在提供示例调用之后，您似乎希望检查字符串类型的所有参数，而不是一个参数。正确的？
你可以用 any 对于该值，如果列表/生成器值中的任何一个为真，则该值为真。在生成器理解中，您可以执行以下操作： isinstance 所有（位置）参数

def speak(*languages):
    """Function takes a set of langs as input  and returns a list of languages"""

    if any(not isinstance(lang, str) for lang in languages):
        raise TypeError('You need to pass in the right datatype')

    return list(languages)

>>> speak('english')
>>> speak('english', 'french')
>>> speak(34)
Traceback (most recent call last):
  File "<string>", line 14, in <module>
  File "<string>", line 5, in speak
TypeError: You need to pass in the right datatype

我认为这里不需要尝试。或许可以尝试以下方式（按照您的推理）：

def speak(*languages):

    """Function takes a set of langs as input  and returns a list of languages"""
    for lang in list(languages):
        if not isinstance(lang, str):
            print(' You need to pass in the right datatype')
            return
    else:
        languages = list(languages)
        return languages

print(speak('english')) # english
print(speak('english', 'french')) # english and french
print(speak(34)) # error

也不 type(languages) == str 也没有 isinstance(languages, str) 将抛出一个 TypeError ，那么你的 except 永远不会被击中。你只需要一个 else 在你 if :

def speak(*languages):
    """Function takes a set of langs as input and returns a list of languages"""

    if isinstance(languages, str):
        return list(languages)
    else:
        raise TypeError('You need to pass in the right datatype')

注：我认为你不想要 list(languages) 以上- list("a str") 给予 ['a', ' ', 's', 't', 'r'] . 但我留下它是因为它与问题无关。

您实际上并没有在这里抛出错误，为此，您需要使用 raise :

def speak(*languages):

    """Function takes a set of langs as input  and returns a list of languages"""

    if  isinstance(languages, str):
        languages = list(languages)

    else:
        raise TypeError('You need to pass in the right datatype')

    return languages