python - 在 2 个字符串中查找唯一字符

Question

我正在编写一个 python 代码来查找 2 个字符串中的常见字符。以下是代码:

class CharactersInString:
    def __init__(self, value1, value2):
        self.value1 = value1
        self.value2 = value2

    def find_chars_order_n(self):
        new_string = [ ]
        new_string1 = " "
        new_value1 = list(self.value1)
        new_value2 = list(self.value2)
        print( "new_value1: ", new_value1)
        print( "new_value2: ", new_value2)
        for i in new_value1:
            if i in new_value2:
                new_string.append(i)
        print(new_string)
        new_list = set(new_string)
        final_list = list(new_list)
        print(final_list)
        print(''.join(final_list))

if __name__ == "__main__":
     obj = CharactersInString("hello manzer", "helzo killmse")
     print(obj.find_chars_order_n())

期望的输出是 helo mz。上面的代码给我的输出为 elzohm。我想获取唯一字符，而且字符的顺序应按照值 1 进行维护。使用 SET 不提供有序集。我是 python 的新手。你能帮我如何获得所需的输出并保持字符串的顺序吗

Answer 1

您可以添加条件以在附加字母之前查看该字母是否在列表中。

试试下面的代码:

class CharactersInString:
    def __init__(self, value1, value2):
        self.value1 = value1
        self.value2 = value2

    def find_chars_order_n(self):
        new_string = []
        # new_value1 = list(self.value1)
        # new_value2 = list(self.value2)
        # print( "new_value1: ", new_value1)
        # print( "new_value2: ", new_value2)
        for i in self.value1:
            if i in self.value2 and i not in new_string:
                new_string.append(i)
        # print(new_string)
        # new_list = set(new_string)
        final_list = list(new_string)
        # print(final_list)
        return ''.join(final_list)

if __name__ == "__main__":
     obj = CharactersInString("hello manzer", "helzo killmse")
     print(obj.find_chars_order_n())

输出:

helo mz

由于我们是在 value1 上迭代，所以我们确信顺序将根据 value1。此外，我们根本没有通过检查 i not in new_string 来添加重复值，因此它应该比“附加”然后使用集合删除更好。