搜索字典键，嵌套字典中匹配的值，包装在列表中

Question

考虑这个列表：

d1  = [{'data': {'id': '1', 'label': 'ID #1', 'expanded': True}, 'classes': 'genesis'}, {'data': {'id': '2', 'label': 'ID #2', 'expanded': True}, 'classes': 'followerNode'}, {'data': {'id': '3', 'label': 'ID #3'}, 'classes': 'followerNode'}, {'data': {'id': '4', 'label': 'ID #4'}, 'classes': 'followerNode'}, {'data': {'id': '5', 'label': 'ID #5'}, 'classes': 'followerNode'}, {'data': {'id': '6', 'label': 'ID #6'}, 'classes': 'followerNode'}, {'data': {'id': '7', 'label': 'ID #7'}, 'classes': 'followerNode'}, {'data': {'id': '21', 'source': '2', 'target': '1'}, 'classes': 'followerEdge'}, {'data': {'id': '31', 'source': '3', 'target': '1'}, 'classes': 'followerEdge'}, {'data': {'id': '41', 'source': '4', 'target': '1'}, 'classes': 'followerEdge'}, {'data': {'id': '51', 'source': '5', 'target': '1'}, 'classes': 'followerEdge'}, {'data': {'id': '61', 'source': '6', 'target': '1'}, 'classes': 'followerEdge'}, {'data': {'id': '71', 'source': '7', 'target': '1'}, 'classes': 'followerEdge'}, {'data': {'id': '8', 'label': 'ID #8'}, 'classes': 'followerNode'}, {'data': {'id': '13', 'label': 'ID #13'}, 'classes': 'followerNode'}, {'data': {'id': '14', 'label': 'ID #14'}, 'classes': 'followerNode'}, {'data': {'id': '15', 'label': 'ID #15'}, 'classes': 'followerNode'}, {'data': {'id': '18', 'label': 'ID #18'}, 'classes': 'followerNode'}, {'data': {'id': '19', 'label': 'ID #19'}, 'classes': 'followerNode'}, {'data': {'id': '20', 'label': 'ID #20'}, 'classes': 'followerNode'}, {'data': {'id': '82', 'source': '8', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '132', 'source': '13', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '142', 'source': '14', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '152', 'source': '15', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '182', 'source': '18', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '192', 'source': '19', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '202', 'source': '20', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '8', 'label': 'ID #8'}, 'classes': 'followerNode'}, {'data': {'id': '13', 'label': 'ID #13'}, 'classes': 'followerNode'}, {'data': {'id': '14', 'label': 'ID #14'}, 'classes': 'followerNode'}, {'data': {'id': '15', 'label': 'ID #15'}, 'classes': 'followerNode'}, {'data': {'id': '18', 'label': 'ID #18'}, 'classes': 'followerNode'}, {'data': {'id': '19', 'label': 'ID #19'}, 'classes': 'followerNode'}, {'data': {'id': '20', 'label': 'ID #20'}, 'classes': 'followerNode'}, {'data': {'id': '82', 'source': '8', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '132', 'source': '13', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '142', 'source': '14', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '152', 'source': '15', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '182', 'source': '18', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '192', 'source': '19', 'target': '2'}, 'classes': 'followerEdge'}, {'data': {'id': '202', 'source': '20', 'target': '2'}, 'classes': 'followerEdge'}]

当存在其他键类型（如“源”或“目标”）时，如何搜索say“1”：

我想删除与我的任何搜索条件匹配的子目录。。。例如，val为1的实例id。。。它应该删除以下内容：

{'data': {'id': '1', 'label': 'ID #1', 'expanded': True}, 'classes': 'genesis'}

以下是我尝试过的：

[{k:v} for lst in d1 for k, v in lst.items() if (("1" not in (v.get('source', None), (v!=None))) and 
            ("1" not in (v.get('target', None), (v!=None))) and
            ("1" not in (v.get('id', None), (v!=None))))]

但是我越来越

AttributeError: 'str' object has no attribute 'get'

Answer 1

在您的代码中，“for k，v In lst.items（）”将k，v作为“数据”、“类”给出，因此它给出了属性错误

要获取具有所需Id的词典，请执行以下操作：

ans = []
ID = '1'
for k in d1:
    if k['data']['id'] == ID:
        ans.append(k)

Answer 2

列表中的每个项都有两个键值对的字典，第一个有一个字典作为值，第二个有一个字符串。当您在列表中的每个项目上循环for k, v in lst.items()时，它将在第一个键值对上工作，但对于第二个键值对，它将看到一个字符串作为值，而不是导致问题的字典

我认为这应该起作用：

[lst for lst in d1 if "1" not in lst["data"].get('source', "None")
    and "1" not in lst["data"].get('target', "None")
    and "1" not in lst["data"].get('id', "None")]

或

for lst in d1:
    if "1" not in lst["data"].get('source', "None")\
            and "1" not in lst["data"].get('target', "None")\
            and "1" not in lst["data"].get('id', "None"):
        lst