Flatten nested json into multiple rows
0
I am working on a heavily nested json file that contains multiple dictionaries and lists. Some of the keys are similar and I am wondering if I can put each dictionary into different row. It doesn't matter, if some columns are not the same, the values can be left blank.
Json file (in similar form to the original)
"ID": "01",
"session": [
"A": [
"abc",
"cde",
"efj"
],
"B": 14,
"C": 14,
"D": [
"F":
"a": "ldjf",
"b": "kdj",
"ID": "01",
"c": "kjasgfk",
"d": [
"pw"
],
"e": "dsg"
,
"F":
"a": "ldjewiorf",
"ID": "01",
"c": "kjasnbgfk",
"d": "mbxzc" ,
"e": "dsg"
,
"F":
"f": "1232",
"g": "rege",
"h": "en-gb",
"i": "dfkj34",
"j": "iyt658"
],
"properties":
"AA":"esg",
"BB": "skdjghk",
"CC": "adfkh",
"DD": "sdlkfh"
,
"A": [
"abc",
"cde",
"efj"
],
"B": 16,
"C": 14,
"D": [
"F":
"a": "sdg",
"b": "sg",
"ID": "01",
"c": "sg",
"d": "shfh",
"e": "weitu"
,
"F":
"f": "1232",
"m": "sdg",
"n": "en-sdg",
"o": "eqe",
"p": "sdg"
],
"properties":
"AA":"ekjhsg",
"BB": "skkldjghk",
"CC": "adfyurkh",
"DD": "sdlkfmlh"
],
"G":
"A1":
"year": 2016,
"month": 5,
"dayOfMonth": 1,
"hourOfDay": 0,
"minute": 0,
"second": 0
,
"A2": "ksjdf",
"A3": "s38764",
"A4": [
"year": 2016,
"month": 5,
"dayOfMonth": 1,
"hourOfDay": 0,
"minute": 0,
"second": 0
]
I tried this code, but puts the whole file in 1 row with multiple columns:
def flatten_json(y):
out =
def flatten(x, name=''):
if type(x) is dict:
for a in x:
flatten(x[a], name + a + '_')
elif type(x) is list:
i = 0
for a in x:
flatten(a, name + str(i) + '_')
i += 1
else:
out[name[:-1]] = x
flatten(y)
return out
Is there a solution of what I am trying to do or is it impossible?
python json
edited Nov 13 '18 at 18:16
user2722968
2,69911637
asked Nov 13 '18 at 18:11
tpklaratpklara
84
add a comment |
0
I am working on a heavily nested json file that contains multiple dictionaries and lists. Some of the keys are similar and I am wondering if I can put each dictionary into different row. It doesn't matter, if some columns are not the same, the values can be left blank.
Json file (in similar form to the original)
"ID": "01",
"session": [
"A": [
"abc",
"cde",
"efj"
],
"B": 14,
"C": 14,
"D": [
"F":
"a": "ldjf",
"b": "kdj",
"ID": "01",
"c": "kjasgfk",
"d": [
"pw"
],
"e": "dsg"
,
"F":
"a": "ldjewiorf",
"ID": "01",
"c": "kjasnbgfk",
"d": "mbxzc" ,
"e": "dsg"
,
"F":
"f": "1232",
"g": "rege",
"h": "en-gb",
"i": "dfkj34",
"j": "iyt658"
],
"properties":
"AA":"esg",
"BB": "skdjghk",
"CC": "adfkh",
"DD": "sdlkfh"
,
"A": [
"abc",
"cde",
"efj"
],
"B": 16,
"C": 14,
"D": [
"F":
"a": "sdg",
"b": "sg",
"ID": "01",
"c": "sg",
"d": "shfh",
"e": "weitu"
,
"F":
"f": "1232",
"m": "sdg",
"n": "en-sdg",
"o": "eqe",
"p": "sdg"
],
"properties":
"AA":"ekjhsg",
"BB": "skkldjghk",
"CC": "adfyurkh",
"DD": "sdlkfmlh"
],
"G":
"A1":
"year": 2016,
"month": 5,
"dayOfMonth": 1,
"hourOfDay": 0,
"minute": 0,
"second": 0
,
"A2": "ksjdf",
"A3": "s38764",
"A4": [
"year": 2016,
"month": 5,
"dayOfMonth": 1,
"hourOfDay": 0,
"minute": 0,
"second": 0
]
I tried this code, but puts the whole file in 1 row with multiple columns:
def flatten_json(y):
out =
def flatten(x, name=''):
if type(x) is dict:
for a in x:
flatten(x[a], name + a + '_')
elif type(x) is list:
i = 0
for a in x:
flatten(a, name + str(i) + '_')
i += 1
else:
out[name[:-1]] = x
flatten(y)
return out
Is there a solution of what I am trying to do or is it impossible?
python json
edited Nov 13 '18 at 18:16
user2722968
2,69911637
asked Nov 13 '18 at 18:11
tpklaratpklara
84
1
Please post your desired output.
– Ajax1234
Nov 13 '18 at 18:13
add a comment |
0
0
0
1
I am working on a heavily nested json file that contains multiple dictionaries and lists. Some of the keys are similar and I am wondering if I can put each dictionary into different row. It doesn't matter, if some columns are not the same, the values can be left blank.
Json file (in similar form to the original)
"ID": "01",
"session": [
"A": [
"abc",
"cde",
"efj"
],
"B": 14,
"C": 14,
"D": [
"F":
"a": "ldjf",
"b": "kdj",
"ID": "01",
"c": "kjasgfk",
"d": [
"pw"
],
"e": "dsg"
,
"F":
"a": "ldjewiorf",
"ID": "01",
"c": "kjasnbgfk",
"d": "mbxzc" ,
"e": "dsg"
,
"F":
"f": "1232",
"g": "rege",
"h": "en-gb",
"i": "dfkj34",
"j": "iyt658"
],
"properties":
"AA":"esg",
"BB": "skdjghk",
"CC": "adfkh",
"DD": "sdlkfh"
,
"A": [
"abc",
"cde",
"efj"
],
"B": 16,
"C": 14,
"D": [
"F":
"a": "sdg",
"b": "sg",
"ID": "01",
"c": "sg",
"d": "shfh",
"e": "weitu"
,
"F":
"f": "1232",
"m": "sdg",
"n": "en-sdg",
"o": "eqe",
"p": "sdg"
],
"properties":
"AA":"ekjhsg",
"BB": "skkldjghk",
"CC": "adfyurkh",
"DD": "sdlkfmlh"
],
"G":
"A1":
"year": 2016,
"month": 5,
"dayOfMonth": 1,
"hourOfDay": 0,
"minute": 0,
"second": 0
,
"A2": "ksjdf",
"A3": "s38764",
"A4": [
"year": 2016,
"month": 5,
"dayOfMonth": 1,
"hourOfDay": 0,
"minute": 0,
"second": 0
]
I tried this code, but puts the whole file in 1 row with multiple columns:
def flatten_json(y):
out =
def flatten(x, name=''):
if type(x) is dict:
for a in x:
flatten(x[a], name + a + '_')
elif type(x) is list:
i = 0
for a in x:
flatten(a, name + str(i) + '_')
i += 1
else:
out[name[:-1]] = x
flatten(y)
return out
Is there a solution of what I am trying to do or is it impossible?
python json
edited Nov 13 '18 at 18:16
user2722968
2,69911637
asked Nov 13 '18 at 18:11
tpklaratpklara
84
I am working on a heavily nested json file that contains multiple dictionaries and lists. Some of the keys are similar and I am wondering if I can put each dictionary into different row. It doesn't matter, if some columns are not the same, the values can be left blank.
Json file (in similar form to the original)
"ID": "01",
"session": [
"A": [
"abc",
"cde",
"efj"
],
"B": 14,
"C": 14,
"D": [
"F":
"a": "ldjf",
"b": "kdj",
"ID": "01",
"c": "kjasgfk",
"d": [
"pw"
],
"e": "dsg"
,
"F":
"a": "ldjewiorf",
"ID": "01",
"c": "kjasnbgfk",
"d": "mbxzc" ,
"e": "dsg"
,
"F":
"f": "1232",
"g": "rege",
"h": "en-gb",
"i": "dfkj34",
"j": "iyt658"
],
"properties":
"AA":"esg",
"BB": "skdjghk",
"CC": "adfkh",
"DD": "sdlkfh"
,
"A": [
"abc",
"cde",
"efj"
],
"B": 16,
"C": 14,
"D": [
"F":
"a": "sdg",
"b": "sg",
"ID": "01",
"c": "sg",
"d": "shfh",
"e": "weitu"
,
"F":
"f": "1232",
"m": "sdg",
"n": "en-sdg",
"o": "eqe",
"p": "sdg"
],
"properties":
"AA":"ekjhsg",
"BB": "skkldjghk",
"CC": "adfyurkh",
"DD": "sdlkfmlh"
],
"G":
"A1":
"year": 2016,
"month": 5,
"dayOfMonth": 1,
"hourOfDay": 0,
"minute": 0,
"second": 0
,
"A2": "ksjdf",
"A3": "s38764",
"A4": [
"year": 2016,
"month": 5,
"dayOfMonth": 1,
"hourOfDay": 0,
"minute": 0,
"second": 0
]
I tried this code, but puts the whole file in 1 row with multiple columns:
def flatten_json(y):
out =
def flatten(x, name=''):
if type(x) is dict:
for a in x:
flatten(x[a], name + a + '_')
elif type(x) is list:
i = 0
for a in x:
flatten(a, name + str(i) + '_')
i += 1
else:
out[name[:-1]] = x
flatten(y)
return out
Is there a solution of what I am trying to do or is it impossible?
python json
python json
edited Nov 13 '18 at 18:16
user2722968
2,69911637
asked Nov 13 '18 at 18:11
tpklaratpklara
84
edited Nov 13 '18 at 18:16
user2722968
2,69911637
asked Nov 13 '18 at 18:11
tpklaratpklara
84
edited Nov 13 '18 at 18:16
user2722968
2,69911637
edited Nov 13 '18 at 18:16
user2722968
2,69911637
edited Nov 13 '18 at 18:16
user2722968
2,69911637
2,69911637
asked Nov 13 '18 at 18:11
tpklaratpklara
84
asked Nov 13 '18 at 18:11
tpklaratpklara
84
asked Nov 13 '18 at 18:11
tpklaratpklara
84
84
1
Please post your desired output.
– Ajax1234
Nov 13 '18 at 18:13
add a comment |
1
Please post your desired output.
– Ajax1234
Nov 13 '18 at 18:13
1
1
Please post your desired output.
– Ajax1234
Nov 13 '18 at 18:13
Please post your desired output.
– Ajax1234
Nov 13 '18 at 18:13
add a comment |
0
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Please post your desired output.
– Ajax1234
Nov 13 '18 at 18:13