如何计算调用此函数的次数？

Question

因此，我的任务（我必须使用while语句）是制作数字猜谜游戏，其中一部分是显示玩家获得数字后的猜测数量。我发现我读过的东西应该可以工作但不会。这是我的代码。

#A text program that is a simple number guessing game.
import time
import random
#Setting up the A.I.
Number = random.randint(1,101)
def AI():
    B = AI.counter =+ 1
    Guess = int(input("Can you guess what number I'm Thinking of?: "))
    while Guess > Number:
        print("nope, to high.")
        return AI()
    while Guess < Number:
        print("Sorry, thats to low. try again!")
        return AI()
    while Guess == Number:
        print("Congragulations! you win! You guessed " + str(B) + " times")
        time.sleep(60)
        quit()
AI.counter = 0
AI()

虽然当玩家获得正确的数字时，它表示即使不是这样，玩家也会猜到这一点。

Answer 1

使用默认参数：

def AI(B=1):
    Guess = int(input("Can you guess what number I'm Thinking of?: "))
    while Guess > Number:
        print("nope, to high.")
        return AI(B + 1)
    while Guess < Number:
        print("Sorry, thats to low. try again!")
        return AI(B + 1)
    while Guess == Number:
        print("Congragulations! you win! You guessed " + str(B) + " times")
        time.sleep(60)
        return

然后调用函数：

AI()

另外，你应该在这里使用ifs而不是whiles，因为循环只运行一次，但是whiles也可以工作，所以没关系。此外，递归可能会占用你的RAM，这只会浪费资源，只要你可以实现与循环相同的东西，但你的方法是有效的，所以这也很好。

Answer 2

你非常接近@Simpson！这是一个稍微改变的版本，可以给你你想要的:)

如果您有任何疑问，请告诉我！

#A text program that is a simple number guessing game.
import random
#Setting up the A.I.
def AI():
    counter = 1
    number = random.randint(1,101)
    guess = int(input("Can you guess what number I'm Thinking of?: "))
    while guess != number:
        if guess < number:
            print("Sorry, thats to low. try again!")
        else:
            print("nope, too high")
        counter += 1
        guess = int(input("Can you guess what number I'm Thinking of?: "))

    print("Congragulations! you win! You guessed " + str(counter) + " times")
    time.sleep(60)
    quit()

AI()

Answer 3

没有递归 - 将whiles更改为ifs并在方法内添加了counter。

import time
import random

Number = random.randint(1,101)

def AI():
    B = 1
    Guess = int(input("Can you guess what number I'm Thinking of?: "))
    while True:
        if Guess > Number:
            print("nope, to high.")
        elif Guess < Number:
            print("Sorry, thats to low. try again!")

        if Guess == Number:
            print("Congragulations! you win! You guessed " + str(B) + " times")
            time.sleep(2)
            break # leave the while true

        # increment number
        B += 1   
        Guess = int(input("Can you guess what number I'm Thinking of?: "))
AI()

Answer 4

计算函数调用是function decorators的一个完美案例，

def call_counted(funct):
    def wrapped(*args, **kwargs):
        wrapped.count += 1  # increase on each call
        return funct(*args, **kwargs)
    wrapped.count = 0  # keep the counter on the function itself
    return wrapped

是Python的一个非常有用的功能。您可以将装饰器定义为：

import time
import random

secret_number = random.randint(1, 101)

@call_counted  # decorate your AI function with the aforementioned call_counted
def AI():
    current_guess = int(input("Can you guess what number I'm thinking of?: "))
    while current_guess > secret_number:
        print("Nope, too high. Try again!")
        return AI()
    while current_guess < secret_number:
        print("Sorry, that's too low. Try again!")
        return AI()
    while current_guess == secret_number:
        print("Congratulations! You win! You guessed {} times.".format(AI.count))
        time.sleep(60)
        quit()

AI()

然后，您可以使用它来装饰您希望计算调用的函数，而无需在流程流中处理计数器本身：

import time
import random

secret_number = random.randint(1, 101)

def AI():
    counter = 0
    while True:
        counter += 1
        current_guess = int(input("Can you guess what number I'm thinking of?: "))
        if current_guess > secret_number:
            print("Nope, too high. Try again!")
        elif current_guess < secret_number:
            print("Sorry, that's too low. Try again!")
        else:
            break
    print("Congratulations! You win! You guessed {} times.".format(counter))
    time.sleep(60)
    quit()

AI()

我重新调整了你的代码，但它基本上是一样的。

我会避免递归，因为这可以写得更简单，而且不需要计算函数调用：

PEP 8 - Style Guide for Python Code

Answer 5

您可以使用函数装饰器进行基因操作，该函数装饰器将调用计数器添加到应用它的任何函数：

（注意我也修改了你的代码，所以它更接近

""" A text program that is a simple number guessing game. """
import functools
import time
import random

def count_calls(f):
    """ Function decorator that adds a call count attribute to it and counts
        the number of times it's called.
    """
    f.call_count = 0
    @functools.wraps(f)
    def decorated(*args, **kwargs):
        decorated.call_count += 1
        return f(*args, **kwargs)

    return decorated

# Setting up the A.I.
number = random.randint(1, 101)
# print('The number is:', number)  # For testing.

@count_calls
def AI():
    guess = int(input("Can you guess what number I'm thinking of?: "))
    while guess > number:
        print("Nope, too high.")
        return AI()
    while guess < number:
        print("Sorry, that's too low. Try again!")
        return AI()
    while guess == number:
        print("Congragulations! You win! You guessed " + str(AI.call_count) + " times")
        time.sleep(10)
        quit()

AI()

。）

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