Python;基於用戶輸入的打印輸出

Question

這是一個簡單的代碼，腳本會詢問用戶的姓名，年齡，性別和身高，並根據該代碼給出輸出。我當前的代碼如下：

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) 

我收到掛了，如果，elif的，else語句：

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" 

如果性別是男是女的代碼將分化，但總會迴歸「你爲xxx高個子」。我無法弄清楚如何得到「你低於平均水平」的回報。

Answer 1

這是因爲raw_input()返回一個字符串，而不是浮動，並且比較始終是一個字符串，在Python 2

>>> "1.0" > 6.0 
True 

做這一個浮子之間的相同的方式：

height = float(raw_input()) 

height = input()本來可以工作，但不鼓勵（因評估而導致的安全問題）

注意：這已在Python 3中修復（可能是因爲這是不是非常有用，而且容易出錯）：試圖做到這一點在

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

這是明確的，將允許實現自己的錯誤造成的。

注：同樣的問題，如果你嘗試比較age（age = raw_input()應該age = int(raw_input())）