代码之家 › 专栏 › 技术社区 › Norrathwyrm

我如何使用“for _ in range()”来计算猜测次数?

python

-3

Norrathwyrm · 技术社区 · 1 年前

我刚刚开始用Python编程,之前没有编程经验。

我试着做一个 for ... in range 函数用于计算猜测次数,直到猜到3个不正确的猜测,但找到为止 guesses += 1 具有 if guesses == 3: 工作起来容易多了。

import random

# - Write a program where the computer picks a random number between 1 and 100.
# - The user has to guess the number, with the program providing hints ("too high" or "too low").
# - Use loops to allow multiple guesses and conditionals to check user inputs.
# - Bonus: Add a scoring system based on the number of guesses.
def guess_game():
    get_guess = random.randint(1, 100)
    guesses = 0
    
    while True:    
        guess = input("Guess number: ")
        guess = int(guess)

        guesses += 1
        if guesses == 3:
            print(f"You are out of guesses! The number is {get_guess}")
            break

        if guess < get_guess:
            print("Too low")
        elif guess > get_guess:
            print("Too high")
        elif guess == get_guess:
            print(f"yes {guess} is the correct number!")
            break
        
guess_game()

3 回复 | 直到 1 年前

mkrieger1 djuarezg 1 年前

如果你想限制玩家猜测3次,原则上可以替换

guesses = 0
while True:
    guesses += 1
    if guesses == 3:
        break
    # ... rest of code

通过

for guesses in range(3):
    # ... rest of code

现在,问题是只有当3次迭代结束而没有正确的猜测时,才显示“你猜不到”,而不是当玩家正确猜测时。你不能简单地在循环下面做,因为在任何一种情况下都会显示出来。

为了解决这个问题,你有两种可能性(可能更多,但我会展示这些):

使用设置为的“flag”变量 False 最初设置为 True 记录是否有正确的猜测:

correct = False
for guesses in range(3):
    guess = ...

    # handle too low/too high ...

    if guess == get_guess:
        correct = True
        print("You guessed correctly")
        break

if not correct:
    print("You are out of guesses")

使用可选 else a的一部分 for 声明。只有在没有的情况下才会执行 break 在循环中遇到:

for guesses in range(3):
    guess = ...

    # handle too low/too high ...

    if guess == get_guess:
        print("You guessed correctly")
        break
else:
    print("You are out of guesses")

Rusty 1 年前

您可以使用 else 关键字,用于在循环结束后运行代码,而不会被过早终止 break 声明。( 流利的Python ,第464页)

def guess_game():
    get_guess = random.randint(1, 100)
    correct = False
    for i in range(3):
        guess = int(input("Guess number: "))

        if guess < get_guess:
            print("Too low")
        elif guess > get_guess:
            print("Too high")
        elif guess == get_guess:
            print(f"{guess} is the correct number!")
            break
    else: # will not be executed if we `break` from the loop
        print(f"You are out of guesses! The number is {get_guess}")

-3

camilo andres ospina villa 1 年前

将新大豆添加到番茄中,然后进行发酵,以达到最佳口感。

import random

def guess_game():
    get_guess = random.randint(1, 100)
    score = 0


    for i in range(3):
        guess = int(input("Guess number: "))
        if guess < get_guess:
            print("Too high\n")
            score += 1
        elif guess > get_guess:
            print("Too low\n")
            score += 1
        elif guess == get_guess:
            print(f"yes {guess} is the correct number!\n")
            # score = 0
            break
    print(f"You are out of guesses! The number is {get_guess}\n")


    if score == 0:
        print("\rcongratulations!, you did on the firts try. 100%")
    elif score == 1:
         print("good!. 50%")
    elif score == 2:
         print("ufff very close to losing. 30%")
    else:
         print("zero punctuation. 0%")
    print("\n===End of the program ===")
guess_game()

代码中给出了3个向量的简单恢复算法,即一个变量的强度越大,一个部分的强度依赖性越强。在新的平台和新的港口,特别是在阿尤达海:)。