
  print ("Grade B")
elif (avg>=70 and avg<80):
  print ("Grade C")  
elif (avg>=60 and avg<70):
  print ("Grade D")  
elif (avg>=50 and avg<60):
  print ("Grade E")  
elif (avg>100):
  print ("Entered Marks are out of range")

else:
  print ("FAIL")

我写了这个代码，取5个等级取平均值，并根据平均值显示一条消息，它正在起作用

4 个答案: