本文介绍了Python整数到字母等级问题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试解决这段代码中的错误:
I am attempting to troubleshoot the errors in this piece of code:
import time
while1 = True
def grader (z):
if z >= 0 or z <= 59:
return "F"
elif z >= 60 or z <= 62:
return "D-"
elif z >= 62 or z <= 66:
return "D"
elif z >= 67 or z <= 69:
return "D+"
elif z >= 70 or z <= 62:
return "C-"
elif z >= 73 or z <= 76:
return "C"
elif z >= 77 or z <= 79:
return "C+"
elif z >= 80 or z <= 82:
return "B-"
elif z >= 83 or z <= 86:
return "B"
elif z >= 87 or z <= 89:
return "B+"
elif z >= 90 or z <= 92:
return "A-"
else:
return "A"
while while1:
z = int(input("I will tell you the grade of this number, enter from 1 - 100\n"))
if z < 0 or z > 100:
print "Between 1 and 100 PLEASE!\n"
while1 = True
print grader(z)
print "New number now\n"
time.sleep(100)
while1 = True
这种情况下的论点是整数 z
。 z
的值由用户设置,然后该函数应该摆入并确定哪个字母等级 z
值得,无论如何它总是返回'F。'
The argument in this situation is the integer z
. z
's value is set by the user and then the function should swing in and determine what letter grade z
is worth, no matter what though it always returns 'F.'
这对我来说相当迷惑(我是新手)我可以使用一些帮助。
This is rather befuddling to me (I am a novice) and I could use some assistance.
推荐答案
你的问题是:
if z >= 0 or z <= 59:
使用:
if 0 <= z <= 59:
这可以缓解你使用或
代替和
的问题,而且更多可读。
This alleviates the problem you're having using or
instead of and
and is more readable.
但你应该看一下模块:
>>> def grade(score, breakpoints=[60, 70, 80, 90], grades='FDCBA'):
i = bisect(breakpoints, score)
return grades[i]
>>> [grade(score) for score in [33, 99, 77, 70, 89, 90, 100]]
['F', 'A', 'C', 'C', 'B', 'A', 'A']
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