Python Pandas: How to split a sorted dictionary in a column of a dataframe

Question

I have a dataFrame like this:

id  asn      orgs
0   3320    {\'Deutsche Telekom AG\': 2288}
1   47886   {\'Joyent\': 16, \'Equinix (Netherlands) B.V.\': 7}
2         


        

Answer 1

Another approach define a function that just calls min on the dict and return a Series so you can assign to multiple columns (function body taken from @Alex Martelli's answer):

In [17]:

def func(x):
    k = min(x, key=x.get)
    return pd.Series([k, x[k]])
df[['orgs', 'value']] = df['orgs'].apply(func)
df

Out[17]:
     asn  id                        orgs  value
0   3320   0         Deutsche Telekom AG   2288
1  47886   1  Equinix (Netherlands) B.V.      7
2  47601   2             fusion services   1024
3  33438   3     Highwinds Network Group    893

EDIT

If your data has empty dicss, then you can just test the len:

In [34]:

df = pd.DataFrame({'id':[0,1,2,3,4],
                   'asn':[3320,47886,47601,33438,56],
                   'orgs':[{'Deutsche Telekom AG': 2288},
                           {'Joyent': 16, 'Equinix (Netherlands) B.V.': 7},
                           {'fusion services': 1024, 'GCE Global Maritime':16859},
                           {'Highwinds Network Group': 893},{}]})
df
Out[34]:
     asn  id                                               orgs
0   3320   0                      {'Deutsche Telekom AG': 2288}
1  47886   1    {'Equinix (Netherlands) B.V.': 7, 'Joyent': 16}
2  47601   2  {'GCE Global Maritime': 16859, 'fusion service...
3  33438   3                   {'Highwinds Network Group': 893}
4     56   4                                                 {}
In [36]:

def func(x):
    if len(x) > 0:
        k = min(x, key=x.get)
        return pd.Series([k, x[k]])
    return pd.Series([np.NaN, np.NaN])

df[['orgs', 'value']] = df['orgs'].apply(func)
df

Out[36]:
     asn  id                        orgs  value
0   3320   0         Deutsche Telekom AG   2288
1  47886   1  Equinix (Netherlands) B.V.      7
2  47601   2             fusion services   1024
3  33438   3     Highwinds Network Group    893
4     56   4                         NaN    NaN

Answer 2

This should work:

In [1]: import pandas as pd  
In [2]: import operator
In [3]: df = pd.DataFrame({ 'id' : [0,1,2,3],
   ...:                      'asn' : [3320, 47886, 47601, 33438],
   ...:                      'orgs' : [{'Deutsche Telekom AG': 2288}, {'Joyent': 16, 'Equinix (Netherlands) B.V.': 7}, {'fusion services': 1024, 'GCE Global Maritime':16859}, {'Highwinds Network Group': 893}]
   ...:                    })

In [4]: df.orgs, df['value'] = zip(*df.orgs.apply(lambda x : sorted(x.items(),key=operator.itemgetter(1),reverse=True)[0]))

In [5]: df
Out[5]:
     asn  id                     orgs  value
0   3320   0      Deutsche Telekom AG   2288
1  47886   1                   Joyent     16
2  47601   2      GCE Global Maritime  16859
3  33438   3  Highwinds Network Group    893

I used zip(* <first element of sorted dict items>) and assigned them to df.orgs and df.value.

For empty dictionaries:

In [3]: df = pd.DataFrame({ 'id' : [0,1,2,3],
   ...:                      'asn' : [3320, 47886, 47601, 33438],
   ...:                      'orgs' : [{'Deutsche Telekom AG': 2288}, {'Joyent': 16, 'Equinix (Netherlands) B.V.': 7}, {'fusion services': 1024, 'GCE Global Maritime':16859}, {}]
   ...:                    })
In [4]: df.orgs.apply(lambda x : sorted(x.items(),key=operator.itemgetter(1),reverse=True)[0] if len(x) else ('',''))
Out[4]:
0     (Deutsche Telekom AG, 2288)
1                    (Joyent, 16)
2    (GCE Global Maritime, 16859)
3                            (, )
Name: orgs, dtype: object

In [5]: df.orgs, df['value'] = zip(*df.orgs.apply(lambda x : sorted(x.items(),key=operator.itemgetter(1),reverse=True)[0] if len(x) else ('','')))

In [6]: df
Out[6]:
     asn  id                 orgs  value
0   3320   0  Deutsche Telekom AG   2288
1  47886   1               Joyent     16
2  47601   2  GCE Global Maritime  16859
3  33438   3