Question

是否有办法使这项工作奏效，或者我将不得不找到另一种逻辑方式？我写得很快，以作为我遇到的问题的一个例子。每次运行guess函数时，我想添加一个猜测。问题在于，当我退出该函数然后重新输入时，猜测计数器将重置为0。而且我无法在已定义的函数之外获取变量“ guesses”，这就是我的困惑。

什么是正确的方法？

this.props.likedJobs.map()

我希望能够不重新设置guesss变量而离开函数并重新输入。

Answer 1

一种常见的处理方法是将您的信息封装在一个类的实例中。该实例将包含属性，例如猜测数和您要维护的任何其他状态。这些方法将创建包括操纵这些属性的行为。

您可以在此处创建一个新实例，包括传递初始猜测数和答案的选项：

class Guess:
    def __init__(self, answer = 2, guesses = 5): # 2 and 5 are deafaults if nothing is passed in
        self.guesses = guesses
        self.answer = answer

    def guess(self):
        guess = input("Guess: ")
        if guess == self.answer:
            print("you win")
        else:
            self.guesses -= 1
            print(f"You have {self.guesses} left")
            self.movement()

    def movement(self):
        choice = input("left or guess? ")
        if choice == "left":
            self.movement()
        if choice == "guess":
            self.guess()


g = Guess(answer = 5, guesses = 2) # make a new game using the passed in values 
g.movement() # start game

Answer 2

我喜欢OOP（面向对象编程），所以我会选择另一个答案。话虽这么说，Python对此有一个有用的东西称为generator。将其视为记住状态的函数。

def my_gen():
    x = 0
    while True:
        # Yield acts just like a normal return, the functions stops
        # after returning x.
        yield x
        # Next time the generator is called it will resume
        # right here, add it will remember all the values
        # it previously had (i.e. it remembers the last value for x.
        x += 1
# Note a generator is called differently then a normal function
# then a normal function
g = my_gen()
print(next(g)) # prints 0
print(next(g)) # prints 1
print(next(g)) # prints 2

还介绍了发电机如何停止：

def my_gen2():
    x = 2
    while x > 0:
        yield x
        x -= 1
# Note that when a generator function
# has no more yields it will throw a
# StopIteration Exception
g = my_gen2()
print(next(g)) # prints 2
print(next(g)) # prints 1
print(next(g)) # This will cause an StopIteration Exception

# you can get around this either by:
g = my_gen2()
for x in g: # A for loop automatically stops on StopIteration
    print(x)

# Or catch the Exception
try:
    g = my_gen2()
    for _ in range(5): # Just calling next enough times for it to break
        print(next(g))
except StopIteration:
    print("can't call generator any more")

您的代码：

def guess():
    x = 5
    guesses = 0
    num_tries = 1
    while guesses < num_tries:
        guess = input("Guess: ")
        if guess == x:
            print("You win")
            yield 0
        else:
            guesses += 1
        if guesses == num_tries:
            yield 0 # Game over
        else:
            print("Try again")
            yield 1 # Game can continue


# Don't add unneeded recusion.
# Python has a limited stack. Don't consume it
# for something you should do in a loop.
def play():
    g = guess()
    while True:
        choice = input("left or guess: ")
        print(choice)
        if choice == "left":
            continue
        elif choice == "guess":
            try:
                if not next(g):
                    print("Game over")
                    break
            except StopIteration:
                print("Somehow called to many times")
                break
        else:
            print("Invalid Entry")
play()

Answer 3

def guess(guesses = 0):
    x = 5
    choice = input("left or guess")
    guesses = guesses + 1
    if choice == "left":
        return guess(guesses=guesses)
    elif choice == "guess":
        guessint  = int(input("Guess(int): "))
        if guessint  == x:
            print("You win")
            print('number:', guesses)
            return guesses
        else:
            print("try again")
            return guess(guesses=guesses)
guess()

在函数中减少变量的正确方法是什么？

3 个答案: