Python3：如何在需要整数的while循环中检查字符串

Question

在 Python3 中：如何在需要整数的 while 循环中检查字符串...

这是我目前的代码：我想用“quit”或“q”替换标记值“0”。我该怎么做？

import random

highest = 10
lowest = 1
counter = 0

random_number = random.randint(lowest, highest)
# The randint function is in the random function within the random MODULE imported above.

guess = int(input(f"Guess a number between {lowest} and {highest}. Enter '0' at any time to quit: "))

while guess is not random_number:
    if guess == 0:
        print("Quitting game.")
        break
    elif guess < random_number:
        counter += 1
        guess = int(input("Guess higher: "))
    elif guess > random_number:
        counter += 1
        guess = int(input("Guess lower: "))
    if guess == random_number:
        print(f"Correct! Well done!! You got it right after {counter} guess(es)!")

我想把sentinel值“0”改成“q/quit”，但不知道怎么...

Answer 1

guess = int(in.... 不要将其转换为 int，而是转换为字符串。

然后看看字符串是“q”还是“quit”

如果不是 q 或退出，则将其转换为 int ： elif int(guess)

Answer 2

在将其转换为整数之前，您必须检查输入是否为q / quit。使用int(input(..))，您可以将输入转换为整数。如果用户输入一个字符串，这将失败。

...
while guess is not str(random_number):
    if guess == "q" or guess == "quit":
        print("Quitting game.")
        break
    guess = int(guess)
    # this might fail if the user did not enter a number
    if guess < random_number:
        counter += 1
        guess = input("Guess higher: ")
    elif guess > random_number:
        counter += 1
        guess = input("Guess lower: ")
    if guess == random_number:
        print(f"Correct! Well done!! You got it right after {counter} guess(es)!")

Answer 3

你可以做一件事：- 一种方法是您可以在循环内输入并检查输入，当用户的猜测与实际数字匹配时使用中断：-

while 1:
 guess = input()
 if guess == 'q' or guess == 'quit':
  print("Quitting game")
  break
 elif int(guess) < random_numer:

然后继续你的程序的其余部分。

Answer 4

不妨试试这个：

import random

highest = 10
lowest = 1
counter = 0

random_number = random.randint(lowest, highest)

guess = input(f"Guess a number between {lowest} and {highest}. Enter 'quit' at any time to quit: ")

while guess is not random_number:
    if guess == "quit":
        print("Quitting game.")
        break
    guess = int(guess)

    if guess < random_number:
        counter += 1
        guess = int(input("Guess higher: "))
    elif guess > random_number:
        counter += 1
        guess = int(input("Guess lower: "))
    if guess == random_number:
        print(f"Correct! Well done!! You got it right after {counter} guess(es)!")

首先，将输入存储为字符串，然后检查它是否等于'quit'。如果不是这样，将字符串转换为整数并正常进行。

我试过了，效果很好，但也许有更优雅的方法。

Answer 5

谢谢大家。这是完全有效的代码副本，也是我正在寻找的。关键是（就像你们都提到的）

1. 让输入最初是一个字符串
2. 立即检查“q”或“退出”。
3. 然后将其余代码的猜测转换为 int
4. 让 if/elif 语句中的后续输入为字符串。

guess = input(f"Guess a number between {lowest} and {highest}. Enter 'quit' at any time to quit: ")

while guess is not random_number:
    if guess.casefold() == "quit" or guess.casefold() == 'q':
        print("Quitting game.")
        break
    guess = int(guess)
    if guess < random_number:
        counter += 1
        guess = input("Guess higher: ")
    elif guess > random_number:
        counter += 1
        guess = input("Guess lower: ")
    if guess == random_number:
        print(f"Correct! Well done!! You got it right after {counter} guess(es)!")

再次感谢！

下面是带有“无效输入”检查器的相同代码，用于继续循环：

guess = input(f"Guess a number between {lowest} and {highest}. Enter 'quit' at any time to quit: ")

while guess is not random_number:
    if guess.casefold() == "quit" or guess.casefold() == 'q':
        print("Quitting game.")
        break

    if guess.isnumeric():
        guess = int(guess)
        if guess < random_number:
            counter += 1
            guess = input("Guess higher: ")
        elif guess > random_number:
            counter += 1
            guess = input("Guess lower: ")
        if guess == random_number:
            print(f"Correct! Well done!! You got it right after {counter} guess(es)!")
    else:
        print("Invalid input. Try again...")
        guess = input(f"Guess a number between {lowest} and {highest}. Enter 'quit' at any time to quit: ")