一道很简单的练习题，谢谢谢谢

S = list(eval(input()))
s = sorted(S)
a = s.index(60)
b = s.index(85)
c = s[:a]
d = s[b+1:]
print('%.2f' % (sum(c)/len(c))
print('%.2f' % (sum(d)/len(d))

这样写究竟错哪了/(ㄒoㄒ)/~~
救救孩子吧，谢谢！！

如果有多个85咋办呢

import numpy as np

score=[85,55,93,75,56,47,67,90,85,85,24,88,60]
#score=sorted(score)

flunk=[x for x in score if x<60]
excellent=[x for x in score if x>85]

print("%.2f" % np.mean(flunk))
print("%.2f" % np.mean(excellent))

同意2楼意见，有多个85时，d = s[b+1:]只忽略了第一个85，但将后面的85计入了优秀统计数

1. # coding: utf-8
   
2. # filename: magic_method.py
   
3. 
   
4. li = [85, 93, 55, 75, 56, 47, 67, 90, 24, 88, 60]
   
5. 
   
6. 
   
7. li1 = list(filter(lambda x: x < 60, li))
   
8. li2 = list(filter(lambda x: x > 85, li))
   
9. result1 = sum(li1)/(len(li1)*1.0)
   
10. result2 = sum(li2)/(len(li2)*1.0)
    
11. print(format(result1, '0.2f'))
    
12. print(format(result2, '0.2f'))
    

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我的思想是遍历整个列表，寻找其中符合条件的数再形成一个列表来求值

1. #Python 3.7
   
2. S = [85,55,93,75,56,47,67,90,24,88,60]
   
3. print (S)
   
4. s = sorted(S)
   
5. a = s.index(60)
   
6. c = s[:a]
   
7. s.reverse()
   
8. b = s.index(85)
   
9. d = s[:b]
   
10. print (c)
    
11. print (d)
    
12. print('%.2f' % (sum(c)/len(c)))
    
13. print('%.2f' % (sum(d)/len(d)))

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1. # Author: 傻眼貓咪
   
2. results = [85, 93, 55, 75, 56, 47, 67, 90, 24, 88, 60]
   
3. gradeA = [i for i in results if i > 85]
   
4. fail = [i for i in results if i < 60]
   
5. print("{:.2f}".format(sum(fail)/len(fail)))
   
6. print("{:.2f}".format(sum(gradeA)/len(gradeA)))

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