你写的代码并没有 * 错误 *，但是它在Python中不是很符合习惯，因此你不得不与语言斗争来添加“parse negative”功能。考虑一下你可以写这样的代码：

user_guess = input("Guess a number: ")
if is_positive_or_negative_number(user_guess):
    user_guess = int(user_guess)
# continue as before

def is_positive_or_negative_number(s: str) -> bool:
    """Checks if a given string represents a positive or negative number"""
    if s.startswith('-'):
        s = s[1:]  # strip off the optional leading unary negation
    return s.isdigit()  # do not allow decimals, so no need to worry
                        # about allowing a "."

然而，如果你只写惯用的Python，这会更容易！你的代码是用一种被亲切地称为LBYL（三思而后行）的风格编写的，代码检查是为了确保一件事在做之前可以完成。Python更喜欢EAFP（比请求许可更容易请求原谅），它让你尝试做一件事，并在抛出错误时捕捉错误。
然后，惯用代码只尝试将输入转换为int，如果失败，则会引起注意。

while True:
    user_guess = input("Guess a number: ")
    try:
        user_guess = int(user_guess)
    except ValueError:
        print("It is not a number.")
        break
    # if we get here, user_guess is guaranteed to be an int
    # and int(user_guess) knows how to parse positive and
    # negative numbers
    if user_guess < 0:
        print("Too low, guess a number between 0 and 10.")
    elif user_guess > 10:
        print("Too high, guess a number between 0 and 10.")

对于负数，它返回“It is not a number”的原因是因为user_guess.isdigit()将负数视为字符串（或非数字）。
下面是一段代码，可以按您的预期工作：

while True:
    user_guess = input("Guess a number: ")
    try:
        user_guess = int(user_guess)
        if user_guess < 0:
            print("Too low, guess a number between 0 and 10.")
        if user_guess > 10:
            print("Too high, guess a number between 0 and 10.")
    except ValueError:
        print("It is not a number.")
        break

由于int()函数可以识别负数，因此使用try-except可以帮助您捕获ValueError异常，该异常在您尝试对非整数使用int()函数时引发。

isdigit（）有问题。如果是减号，isdigit（）将返回False。一个解决方法是让int（）验证user_guess。

while True:
  try:
    user_guess = int( input( "Guess a number: "))
  except ValueError:
    print( "It is not a number.")
    break # exit loop
  # validate user entry
  if user_guess < 0:
    print("Too low...")
    continue
  elif user_guess > 10:
    print("Too high...")
    continue
  # do processing
  ...

def input_number(message):
    while True:
        user_guess = input(message)
        try:
            n = int(user_guess)
            if n < 0:
                print("Too low, guess a number between 0 and 10.")
            elif n > 10:
                print("Too high, guess a number between 0 and 10.")
            else:
                return n
        except ValueError:
            print("It is not a number. Try again")
            continue

if __name__ == '__main__':
    number = input_number("Guess a number.")
    print("Your number", number)

编辑：我将解释我的代码以及为什么它能解决你的问题。你使用的isdigit方法只会检查字符串中的字符是否由数字组成。减号不是数字，所以它返回False。
相反，我尝试将字符串转换为数字，如果python失败，我就再次循环（继续）并要求输入一个新的数字。如果输入确实是一个数字，代码的下半部分会检查有效的区间。只有当数字在区间内时，控制循环的变量才会被设置，循环退出。
我的代码不依赖于isdigit，因此避免了您的问题。希望这有帮助，并提供洞察力。

user_guess = None
while user_guess is None:
    inp = input("Guess a number: ")

    try:
        nr_inp = int(inp)
    except ValueError:
        print("It is not a number.")
        continue

    if nr_inp < 0:
        print("Too low, guess a number between 0 and 10.")
    elif nr_inp > 10:
        print("Too high, guess a number between 0 and 10.")
    else:
        user_guess = nr_inp

print("Done:", user_guess)