Python3如何按嵌套列表中子列表的首元素去重并保留首个出现项

实现思路 #

遍历嵌套列表时用集合记录已经出现过的首元素值，仅将首元素未出现过的子列表加入结果集，保证仅保留第一个出现的同首元素子列表，时间复杂度为O(n)，同时保留原列表的元素顺序。

完整实现代码 #

def deduplicate_by_first_element(input_list):
    seen = set()
    output = []
    for sub_lst in input_list:
        # 兼容空子列表的边界场景，不需要可删除该判断
        if not sub_lst:
            continue
        first_key = sub_lst[0]
        if first_key not in seen:
            seen.add(first_key)
            output.append(sub_lst)
    return output

效果验证 #

用给出的三个输入测试，结果完全符合预期：

# 测试输入
input1 = [
    ['nom', 'N', 'eye'],
    ['nom', 'N', 'face'],
    ['acc', 'E', 'computer'],
    ['dat', 'C', 'screen']
]
input2 = [
    ['nom', 'O', 'heart'],
    ['acc', 'O', 'root'],
    ['acc', 'C', 'life'],
    ['dat', 'E', 'evidence']
]
input3 = [
    ['nom', 'O', 'author'],
    ['acc', 'O', 'tear'],
]

# 输出结果
print(deduplicate_by_first_element(input1))
# [['nom', 'N', 'eye'], ['acc', 'E', 'computer'], ['dat', 'C', 'screen']]
print(deduplicate_by_first_element(input2))
# [['nom', 'O', 'heart'], ['acc', 'O', 'root'], ['dat', 'E', 'evidence']]
print(deduplicate_by_first_element(input3))
# [['nom', 'O', 'author'], ['acc', 'O', 'tear']]

内容的提问来源于stack exchange，提问作者Carlos Luis Rivera