如何合并具有相同镜头的二维列表

def merge_lists(lists):
    merged = {}  # 或 merged = []

    for sublist in lists:
        lens = len(sublist)
        if lens > 0:
            key = sublist[0]  # 假设镜头在子列表的第一个元素
            if key in merged:
                merged[key].extend(sublist[1:])  # 合并到已存在的镜头对应的值中
            else:
                merged[key] = sublist[1:]  # 添加到结果字典或列表中

    return merged

# 测试示例
lists = [['A', '1', '2'], ['B', '3', '4'], ['A', '5', '6'], ['C', '7', '8']]
result = merge_lists(lists)
print(result)

{'A': ['1', '2', '5', '6'], 'B': ['3', '4'], 'C': ['7', '8']}