Python 列表list中元素去重的6种方式总结

曾经终败给现在 2023-06-08 03:19 23阅读 0赞

*相关文章：[https://blog.csdn.net/sinat\_26811377/article/details/102568410][https_blog.csdn.net_sinat_26811377_article_details_102568410]*

**输入：**

s = [1, 3, 2, 34, 4, 6, 6, 7, 1, 4, 8, 98]

**输出：**

string_duplicate_1 [1, 3, 2, 34, 4, 6, 7, 8, 98]
    string_duplicate_2 dict_keys([1, 3, 2, 34, 4, 6, 7, 8, 98])
    string_duplicate_3 dict_keys([1, 3, 2, 34, 4, 6, 7, 8, 98])
    string_duplicate_4 [1, 3, 2, 34, 4, 6, 7, 8, 98]
    string_duplicate_5 [1, 2, 3, 4, 6, 7, 8, 34, 98]
    string_duplicate_6 [1, 2, 3, 4, 6, 7, 8, 34, 98]

**Python 实现list元素去重：**

from functools import reduce
    
    
    class StringReverse(object):
        '''
        列表去重，并按照原来的顺序排序
        '''
    
        # 1.利用set方法和sort方法，原序
        def string_duplicate_1(self, s):
            new_s = list(set(s))  # set无序
            new_s.sort(key=s.index)
            return new_s
    
        # 2.用列表中的元素作为字典中的key生成一个新字典，然后获取字典的key，非原序
        def string_duplicate_2(self, s):
            a = {}
            # fromkeys(s,v)该方法的功能是生成一个字典，字典的key是 s中的值，s为可迭代对象，可以为str,tuple,list,set,dict，v为每一个key的值，默认为None
            return a.fromkeys(s).keys()
    
        # 3.利用defaultdict, 非原序
        def string_duplicate_3(self, s):
            # 按照之前的顺序
            from collections import defaultdict
            a = defaultdict()
            for x in s:
                a[x] = 0
            return a.keys()
    
        # 4.最简单的循环，添加入新的列表，如果新列表中没有相同元素，则加入。原序
        def string_duplicate_4(self, s):
            new_s = []
            for x in s:
                if x not in new_s:
                    new_s.append(x)
            return new_s
    
        # 5.利用itertools的groupby方法。非原序
        def string_duplicate_5(self, s):
            from itertools import groupby
            s.sort()
            new_groupby = groupby(s)
            new_s = []
            for x, y in new_groupby:
                new_s.append(x)
            return new_s
    
        # 6.reduce方法。非原序
        def string_duplicate_6(self, s):
            return reduce(lambda x, y: x if y in x else x + [y], [[], ] + s)
    
    
    if __name__ == "__main__":
        stringReverse = StringReverse()
        s = [1, 3, 2, 34, 4, 6, 6, 7, 1, 4, 8, 98]
        print("string_duplicate_1", stringReverse.string_duplicate_1(s))
        print("string_duplicate_2", stringReverse.string_duplicate_2(s))
        print("string_duplicate_3", stringReverse.string_duplicate_3(s))
        print("string_duplicate_4", stringReverse.string_duplicate_4(s))
        print("string_duplicate_5", stringReverse.string_duplicate_5(s))
        print("string_duplicate_6", stringReverse.string_duplicate_6(s))

[https_blog.csdn.net_sinat_26811377_article_details_102568410]: https://blog.csdn.net/sinat_26811377/article/details/102568410