I've got a nested dictionary like that:
d={'a1': {'b': ['x', 1]}, 'a2': {'b1': ['x1', 2]}}
Expected result:
[
{
"measurements": "XXXXX",
"tags": {
"MPC": b,
"host": a1
},
"time": "timexxxxx",
"fields": {
x: 1
}
},
{
"measurements": "XXXXX",
"tags": {
"MPC": b,
"host": a2
},
"time": "timexxxxx",
"fields": {
x: 1
}
}
]
that is what I'm trying, however it's being overwritten
for k,v in d.items():
metrics['measurements'] = "XXXXX"
if isinstance(v,dict):
for j,h in v.items():
metrics['tags'] = {'MPC':j,'host':k}
metrics['time'] = "timexxxxx"
for value in h:
metrics['fields'] = {j:h}
and I'm getting:
{'fields': {'b1': ['x1', 2]},
'measurements': 'XXXXX',
'tags': {'MPC': 'b1', 'host': 'a2'},
'time': 'timexxxxx'}
Could you give me some pointers on how to deal with this?
Thanks
see below
import pprint
d = {'a1': {'b': ['x', 1]}, 'a2': {'b1': ['x1', 2]}}
data = []
for k, v in d.items():
entry = {"measurements": "XXXXX"}
entry['tags'] = {'MPC': list(v.keys())[0],"host": k}
entry["time"] = "timexxxxx"
values= list(v.values())
entry["fields"] = {values[0][0]:values[0][1]}
data.append(entry)
pprint.pprint(data)
output
[{'fields': {'x': 1},
'measurements': 'XXXXX',
'tags': {'MPC': 'b', 'host': 'a1'},
'time': 'timexxxxx'},
{'fields': {'x1': 2},
'measurements': 'XXXXX',
'tags': {'MPC': 'b1', 'host': 'a2'},
'time': 'timexxxxx'}]
This code can help you:
d={'a1': {'b': ['x', 1]}, 'a2': {'b1': ['x1', 2]}}
def convert(dictionary):
return [
{
"measurements": "XXXXX",
"tags": {
"MPC": list(value.keys())[0],
"host": key
},
"time": "timexxxxx",
"fields": dict(value.values())
} for key, value in dictionary.items()
]
print(convert(d))
Results in [{'measurements': 'XXXXX', 'tags': {'MPC': 'b', 'host': 'a1'}, 'time': 'timexxxxx', 'fields': {'x': 1}}, {'measurements': 'XXXXX', 'tags': {'MPC': 'b1', 'host': 'a2'}, 'time': 'timexxxxx', 'fields': {'x1': 2}}]
You can do it like this
#Empty List
li=[]
#Add Items in list
for i in range(2):
d = {}
d["measurment"] = "XXXXX"
d["tags"] = {1: "x"}
d["time"] = "timexxx"
d["field"] = {2: "y"}
li.append(d)
#Print list elements
for i in li:
for key, value in i.items():
print(key, ":", value)
print()
Related
I need to get the count of groups which is same 'id' and 'name'
Input:
myd = {
"Items": [
{
"id": 1,
"name": "ABC",
"value": 666
},
{
"id": 1,
"name": "ABC",
"value": 89
},
{
"id": 2,
"name": "DEF",
"value": 111
},
{
"id": 3,
"name": "GHI",
"value": 111
}
]
}
Expected output:
The count of {'id':1, 'name': 'ABC' } is 2
The count of {'id':2, 'name': 'DEF' } is 1
The count of {'id':3, 'name': 'GHI' } is 1
for total length we can get by len(myd) for single key its len(myd['id'])
How to get the count for the combination of id and name
You can use collections.OrderedDict and set both 'id' and 'name' as tuple keys. In this way, the OrderedDict automatically groups the dictionaries with same 'id' and 'name' values in order:
myd = {'Items': [
{'id':1, 'name': 'ABC', 'value': 666},
{'id':1, 'name': 'ABC', 'value': 89},
{'id':2, 'name': 'DEF', 'value': 111 },
{'id':3, 'name': 'GHI', 'value': 111 }]
}
from collections import OrderedDict
od = OrderedDict()
for d in myd['Items']:
od.setdefault((d['id'], d['name']), set()).add(d['value'])
for ks, v in od.items():
print("The count of {{'id': {}, 'name': {}}} is {}".format(ks[0], ks[1], len(v)))
Output:
The count of {'id': 1, 'name': ABC} is 2
The count of {'id': 2, 'name': DEF} is 1
The count of {'id': 3, 'name': GHI} is 1
This is a good candidate for groupby and itemgetter usage:
from itertools import groupby
from operator import itemgetter
myd = {'Items': [
{'id': 1, 'name': 'ABC', 'value': 666},
{'id': 1, 'name': 'ABC', 'value': 89},
{'id': 2, 'name': 'DEF', 'value': 111},
{'id': 3, 'name': 'GHI', 'value': 111}]
}
grouper = itemgetter('id', 'name')
for i, v in groupby(sorted(myd['Items'], key=grouper), key=grouper):
print(f"the count for {dict(id=i[0], name=i[1])} is {len(list(v))}")
I am working with a nested data structure which needs to be flattened. The values need to be aggregated so totals are produced across each level of the nested data. I'm trying to do this recursively but it's not clear how best to achieve this?
The following is an example of the data I'm working with.
def get_result():
return {
"a1": {
"b1": {
"c1": {
"d1": 1,
"d2": 1,
},
"c2": {
"d3": 1,
}
},
"b2": {
"c3": {
"d4": 1
}
}
},
"a2": {}
}
The data I'd like to produce would be as follows:
[
{
"key": "a1",
"total": 4
},
{
"key": "b1",
"total": 3
},
{
"key": "c1",
"total": 2
},
{
"key": "d1",
"total": 1
},
{
"key": "d2",
"total": 1
}
{
"key": "c2",
"total": 1
},
{
"key": "d3",
"total": 1
},
{
"key": "b2",
"total": 1
},
{
"key": "c3",
"total": 1
},
{
"key": "d4",
"total": 1
}
]
You can use recursion
from collections import defaultdict
def agg(data):
result = defaultdict(int)
agg_sum = 0
for k, v in data.items():
if isinstance(v, dict):
d, sub = agg(v)
if sub:
result.update(d)
result[k] += sub
agg_sum += sub
else:
result[k] += v
agg_sum += v
return result, agg_sum
You can use a recursive generator function for a shorter solution:
d = {'a1': {'b1': {'c1': {'d1': 1, 'd2': 1}, 'c2': {'d3': 1}}, 'b2': {'c3': {'d4': 1}}}, 'a2': {}}
def get_aggr(d):
return d if not isinstance(d, dict) else sum(map(get_aggr, d.values()))
def aggr_keys(d):
for a, b in d.items():
yield {'key':a, 'total':get_aggr(b)}
yield from (() if not isinstance(b, dict) else aggr_keys(b))
print(list(aggr_keys(d)))
Output:
[{'key': 'a1', 'total': 4},
{'key': 'b1', 'total': 3},
{'key': 'c1', 'total': 2},
{'key': 'd1', 'total': 1},
{'key': 'd2', 'total': 1},
{'key': 'c2', 'total': 1},
{'key': 'd3', 'total': 1},
{'key': 'b2', 'total': 1},
{'key': 'c3', 'total': 1},
{'key': 'd4', 'total': 1},
{'key': 'a2', 'total': 0}]
I need to convert this DataFrame to json file.
Code:
def new_json(df):
drec = dict()
ncols = df.values.shape[1]
for line in df.values:
d = drec
for j, col in enumerate(line[:-1]):
if not col in d.keys():
if j != ncols-2:
d[col] = {}
d = d[col]
else:
d[col] = line[-1]
else:
if j!= ncols-2:
d = d[col]
return drec
a=new_json(df)
print(a)
result:
{'a': {'a2': {'a21': 'new', 'a22': 'old'}, 'a3': {'a31': 'content'}, 'a4': {'a41': 'old'}}, 'b': {'b1': {'b11': 'content', 'b12': 'new', 'b13': 'new'}}, 'c': {'c1': {'c11': 'content'}, 'c2': {'c21': 'content'}, 'c3': {'c31': 'old'}}}
Is it possible to modify the result in this json format?
{
'a': {
'a2': {
'a21': 'new',
'a22': 'old'
},
'a3': {
'a31': 'content'
},
'a4': {
'a41': 'old'
}
},
'b': {
'b1': {
'b11': 'content',
'b12': 'new',
'b13': 'new'
}
},
'c': {
'c1': {
'c11': 'content'
},
'c2': {
'c21': 'content'
},
'c3': {
'c31': 'old'
}
}
}
I have a dataframe like below. Each topic has several sub-topics.
pd.DataFrame({'topic': ['A', 'A', 'A', 'B', 'B'],
'sub-topic': ['A1', 'A2', 'A3', 'B1', 'B3' ],
'value': [2,12,44,21,1]})
topic sub-topic value
0 A A1 2
1 A A2 12
2 A A3 44
3 B B1 21
4 B B3 1
I need to convert it to Json format like below. Within first layer, for example topic A, the value is the sum of all its sub-topics.
{'A': {
'value': 58,
'children': {
'A1': {'value': 2},
'A2': {'value': 12},
'A3': {'value': 44}
},
},
'B': {
'value': 22,
'children': {
'B1': {'value': 21},
'B3': {'value': 1}
}
}
}
Does anyone know how I can convert the data to this specific json? I have no clue how I should approach that. Thanks a lot in advance.
Use cusom function in GroupBy.apply, last use Series.to_dict or Series.to_json:
def f(x):
d = {'value': x['value'].sum(),
'children': x.set_index('sub-topic')[['value']].to_dict('index')}
return (d)
#for dictonary
out = df.groupby('topic').apply(f).to_dict()
#for json
#out = df.groupby('topic').apply(f).to_json()
print (out)
{
'A': {
'value': 58,
'children': {
'A1': {
'value': 2
},
'A2': {
'value': 12
},
'A3': {
'value': 44
}
}
},
'B': {
'value': 22,
'children': {
'B1': {
'value': 21
},
'B3': {
'value': 1
}
}
}
}
How do I convert the following defaultdict()?
defaultdict(<class 'dict'>, {
'key1_A': {
'id': 'key1',
'length': '663',
'type': 'A'},
'key1_B': {
'id': 'key1',
'length': '389',
'type': 'B'},
'key2_A': {
'id': 'key2',
'length': '865',
'type': 'A'},
'key2_B': {
'id': 'key2',
'length': '553',
'type': 'B' ........}})
the value of the id i.e key1 becomes the key, and the key called length is changed to length_A or B with corresponding values belonging in the earlier type.
defaultdict(<class 'dict'>, {
'key1': {
'length_A': '663',
'length_B': '389'},
'key2': {
'length_A': '865',
'length_B': '553'}})
Thanks,
I think this does what you want:
from collections import defaultdict
import pprint
d = {
'key1_A': {
'id': 'key1',
'length': '663',
'type': 'A',
},
'key1_B': {
'id': 'key1',
'length': '389',
'type': 'B',
},
'key2_A': {
'id': 'key2',
'length': '865',
'type': 'A',
},
'key2_B': {
'id': 'key2',
'length': '553',
'type': 'B',
},
}
transformed = defaultdict(dict)
for v in d.values():
transformed[v["id"]]["length_{}".format(v["type"])] = v["length"]
pprint.pprint(transformed)
# Output:
# defaultdict(<class 'dict'>,
# {'key1': {'length_A': '663', 'length_B': '389'},
# 'key2': {'length_A': '865', 'length_B': '553'}})