Issue
I have all_writing_test_name_data
all_writing_test_name_data=
[
{
"id": 1,
"book_name": Math,
"writing_test_description": "string",
"subject_id": 1,
"book_id": 2,
"writing_test": "string"
},
{
"id": 2,
"book_name": Math-1,
"writing_test_description": "string-1",
"subject_id": 1,
"book_id": 2,
"writing_test": "string-1"
}
]
and I want to Concatenate all_writing_test_name_data like this
[
{
"subject_id": 1,
"writing_items": [
{
"id": 1,
"book_name": Math,
"writing_test_description": "string",
"book_id": 2,
"writing_test": "string"
},
{
"id": 2,
"book_name": Math-1,
"writing_test_description": "string-1",
"book_id": 2,
"writing_test": "string-1"
}
]
}
]
i have tried this but i think there is some lacking in the code for this i can't get the desire data
x=all_writing_test_name_data
# printing original list
print("The original list is : " + str(x))
import operator
from functools import reduce
all_keys = reduce(operator.or_, (d.keys() for d in x))
bar = {key: [d.get(key) for d in x] for key in all_keys}
print('bar',bar['writing_test']+bar['writing_test_description'])
from collections import Counter
result = Counter()
for d in x:
result[d['writing_test']] = d['writing_test']
result[d['writing_test_description']] = d['writing_test_description']
print(result)
z=bar['writing_test']+bar['writing_test_description']
print bar
but I can't get my desire data. how can i get the exact same data,what is the mistake
Solution
You can group the input data by the subject_id
property of each dict
, then re-format that data into the value you want:
from collections import defaultdict
groups = defaultdict(list)
for test in all_writing_test_name_data:
subject_id = test.pop('subject_id')
groups[subject_id].append(test)
result = [ { 'subject_id' : k, 'writing_items' : v } for k, v in groups.items() ]
Output:
[
{
"subject_id": 1,
"writing_items": [
{
"id": 1,
"book_name": "Math",
"writing_test_description": "string",
"book_id": 2,
"writing_test": "string"
},
{
"id": 2,
"book_name": "Math-1",
"writing_test_description": "string-1",
"book_id": 2,
"writing_test": "string-1"
}
]
}
]
Note that this will alter the value of all_writing_test_name_data
(due to the test.pop()
). If this is not desired, add
test = test.copy()
prior to the pop
.
Answered By - Nick
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