Question

我知道为了获得列表列表的组合必须使用笛卡尔积，但在我的情况下，我需要将结果作为字典，其中组合子组嵌套在里面，所以我可以将结果用作JSON用于递归角度指令，显示嵌套在N个深度级别的元素。

这是一个数据输入示例：

[
    [{"cluster_id":1,"id":1,"name":"A"}, {"cluster_id":1,"id":2,"name":"B"}, {"cluster_id":1,"id":3,"name":"C"}],
    [{"cluster_id":2,"id":1,"name":"D"}, {"cluster_id":2,"id":2,"name":"E"}, {"cluster_id":2,"id":3,"name":"F"}],
    [{"cluster_id":3,"id":1,"name":"G"}, {"cluster_id":3,"id":2,"name":"H"}, {"cluster_id":3,"id":3,"name":"I"}]
]

这应该是这样的结果：

[{"cluster_id":1,"id":1,"name":"A", "subcluster": [
        {"cluster_id":2,"id":1,"name":"D", "subcluster": [
                {"cluster_id":3,"id":1,"name":"G"},
                {"cluster_id":3,"id":2,"name":"H"},
                {"cluster_id":3,"id":3,"name":"I"},
            ]
        },
        {"cluster_id":2,"id":2,"name":"E", "subcluster": [
                {"cluster_id":3,"id":1,"name":"G"},
                {"cluster_id":3,"id":2,"name":"H"},
                {"cluster_id":3,"id":3,"name":"I"},
            ]
        },
        {"cluster_id":2,"id":3,"name":"F", "subcluster": [
                {"cluster_id":3,"id":1,"name":"G"},
                {"cluster_id":3,"id":2,"name":"H"},
                {"cluster_id":3,"id":3,"name":"I"},
            ]
        }
    ]
},
{"cluster_id":1,"id":2,"name":"B", "subcluster": [
        {"cluster_id":2,"id":1,"name":"D", "subcluster": [
                {"cluster_id":3,"id":1,"name":"G"},
                {"cluster_id":3,"id":2,"name":"H"},
                {"cluster_id":3,"id":3,"name":"I"},
            ]
        },
        {"cluster_id":2,"id":2,"name":"E", "subcluster": [
                {"cluster_id":3,"id":1,"name":"G"},
                {"cluster_id":3,"id":2,"name":"H"},
                {"cluster_id":3,"id":3,"name":"I"},
            ]
        },
        {"cluster_id":2,"id":3,"name":"F", "subcluster": [
                {"cluster_id":3,"id":1,"name":"G"},
                {"cluster_id":3,"id":2,"name":"H"},
                {"cluster_id":3,"id":3,"name":"I"},
            ]
        }
    ]
},
{"cluster_id":1,"id":3,"name":"C", "subcluster": [
        {"cluster_id":2,"id":1,"name":"D", "subcluster": [
                {"cluster_id":3,"id":1,"name":"G"},
                {"cluster_id":3,"id":2,"name":"H"},
                {"cluster_id":3,"id":3,"name":"I"},
            ]
        },
        {"cluster_id":2,"id":2,"name":"E", "subcluster": [
                {"cluster_id":3,"id":1,"name":"G"},
                {"cluster_id":3,"id":2,"name":"H"},
                {"cluster_id":3,"id":3,"name":"I"},
            ]
        },
        {"cluster_id":2,"id":3,"name":"F", "subcluster": [
                {"cluster_id":3,"id":1,"name":"G"},
                {"cluster_id":3,"id":2,"name":"H"},
                {"cluster_id":3,"id":3,"name":"I"},
            ]
        }
    ]
}]

如果可能的话，我想知道如何使用Pyhton生成此输出，但欢迎任何帮助。

Answer 1

我终于能够使用以下递归函数解决它：

def generate_tree(filters, depth):
    if depth == len(filters)-1:
        return filters[depth]
    else:
        return [{'name': i['name'], 'id': i['id'], 'cycle_cluster_id': i['cycle_cluster_id'], 'subcluster': generate_tree(filters, depth+1)} for i in filters[depth]]

嵌套字典中的Python笛卡尔积

1 个答案: