Question

在domain.xml示例中，我有以下嵌套词典：

json

请注意，{ "DICT": { "List of dict": [ { #first dict inside the outer list "K1": "V1", "K2": "V2", "K3": "V3", "K4": [ { "K4_1_1": "V4_1" }, { "K4_2_1": "V4_2" }, { "K4_3_1": null } ], "K5 is a list of Dict": [ { "K5_1_1": "V5_1", "K5_1_2": "V5_2", "K5_1_3": "V5_3", "K5_1_4": "V5_4" }, { "K5_2_1": "V5_1", "K5_2_2": "V5_2", "K5_2_3": "V5_3", "K5_2_4": "V5_4" } ] }, { #second dict in the outerlist "K1": "V1", "K2": "V2", "K3": "V3", "K4": [ { "K4_1_1": "V4_1_1" }, { "K4_2_1": "V4_2_1" } ], "K5": { "K5_1_1": "V_1_1", "K5_1_2": "V_1_2", "K5_1_3": null, "K5_1_4": null } } ] } }和K4始终是K5中的list。无论它们在字典或列表中有多深，我都需要摆脱所有null。因此，我编写了以下python函数，但是输出是相同的，并且所有dict值仍然存在：

None

我里面的字典不是def RemoveNones(Dict): for k, v in Dict.items(): if type(v) == collections.OrderedDict: RemoveNones(v) elif type(v) == list: for i in v: RemoveNones(i) else: Dict = dict((K,V) for K,V in Dict.items() if V!=None)，而是dict。

Answer 1

type()的语法如下：if type(v) is list:（不是==）

所以你想要这样的东西：

import json
from collections import OrderedDict

raw_text = '{"DICT":{"List of dict":[{"K1":"V1","K2":"V2","K3":"V3","K4":[{"K4_1_1":"V4_1"},{"K4_2_1":"V4_2"},{"K4_3_1":null}],"K5 is a list of Dict":[{"K5_1_1":"V5_1","K5_1_2":"V5_2","K5_1_3":"V5_3","K5_1_4":"V5_4"},{"K5_2_1":"V5_1","K5_2_2":"V5_2","K5_2_3":"V5_3","K5_2_4":"V5_4"}]},{"K1":"V1","K2":"V2","K3":"V3","K4":[{"K4_1_1":"V4_1_1"},{"K4_2_1":"V4_2_1"}],"K5":{"K5_1_1":"V_1_1","K5_1_2":"V_1_2","K5_1_3":null,"K5_1_4":null}}]}}'

raw_json = json.JSONDecoder(object_pairs_hook=OrderedDict).decode(raw_text)

def remove_nulls(x):
    if type(x) is list:
        return [remove_nulls(v) for v in x if v is not None]
    elif type(x) is OrderedDict:
        return OrderedDict((k,remove_nulls(v)) for k,v in x.items() if v is not None)
    else:
        return x

de_nullified_json = remove_nulls(raw_json)
print(json.dumps(de_nullified_json, indent=2))

输出：

{
  "DICT": {
    "List of dict": [
      {
        "K1": "V1",
        "K2": "V2",
        "K3": "V3",
        "K4": [
          {
            "K4_1_1": "V4_1"
          },
          {
            "K4_2_1": "V4_2"
          },
          {}
        ],
        "K5 is a list of Dict": [
          {
            "K5_1_1": "V5_1",
            "K5_1_2": "V5_2",
            "K5_1_3": "V5_3",
            "K5_1_4": "V5_4"
          },
          {
            "K5_2_1": "V5_1",
            "K5_2_2": "V5_2",
            "K5_2_3": "V5_3",
            "K5_2_4": "V5_4"
          }
        ]
      },
      {
        "K1": "V1",
        "K2": "V2",
        "K3": "V3",
        "K4": [
          {
            "K4_1_1": "V4_1_1"
          },
          {
            "K4_2_1": "V4_2_1"
          }
        ],
        "K5": {
          "K5_1_1": "V_1_1",
          "K5_1_2": "V_1_2"
        }
      }
    ]
  }
}

从嵌套字典中删除所有“无”值

1 个答案: