Python 如何创建另一个包含字符串的字典

我正在试图提取包含字符串的词典

字典在下面

[ { "id": "1", "name": "A", "businessArea": [ "Accounting" ], "Designation": [ "L2" ], "Location":"NY" }, 
{ "id": "2", "name": "B", "businessArea": [ "Engineering" ], "Role": [ "Tester","Developer" ], "Designation": [ "L1" ],"Location":"CA" },
 { "id": "3", "name": "C", "businessArea": [ "Engineering" ], "Role": [ "Developer" ], "Designation": [ "L1" ],"Location":"NY" }]

我正在努力提取

• 业务领域是
  工程领域
  ，并且

• 角色

  是测试人员还是开发人员以及

• 名称

  为

  L1

  和

• 位置

  是

  NY

代码如下

def get_set(d,field):
    return {d[field]} if isinstance(d[field], str) else set(d[field])
    
# we use this to filter
def validate(d):
    if 'Role' in d or `businessArea` in d or `Designation` in d or `Location` in d :
        return get_set(d,'Role').intersection({'Developer','Tester'}) and \
               get_set(d,'businessArea').intersection({'Engineering'}) and \
               get_set(d,'Designation').intersection({'L1'}) and \
               get_set(d,'Location').intersection({'NY'})

result = [d for d in test if validate(d)]

• 我得到了空列表

预期输出为

[{“id”：“3”，“name”：“C”，“businessArea”：[“engineering”]，“Role”：[“Developer”]，“designment”：[“L1”]，“Location”：“NY”}]

添加一个以上的字典进行测试

[{'_index': '1',
  '_type': '_doc',
  '_id': '1',
  '_score': 1.0,
  '_source': {'id': '1',
   'name': 'A',
   'businessArea': [{'id': '25', 'name': 'Accounting'}],
   'Role': ['Developer'],
   'Designation': ['L2'],
   'Location': 'NY'}},
 {'_index': '1',
  '_type': '_doc',
  '_id': '2',
  '_score': 1.0,
  '_source': {'id': '2',
   'name': 'B',
   'businessArea': [{'id': '25', 'name': 'Engineering'}],
   'Role': ['Tester', 'Developer'],
   'Designation': ['L1'],
   'Location': 'NY'}},
 {'_index': '1',
  '_type': '_doc',
  '_id': '3',
  '_score': 1.0,
  '_source': {'id': '3',
   'name': 'C',
   'businessArea': [{'id': '25', 'name': 'Engineering'}],
   'Role': ['Tester', 'Developer'],
   'Designation': ['L1'],
   'Location': 'NY'}}]

[{'id'：'1'，'name'：'Group1'，'BusinessArea'：[{'id'：'14'，'name'：'Accounting'}，{'id'：'3'，'name'：'Accounting'}，'designment'：[{'id'：'16'，'name'：'L1'}，{'id'：'20'，'name'：'L2'，'id'：'L2'，'name'：'Group1'，'BusinessArea'：['id'：'id'：'14'，'Research'}，{'id'：'3'，'name'：'Accounting'}]，'Role'：[{'id'：'5032'，'name'：'Tester'}，{'id'：'5033'，'name'：'Developer'}]，'designment'：[{'id'：'16'，'name'：'L1'，'name'：'L1'，{'id'：'20'，'name'：'L2'}，{'id'：'25'，'name'：'L2'}，}，{'id'：'1'，'name'：'Group1'，'BusinessArea id'：'14'，'Research'}，{'id'：'3'，'name'：'Accounting'}]，'Role'：[{'id'：'5032'，'name'：'Developer'}，{'id'：'5033'，'name'：'Developer'}]，'designment'：[{'id'：'16'，'name'：'L1'}，{'id'：'20'，'name'：'L2'}，{'id'：'25'，'name'：'L2'}]

要测试的第二本词典

[{'_index': '1',
  '_type': '_doc',
  '_id': '1',
  '_score': 1.0,
  '_source': {'id': '1',
   'name': 'A',
   'businessArea': [{'id': '25', 'name': 'Accounting'}],
   'Role': ['Developer'],
   'Designation': ['L2'],
   'Location': 'NY'}},
 {'_index': '1',
  '_type': '_doc',
  '_id': '2',
  '_score': 1.0,
  '_source': {'id': '2',
   'name': 'B',
   'businessArea': [{'id': '25', 'name': 'Engineering'}],
   'Role': ['Tester', 'Developer'],
   'Designation': ['L1'],
   'Location': 'NY'}},
 {'_index': '1',
  '_type': '_doc',
  '_id': '3',
  '_score': 1.0,
  '_source': {'id': '3',
   'name': 'C',
   'businessArea': [{'id': '25', 'name': 'Engineering'}],
   'Role': ['Tester', 'Developer'],
   'Designation': ['L1'],
   'Location': 'NY'}}]

试着这样做：

myList = [ { "id": "1", "name": "A", "businessArea": [ "Accounting" ], "Designation": [ "L2" ], "Location":"NY" }, 
{ "id": "2", "name": "B", "businessArea": [ "Engineerring" ], "Role": [ "Tester","Developer" ], "Designation": [ "L1" ],"Location":"CA" },
 { "id": "3", "name": "C", "businessArea": [ "Engineerring" ], "Role": [ "Developer" ], "Designation": [ "L1" ],"Location":"NY" }]
for x in myList:
  if (("Tester" in x['Role'] or "Developer" in x['Role']) and (x["Designation"] == "L1")) and (x["Location"] == "NY"): 
    print(x)
    
    

---

您的示例中有一个打字错误（

engineering

）：

---

修复原始代码

您可以通过在函数validate中的if语句中替换或和来运行代码

def get_set(d,field):
    return {d[field]} if isinstance(d[field], str) else set(d[field])
    
# we use this to filter
def validate(d):
    # change or to and in if statement
    if 'Role' in d and 'businessArea' in d and 'Designation' in d and 'Location' in d :
        return get_set(d,'Role').intersection({'Developer','Tester'}) and \
               get_set(d,'businessArea').intersection({'Engineering'}) and \
               get_set(d,'Designation').intersection({'L1'}) and \
               get_set(d,'Location').intersection({'NY'})

test = [ { "id": "1", "name": "A", "businessArea": [ "Accounting" ], "Designation": [ "L2" ], "Location":"NY" }, 
{ "id": "2", "name": "B", "businessArea": [ "Engineering" ], "Role": [ "Tester","Developer" ], "Designation": [ "L1" ],"Location":"CA" },
 { "id": "3", "name": "C", "businessArea": [ "Engineering" ], "Role": [ "Developer" ], "Designation": [ "L1" ],"Location":"NY" }]

result = [d for d in test if validate(d)]   

print(result)

输出

[{'id': '3',
  'name': 'C',
  'businessArea': ['Engineering'],
  'Role': ['Developer'],
  'Designation': ['L1'],
  'Location': 'NY'}]

处理嵌套字典的新算法

def contains(item, field, values):
    '''
       Search through nested dictionary starting a key field
       to find if a value is in values
    '''
    if field:
        item = item.get(field)      # Get value of field
    if item is None:
        return None                 # Done if field not found
    if isinstance(item, list):
        return any(contains(v, None, values) for v in item)  # recursively check if any value in list in values
    if isinstance(item, dict):
        return any(contains(v, None, values) for v in item.values())  # recursively check if any value in dictionary in values
    return item in values       # Not list or dict, so check if item in values
    
def validate(d):
    return contains(d, 'Role', {'Developer','Tester'}) and \
           contains(d, 'businessArea', {'Engineering'}) and \
           contains(d, 'Designation', {'L1'}) and \
           contains(d, 'Location', {'NY'})

测试

def contains(item, field, values):
    '''
       Search through nested dictionary starting a key field
       to find if a value is in values
    '''
    if field:
        item = item.get(field)      # Get value of field
    if item is None:
        return None                 # Done if field not found
    if isinstance(item, list):
        return any(contains(v, None, values) for v in item)  # recursively check if any value in list in values
    if isinstance(item, dict):
        return any(contains(v, None, values) for v in item.values())  # recursively check if any value in dictionary in values
    return item in values       # Not list or dict, so check if item in values
    
def validate(d):
    return contains(d, 'Role', {'Developer','Tester'}) and \
           contains(d, 'businessArea', {'Engineering'}) and \
           contains(d, 'Designation', {'L1'}) and \
           contains(d, 'Location', {'NY'})

测试1

产出1

测试2

产出2

测试3

产出3

测试4（有问题的第二本词典）

输出

---

您应该在发布时提供您的值和执行跟踪。预期的操作在哪里发生故障？是否与访问值有关？使用

if

语句，使您的函数默认返回

None

？比较结果是否为空？我的代码有什么问题？我将等待您添加请重复我提出的调试要点。为了避免函数get_set中的键错误，请将函数validate中的所有

或
更改为
和
，您当前的代码将成功运行。@DarrylG，如果我给出
和
将正确抛出错误，我需要检查字典中是否有这些键。您可以看到first dictionary没有
角色
，因此它会抛出key error这就是答案。您的代码没有问题。现在尝试运行它。您可以在第一个dictionary中看到没有
角色
您必须考虑某个键不存在的情况。如果我的业务领域是数组列表，那么如何循环。
[{'id'：'1'，'name'：'A'，'businessArea'：[{'id'：'25'，'name'：'Accounting'}]，'Role'：['Developer']，'Assignment'：['L2']，'Location'：'NY'}，{'id'：'2'，'name'：'B'，'businessArea'：[{'id'：'25'，'name'：'Engineerring'}]，'Role'：['Tester'，'Developer']，'Assignment'：['L1']，'Location'：'NY'}，{'id'：'3'，'name'：'C'，'businessArea'：[{'id'：'25'，'name'：'engineering'}]，'Role'：['Tester'，'Developer']，'designment'：['L1']，'Location'：'NY'}]
@Nons在本例中，businessarea是一个包含一个字典的列表。这是最坏的情况，还是更复杂？与businessarea一样，role也是一个包含大量id的字典列表，我需要检查每个属性中的
name
（如businessarea['name']，'role'['name']）我又增加了一本字典test@Nons--添加了一个可以处理任意嵌套字典的新算法。
[{'id': '3', 'name': 'C', 'businessArea': ['Engineering'], 'Role': ['Developer'], 'Designation': ['L1'], 'Location': 'NY'}]

test = [{'id': '1',   'name': 'A',   'businessArea':[{'id': '25', 'name': 'Accounting'}],   'Role': ['Developer'],   'Designation': ['L2'],   'Location': 'NY'},  {'id': '2',   'name': 'B',   'businessArea':[{'id': '25', 'name': 'Engineering'}],   'Role': ['Tester', 'Developer'],   'Designation': ['L1'],   'Location': 'NY'},  {'id': '3',   'name': 'C',   'businessArea':[{'id': '25', 'name': 'Engineering'}],   'Role': ['Tester', 'Developer'],   'Designation': ['L1'],   'Location': 'NY'}]
result = [d for d in t if validate(d)]  
print(result)

[{'id': '2', 'name': 'B', 'businessArea': [{'id': '25', 'name': 'Engineering'}], 'Role': ['Tester', 'Developer'], 'Designation': ['L1'], 'Location': 'NY'}, {'id': '3', 'name': 'C', 'businessArea': [{'id': '25', 'name': 'Engineering'}], 'Role': ['Tester', 'Developer'], 'Designation': ['L1'], 'Location': 'NY'}]

test = [{ 'id': '1', 'name': 'Group1', 'BusinessArea': [ { 'id': '14', 'name': 'Accounting' }, { 'id': '3', 'name': 'Accounting' } ],'Designation': [ { 'id': '16', 'name': 'L1' }, { 'id': '20', 'name': 'L2' }, { 'id': '25', 'name': 'L2' }, ] }, { 'id': '2', 'name': 'Group1', 'BusinessArea': [ { 'id': '14', 'name': 'Research' }, { 'id': '3', 'name': 'Accounting' } ], 'Role': [ { 'id': '5032', 'name': 'Tester' }, { 'id': '5033', 'name': 'Developer' } ], 'Designation': [ { 'id': '16', 'name': 'L1' }, { 'id': '20', 'name': 'L2' }, { 'id': '25', 'name': 'L2' }, ] }, { 'id': '1', 'name': 'Group1', 'BusinessArea': [ { 'id': '14', 'name': 'Research' }, { 'id': '3', 'name': 'Accounting' } ], 'Role': [ { 'id': '5032', 'name': 'Developer' }, { 'id': '5033', 'name': 'Developer' } ], 'Designation': [ { 'id': '16', 'name': 'L1' }, { 'id': '20', 'name': 'L2' }, { 'id': '25', 'name': 'L2' }] }]
result = [d for d in t if validate(d)] 
print(result)

[{'id': '2', 'name': 'B', 'businessArea': [{'id': '25', 'name': 'Engineering'}], 'Role': ['Tester', 'Developer'], 'Designation': ['L1'], 'Location': 'NY'}, {'id': '3', 'name': 'C', 'businessArea': [{'id': '25', 'name': 'Engineering'}], 'Role': ['Tester', 'Developer'], 'Designation': ['L1'], 'Location': 'NY'}]

test = [{'_index': '1',
  '_type': '_doc',
  '_id': '1',
  '_score': 1.0,
  '_source': {'id': '1',
   'name': 'A',
   'businessArea': [{'id': '25', 'name': 'Accounting'}],
   'Role': ['Developer'],
   'Designation': ['L2'],
   'Location': 'NY'}},
 {'_index': '1',
  '_type': '_doc',
  '_id': '2',
  '_score': 1.0,
  '_source': {'id': '2',
   'name': 'B',
   'businessArea': [{'id': '25', 'name': 'Engineering'}],
   'Role': ['Tester', 'Developer'],
   'Designation': ['L1'],
   'Location': 'NY'}},
 {'_index': '1',
  '_type': '_doc',
  '_id': '3',
  '_score': 1.0,
  '_source': {'id': '3',
   'name': 'C',
   'businessArea': [{'id': '25', 'name': 'Engineering'}],
   'Role': ['Tester', 'Developer'],
   'Designation': ['L1'],
   'Location': 'NY'}}]

result = [d for d in t if validate(d)] 
print(result)

[{'id': '2', 'name': 'B', 'businessArea': [{'id': '25', 'name': 'Engineering'}], 'Role': ['Tester', 'Developer'], 'Designation': ['L1'], 'Location': 'NY'}, {'id': '3', 'name': 'C', 'businessArea': [{'id': '25', 'name': 'Engineering'}], 'Role': ['Tester', 'Developer'], 'Designation': ['L1'], 'Location': 'NY'}]