我会用字典来存储计数。但是首先我要剥离所有spaces其他符号，然后再剥离az，我还想将大写和小写字母视为一个相同。

当使用我所有的值构建dict时，我将使用该max函数。的max需要迭代，因此我们将dict作为元组的“列表”传递(key, val)。我们需要告诉我们max如何确定要比较的内容，为此，我们提供了一个lambda函数，该函数将val元组中的第二个元素（）带到key- arg。

作为回报，max将吐出最高的元组val。

class MyString:

    def __init__(self, myString):
        self.__myString = myString

    def countWord(self):
        count = len(self.__myString.split())
        return count

    def findMostFrequentChar(self):
        counter = {}
        # Instead of performing various modifications on the string
        # one can instead filter all the undesired chars.
        # new_string = self.__myString.replace(' ', '').lower()
        new_string = list(filter(lambda x: 'a' >= x <= 'z', self.__myString.lower()))
        for char in new_string:

            if char in counter:
                counter[char] += 1
            else:
                counter[char] = 1

        key, value = max(counter.items(), key=lambda x:x[1])
        return value, key


def main():
    aString = MyString("This is a super long long long string. Please help count me")
    print("There are", aString.countWord(), "words in the string.")

    count, letter = aString.findMostFrequentChar()
    print("The most frequent character is", letter, "which appeared", count, "times")

main()