我有['key','value']格式列表,其中还包含子列表。如何在python
中将嵌套列表转换为JSON格式[[' key ', ' 1542633482511430199'
],
['value=>>>BasicData',
[['isConfirmAndOrder', '0'],['brmRequestId', 'BR-2018-0000124'],
['requestType','batch'],['projectId', 'PRJ-2018-0000477'],
['createdOn', 'Mon Nov 19 18:48:02 IST 2018']]
],
['createdBy=>>>BasicData',
[['userId', '999996279'], ['email', 'ITEST275@ITS.JNJ.com'],
['firstName', 'Iris'], ['lastName', 'TEST275'],
['ntId', 'itest275'], ['region', 'NA'],
[' LastAccessTime ', ' 1542639905785 ']]
]
]
格式为
{
"key": "1542633482511430199",
"value=>>>BasicData": {
"isConfirmAndOrder": "0",
"brmRequestId": "BR-2018-0000124"
.
},
"createdBy=>>>BasicData": {
"userId": "999996279",
"email": "ITEST275@ITS.JNJ.com"
.
}
.
}
大数据的实际格式为:
[
[
['key11','value11']
['key12',['key13','value13']]
['key14',['key15','value15']]
]
[
['key21','value21']
['key22',['key23','value23']]
['key24',['key25','value25']]
]
]
答案 0 :(得分:1)
您可以为此编写一个简单的递归函数:
def to_dict_recursive(x):
d = {}
for key, value in x:
if isinstance(value, list):
value = to_dict_recursive(value)
else:
value = value.strip() # get rid of unnecessary whitespace
d[key.strip()] = value
return d
to_dict_recursive(x)
# {'createdBy=>>>BasicData': {'displayName': 'Iris TEST275',
# 'email': 'ITEST275@ITS.JNJ.com',
# 'firstName': 'Iris',
# 'lastName': 'TEST275',
# 'ntId': 'itest275',
# 'region': 'NA',
# 'roles': '[0]CG510_DHF_AP_Role',
# 'userId': '999996279'},
# 'formulaDetails=>>>BasicData': {'CreationTime': '1542633482512',
# 'LastAccessTime': '1542639905785',
# 'batchSizeUnits': 'kg<<<<<<',
# 'hitCount': '1',
# 'version': '1'},
# 'key': '1542633482511430199',
# 'value=>>>BasicData': {'brmRequestId': 'BR-2018-0000124',
# 'createdMonth': 'Nov',
# 'createdOn': 'Mon Nov 19 18:48:02 IST 2018',
# 'department': 'Global Packaging',
# 'gxp': '1',
# 'id': '1542633482511430199',
# 'isConfirmAndOrder': '0',
# 'isFilling': 'false',
# 'projectId': 'PRJ-2018-0000477',
# 'projectName': 'Automation_Product_By_Admin',
# 'requestType': 'batch',
# 'status': 'New',
# 'statusDescription': 'Batch request created',
# 'updatedOn': 'Mon Nov 19 18:48:02 IST 2018'}}
(我在Python 3.6中运行了此命令,因此字典表示形式中键的顺序与插入顺序不同。在Python 3.7+中,此顺序将有所不同。)
您甚至可以将其理解为字典理解:
def to_dict_recursive(x):
return {key.strip(): to_dict_recursive(value) if isinstance(value, list)
else value.strip
for key, value in x}
由于对象中的某些元素显然不是键和值的两元素列表,因此可以添加一个简单的防范措施:
def to_dict_recursive(x):
d = {}
try:
for key, value in x:
if isinstance(value, list):
value = to_dict_recursive(value)
else:
value = value.strip()
d[key.strip()] = value
except ValueError:
return x
return d
x = [[' key ', ' 1542633482511430199'],
["test", ["a", "b", "c"]]
]
to_dict_recursive(x)
# {'key': '1542633482511430199', 'test': ['a', 'b', 'c']}
答案 1 :(得分:0)
请注意,如果mylist
是键值对列表,那么dict(mylist)
只会返回其字典版本。棘手的部分是遍历这些嵌套列表,以用字典替换它们。这是执行此操作的递归函数:
# Where <kv> is your giant list-of-lists.
def kv_to_dict(kv):
if isinstance(kv, list):
kv = dict(kv)
for k in kv:
if isinstance(kv[k], list):
kv[k] = kv_to_dict(kv[k])
return kv
newdict = kv_to_dict(kvpairs)
一旦您将内容转换为字典,就可以使用json.dumps()
将其格式化为JSON:
import json
as_json = json.dumps(newdict, indent=4)
print(as_json)
我知道您尝试过类似的操作并遇到错误。您确定数据中的所有列表都是真正的键值对,而不是例如3个字符串的列表吗?