如何让变量检测它是大于还是小于另一个? [英] How do I make a variable detect if it is greater than or less than another one?

问题描述

我目前正在学习 Python，我决定构建一个小型的猜数字"类型的游戏.我正在使用随机功能，并试图使其能够检测用户的输入是否等于随机数.这是我的代码:

I am currently learning Python, and I decided to build a small "Guess the Number" type of game. I am using the random feature, and trying to make it so it will detect if the user's input is equal the the random number. This is my code:

irand = randrange(1, 10)
while True:
    number = input ("Pick a number 1-10: ")
    if number < irand:
        print ("    ")
        print ("That's too high, try again.")
    if number > irand:
        print ("    ")
        print ("That's too low, try again.")
    if number == irand:
        print ("    ")
        print ("You got it right! You won!")
        break

我不确定如何让它检测 number 是否等于、小于或大于 irand.有什么帮助吗?谢谢！:)

I am not exactly sure how to make it detect if number is equal to, less than or greater than irand. Any help? Thank you! :)

推荐答案

您检查等于、小于或大于"的代码是正确的，但是您更早遇到了问题在代码中.

Your code to check "equal to, less than, or greater than" is correct, but you've got a problem earlier in the code.

在 Python(3.0 及更高版本)中，input 返回一个字符串.因此，您将字符串 number 与数字 irand 进行比较.某些语言可能会自动转换一种或另一种方式，但这可能会令人困惑，*，因此 Python 拒绝这样做，而是引发 TypeError: unorderable types: int() <;str().

In Python (3.0 and later), input returns a string. So, you're comparing number, a string, to irand, a number. Some languages might convert one way or the other automatically, but that can be confusing,*, so Python refuses to do so, and instead raises a TypeError: unorderable types: int() < str().

解决办法是把输入解析成一个数字，像这样:

The solution is to parse the input into a number, like this:

number = int(number)

如果用户给你无效输入，这将引发一个 ValueError，比如字母 c 而不是 1 到 10 的数字.如果你想处理那个更好的是，您使用 try 语句.

This will raise a ValueError if the user gives you invalid input, like the letter c instead of a number from 1 to 10. If you want to deal with that more nicely, you use a try statement.

作为旁注，您可能需要 elif 而不是 if.如果你做对了一切，这没有任何区别，但如果你犯了错误，它有助于发现错误.

As a side note, you probably want elif instead of if. If you've got everything right, this doesn't make any difference, but if you've made a mistake, it helps catch the mistake.

所以，把它们放在一起:

So, putting it all together:

while True:
    number = input("Pick a number 1-10: ")
    try:
        number = int(number)
    except ValueError:
        print(number, 'is not a number, try again.')
        continue
    if number < irand:
        print("    ")
        print("That's too high, try again.")
    elif number > irand:
        print("    ")
        print("That's too low, try again.")
    else:
        print("    ")
        print("You got it right! You won!")
        break

(注意我在 except 子句中使用了 continue，所以我们跳过循环的其余部分，不必担心 number 不再是一个数字.我们也可以将整个循环移动到 try 上的 else 原因中，或者添加一个 isinstance 检查在每个 if 等上，但这有点笨拙.)

(Notice that I used continue in the except clause, so we skip over the rest of the loop and don't have to worry about number not being a number anymore. We could also move the whole loop into an else cause on the try, or add an isinstance check on each if, etc., but that gets a bit clumsy.)

_{* 考虑将字符串 "2" 与数字 10 进行比较.将字符串转换为数字的语言会说 2 <10;将数字转换为字符串的语言会说 "10" <2".没有一个是对"或错"，但两者都有一半的时间令人惊讶......}

_{* Consider comparing the string "2" to the number 10. A language that converts the string to a number will say that 2 < 10; a language that converts the number to a string will say that "10" < "2". Neither one is "right" or "wrong", but both are surprising half the time…}

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