Question

这是一个简单的代码，脚本会询问用户的姓名，年龄，性别和身高，并根据该代码提供输出。我目前的代码如下：

print "What is your name?"
name = raw_input()
print "How old are you?"
age = raw_input()
print "Are you male? Please answer Y or N"
sex = raw_input()
if sex == "Y" or sex == "y":
    sex = "Male"
else:
    sex = "Female"
print "How tall are you? Please enter in feet such as 5.5"
height = raw_input()

if sex == "Male" and height >= 6.0:
    t = "You are tall for a male"
elif sex == "Male" and height < 6.0:
    t = "You are below average for a male"

elif sex == "Female" and height >= 5.5:
    t = "You are tall for a female"
else:
    t = "You are below average for a female"


print "Hello %s. You are %s years old and %s feet tall. %s." % (name, age, height, t)

我对if，elif，else语句感到很沮丧：

if sex == "Male" and height >= 6.0:
    t = "You are tall for a male"
elif sex == "Male" and height < 6.0:
    t = "You are below average for a male"

elif sex == "Female" and height >= 5.5:
    t = "You are tall for a female"
else:
    t = "You are below average for a female"

如果性别是男性或女性，代码将区分，但总会返回“你是一个高xxx”。我无法弄清楚如何获得“你低于平均水平”的回报。

Answer 1

那是因为raw_input()返回一个字符串，而不是一个浮点数，并且在Python 2中字符串和浮点数之间的比较总是相同的。

>>> "1.0" > 6.0
True

这样做：

height = float(raw_input())

height = input()也会起作用但不鼓励（因为评估而导致安全问题）

注意：这在Python 3中已得到修复（可能因为它不是非常有用且容易出错）：尝试这样做会导致

TypeError: unorderable types: str() > float()

这是明确的，并且可以让你意识到自己的错误。

注意：如果您尝试比较age同样的问题（age = raw_input()应为age = int(raw_input())）

蟒蛇;根据用户输入打印输出

1 个答案: