如何根據級別編號制作嵌套的 python 字典

Question

如何根據級別數構建嵌套的 python 字典，像這樣？

遞歸 function 如何通過 int 級別？

{
"aggs": {
      "cat_level_1": {
            "terms": {
                 "field": "cat_level_1"
            },
              "aggs": {
                  "cat_level_2": {
                      "terms": {
                            "field": "cat_level_2"
                           },
                            "aggs": {
                                 "cat_level_3": {
                                     "terms": {
                                          "field": "cat_level_3"
                                    }
                              }
                       }
                   }
             }
       }
  }
}

Answer 1

我們可以從最深層次構建字典，在每個層次上包裝之前構建的字典。

def dict_with_depth_of(n):
    return {
        "aggs": {
            f"cat_level_{n}": {
                "terms": {
                    "field": f"cat_level_{n}"
                }
            }
        }
    }

def nested_dict(n):
    d = None
    for i in reversed(range(1, n + 1)):
        new_d = dict_with_depth_of(i)
        if d is not None:
            new_d["aggs"][f"cat_level_{i}"] = d
        d = new_d
    return d

print(nested_dict(3))

Output：

{'aggs': {'cat_level_1': {'aggs': {'cat_level_2': {'aggs': {'cat_level_3': {'terms': {'field': 'cat_level_3'}}}}}}}}

Answer 2

這是正確答案

def dict_with_depth_of(n):
    return {
        "aggs": {
            f"cat_level_{n}": {
                "terms": {
                    "field": f"cat_level_{n}"
                }
            }
        }
    }


def nested_dict(n):
    d = None
    for i in reversed(range(1, n + 1)):
        new_d = dict_with_depth_of(i)
        if d is not None:
            new_d["aggs"][f"cat_level_{i}"].update(d)
        d = new_d