将字典转换为具有键值对的 python 数据帧

Question

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

Answer 1

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

Answer 2

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)

Answer 3

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'])

Answer 4

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

Answer 5

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

Answer 6

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)