Reputation: 65
I have my dictionary as
{'id': '6576_926_1',
'name': 'xyz',
'm': 926,
0: {'id': '2896_926_2',
'name': 'lmn',
'm': 926},
1: {'id': '23_926_3',
'name': 'abc',
'm': 928}}
And I want to convert it into dataframe like
Id Name M
6576_926_1 Xyz 926
2896_926_2 Lmn 926
23_926_3 Abc 928
I am fine even if first row is not available as it doesn't have index. There are around 1.3 MN records and so speed is very important. I tried using a for loop and append statement and it takes forever
Upvotes: 1
Views: 11851
Reputation: 836
mydict = {'id': '6576_926_1',
'name': 'xyz',
'm': 926,
0: {'id': '2896_926_2',
'name': 'lmn',
'm': 926},
1: {'id': '23_926_3',
'name': 'abc',
'm': 928}}
import pandas as pd
del mydict['id']
del mydict['name']
del mydict['m']
d = pd.DataFrame(mydict).T
Upvotes: 0
Reputation: 35
import pandas as pd
data={'id': '6576_926_1','name': 'xyz','m': 926,0: {'id': '2896_926_2', 'name': 'lmn', 'm': 926},1: {'id': '23_926_3', 'name': 'abc','m': 928}}
Id=[]
Name=[]
M=[]
for k,val in data.items():
if type(val) is dict:
Id.append(val['id'])
Name.append(val['name'])
M.append(val['m'])
df=pd.DataFrame({'Name':Name,'Id':Id,'M':M}) print(df)
Upvotes: 0
Reputation: 206
As you have mentioned that first row is not mandatory for you. So, here i've tried this. Hope this will solve your problem
import pandas as pd
lis = []
data = {
0: {'id': '2896_926_2', 'name': 'lmn', 'm': 926},
1: {'id': '23_926_3', 'name': 'abc', 'm': 928}
}
for key,val in data.iteritems():
lis.append(val)
d = pd.DataFrame(lis)
print d
Output--
id m name
0 2896_926_2 926 lmn
1 23_926_3 928 abc
And if you want to id as your index then add set_index
for i,j in data.iteritems():
lis.append(j)
d = pd.DataFrame(lis)
d = d.set_index('id')
print d
Output-
m name
id
2896_926_2 926 lmn
23_926_3 928 abc
Upvotes: 1
Reputation: 129
Try this code!! Still, complexity is O(n)
my_dict.pop('id')
my_dict.pop('name')
my_dict.pop('m')
data = [ row.values() for row in my_dict.values()]
pd.DataFrame(data=data, columns=['id','name','m'])
Upvotes: 0
Reputation: 521
Use the following approach
import pandas as pd
data = pd.Dataframe(dict)
data = data.drop(0, axis=1)
data = data.drop(1, axis=1)
You can also try this
import pandas as pd
del dict['id']
del dict['name']
del dict['m']
pd.DataFrame(dict)
Upvotes: 0
Reputation: 1028
You can use a loop to convert each dictionary's entries into a list, and then use panda's .from_dict to convert to a dataframe. Here's the example given:
>>> data = {'col_1': [3, 2, 1, 0], 'col_2': ['a', 'b', 'c', 'd']}
>>> pd.DataFrame.from_dict(data)
col_1 col_2
0 3 a
1 2 b
2 1 c
3 0 d
Upvotes: 0