我正在尝试解决这段代码中的错误:
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
这种情况下的参数是整数z
。 z
的值由用户设置,然后函数应该进入并确定哪个字母等级z
是值得的,无论它总是返回'F'。
这对我来说相当迷惑(我是新手),我可以使用一些帮助。
答案 0 :(得分:8)
你的问题是:
if z >= 0 or z <= 59:
使用:
if 0 <= z <= 59:
这可以缓解您使用or
而不是and
时遇到的问题,并且更具可读性。
但你应该看一下bisect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']
答案 1 :(得分:2)
or
中的grader
应为and
。您输入的任何大于0的内容都将传递第一个条件,因此将是F。
答案 2 :(得分:1)
基于@Jon Clements'(伟大的)答案,但我认为更容易理解:
def grade(score, breakpoints, grades):
for k, v in zip(breakpoints, grades):
if score > k:
return v
return 'Error'
grade(score, breakpoints=(90, 80, 70, 60, 0), grades=('A','B','C','D','E','F'))
答案 3 :(得分:0)
if x>0 or x < 59
将覆盖0以上的所有内容
你和os混淆了
if x>0 and x < 59
的范围仅为0到59
但可以更简洁
0 < x < 59
答案 4 :(得分:0)
这一点早已解决,但仍有几点:
while1 = True
。完整的解决方案与@Jon Clement's great answer相结合,将是
import time
from bisect import bisect
def grader(score,
breakpoints=[60, 62, 67, 70, 73, 77, 80, 83, 87, 90, 93],
grades=['F', 'D-', 'D', 'D+', 'C-', 'C', 'C+', 'B-', 'B', 'B+', 'A-',
'A']):
i = bisect(breakpoints, score)
return grades[i]
while True:
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 0 and 100 PLEASE!\n"
continue
print grader(z)
time.sleep(100)
print "New number now\n"