Question

我正在尝试提取包含字符串的字典

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" }]

我正在尝试提取

businessArea是Engineering，
Role，即Tester or Developer和
Designation是L1和
Location是NY

代码在下面

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)]

我得到空列表

预期为[{ "id": "3", "name": "C", "businessArea": [ "Engineerring" ], "Role": [ "Developer" ], "Designation": [ "L1" ],"Location":"NY" }]

再添加一本字典进行测试

[{ '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' }] }]

第二本要测试的字典

[{'_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'}}]

Answer 1

尝试这样：

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)

Answer 2

修正原始代码

您可以通过在函数validate中的if语句中替换或并使其与之一起运行，从而使代码运行

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'}]

处理嵌套词典的新算法

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

测试1

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" }]

输出1

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

测试2

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)

输出2

[{'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'}]

测试3

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)

输出3

[{'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'}]

测试4 （有问题的第二本词典）

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)

输出

[{'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'}]

Answer 3

您的示例有错字（engineerring）

[ { "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" }]

如何创建另一个包含字符串的字典

3 个答案: