PYTHON-将dict列表分组
在Python3中,有没有一种简单的方法可以按键对dict列表进行分组 我有一个复杂的输入列表,我想格式化 我的输入对象如下所示:PYTHON-将dict列表分组,python,python-3.x,list,group-by,Python,Python 3.x,List,Group By,在Python3中,有没有一种简单的方法可以按键对dict列表进行分组 我有一个复杂的输入列表,我想格式化 我的输入对象如下所示: my_input = [ { 'name': 'nameA', 'departments': [ { 'name': 'dep1', 'details': [ { 'name': 'name_detA', 'tech_name': 't
my_input = [
{
'name': 'nameA',
'departments': [
{
'name': 'dep1',
'details': [
{
'name': 'name_detA',
'tech_name': 'techNameA',
'others': None,
'sub_details': []
},
{
'name': 'name_detB',
'tech_name': 'techNameB',
'others': 22,
'sub_details': [
{
'id': 'idB',
'column2': 'ZZ',
'column3': 'CCC',
'column4': {
'id': 'id2',
'subColumn1': 'HHH',
'subColumn1': 'PPPP',
'subColumn1': 'FFFFFF'
}
}
]
},
{
'name': 'name_detB',
'tech_name': 'techNameB',
'others': 22,
'sub_details': [
{
'id': 'idA',
'column2': 'AA',
'column3': 'BBB',
'column4': {
'id': 'id1',
'subColumn1': 'XXXX',
'subColumn1': 'YYYYY',
'subColumn1': 'DDDDDD'
}
}
]
}
]
}
]
}
]
我的目标是将具有相同详细信息['techName']
的元素分组到一个元素中,并合并它们的子单元详细信息
预期产出:
my_output = [
{
"name": "nameA",
"departments": [
{
"name": "dep1",
"details": [
{
"name": "name_detA",
"tech_name": "techNameA",
"others": None,
"sub_details": []
},
{
"name": "name_detB",
"tech_name": "techNameB",
"others": 22,
"sub_details": [
{
"id": "idB",
"column2": "ZZ",
"column3": "CCC",
"column4": {
"id": "id2",
"subColumn1": "HHH",
"subColumn1": "PPPP",
"subColumn1": "FFFFFF"
}
},
{
"id": "idA",
"column2": "AA",
"column3": "BBB",
"column4": {
"id": "id1",
"subColumn1": "XXXX",
"subColumn1": "YYYYY",
"subColumn1": "DDDDDD"
}
}
]
}
]
}
]
}
]
我试过:
result_list = []
sub = []
for elem in my_input:
for data in elem["departments"]:
for sub_detail, dicts_for_that_sub in itertools.groupby(data["details"], key=operator.itemgetter("sub_details")):
sub.append({"sub_details": sub_detail})
print(sub)
但是我正在努力创建新的输出假设我在这里使用的输入是您真正想要的,那么您就在正确的轨道上了。我将最内层的for循环重新实现为对方法的调用,但这并不是严格需要的 对于使用
setdefault()
的merge\u details()
方法,我可能会采取稍微不同的方法,而不是使用if
/else
的方法,但如果您以前没有使用过setdefault()
,那么这种方法更容易遵循
导入json的import
只是为了让打印效果“好”,而不是解决方案的一部分
import json
my_input = [
{
"name": "nameA",
"departments": [
{
"name": "dep1",
"details": [
{
"name": "name_detB",
"tech_name": "techNameB",
"others": 22,
"sub_details": [
{
"id": "idB",
"column2": "ZZ",
"column3": "CCC",
"column4": {
"id": "id2",
"subColumn1": "HHH",
"subColumn2": "PPPP",
"subColumn3": "FFFFFF"
}
}
]
},
{
"name": "name_detA",
"tech_name": "techNameA",
"others": None,
"sub_details": []
},
{
"name": "name_detB",
"tech_name": "techNameB",
"others": 22,
"sub_details": [
{
"id": "idA",
"column2": "AA",
"column3": "BBB",
"column4": {
"id": "id1",
"subColumn1": "XXXX",
"subColumn2": "YYYYY",
"subColumn3": "DDDDDD"
}
}
]
}
]
}
]
}
]
def merge_details(details):
## --------------------
## dict to hold details by key (tech_name)
keyed_details = {}
## --------------------
## --------------------
## for each each "detail" if we find it in the key_detail merge the
## sub_details lists otherwise add it as the value of the key
## --------------------
for detail in details:
key = detail["tech_name"]
if keyed_details.get(key):
keyed_details[key]["sub_details"].extend(detail["sub_details"])
else:
keyed_details[key] = detail
## --------------------
return list(keyed_details.values())
for elem in my_input:
for department in elem["departments"]:
department["details"] = merge_details(department["details"])
print(json.dumps(my_input, indent=4, sort_keys=True))
在您的sub_details
示例数据中,column4
的关键始终是subColumn1
这就是您想要的吗?这正是我想要实现的!谢谢你,伙计