简单Python "if“、"elif”、“Python”

Question

在Python的课堂上制作一个“疯狂的Libs”风格的东西，这是我的第一天，我已经学到了大部分我需要知道的东西，但我不知道如何使用"if“、"elif”、“thing”这些东西。这是我到目前为止，基本上，当年龄是输入，我希望它选择的人是成年人还是孩子。

print "welcome to your short story"

name = raw_input("Input your name: ")
age = raw_input("Input your age: ")

if age > 21:
    age = "adult"
elif age < 21:
    age = "kid"


print "My name is ",name,"and I am a " ,age,"year old ",age

Answer 1

( 1)你正在用孩子的字符串覆盖输入的年龄。2)当年龄等于21岁时，你必须处理这件事。3)年龄输入需要转换为整数。

让我们重写您的代码，看看我们如何改进：

print "welcome to your short story"

name = raw_input("Input your name: ")
# Convert the input to an integer.
age = int(raw_input("Input your age: "))

# This is the status variable being either adult or child
# before you were overriding age variable with adult or kid
status = ""

# Also, you have to handle the case where the age equals 21, before 
# you were just checking if it is less or greater than 21
if age >= 21:
    status = "adult"
elif age < 21:
    status = "kid"

print "My name is ", name ," and I am a " , age ," year old " , status

Answer 2

正如from所建议的那样，要将输入从字符串转换为int，就需要int函数。

age = int(raw_input("Input your age: "))

字符" 1“、" 2”、" 3“等与数字1、2、3等有区别。有些语言试图为您转换，但python不会这样做，因此您最终会将"32“与”21“进行比较，这是一种苹果与橘子的比较。

此外，虽然您在这个特定的实例中是安全的，但是您将注意到如何在仍然基于原始值进行计算的情况下重新分配。这通常不安全，最好将其赋值给一个新变量：

if age > 21:
    age_label = "adult"
elif age < 21:
    age_label = "kid"
else:
    age_label = "person" # in case a 21 year old uses your program

在这种情况下，您是安全的，因为if/ get结构将只执行一个分支，但最好不要养成在仍在查阅输入值时重写输入值的习惯。

Answer 3

我意识到我需要一个不同的价值，并改变了它。

print "welcome to your short story"

name = raw_input("What is your name?: ")
age = raw_input("How old are you?: ")
sex = raw_input("Are you a boy or a girl?: ")

if age > 21:
    targetAge = "adult"
elif age < 21:
    targetAge = "kid"


print "My name is ",name,"and I am a " ,age,"year old ",targetAge,"."

所以，基本上，当它被打印出来时，它应该是“我的名字是_____，我是一个__的孩子/成年人。”取决于他们输入的数字。我之所以不使用int()函数，是因为从来没有提到过，这实际上是上课的第一天，所以我将按照讲师所做的进行讨论。