How do I write dictionary inside a DataFrame as an element?-CodePudding

Is it possible to write a dictionary into Data Frame? I have created the following DataFrame with my structure:

df =pd.DataFrame({'name': ['L1', 'L2'], 'DEF': [None, None]})

df
Out[70]: 
  name   DEF
0   L1  None
1   L2  None

I also have a dictionary

dict1={'DEF120':50}

If try to write dictionary into df as

df.loc[df.name=='L2', 'DEF'] = dict1

I am getting NaN as

df
Out[76]: 
  name   DEF
0   L1  None
1   L2   NaN

But! If I write a random number, then it works!

df.loc[df.name=='L2', 'DEF'] = 1323213

df
Out[78]: 
  name      DEF
0   L1     None
1   L2  1323213

Can someone please explain what is the problem here? Why does writing dictionary not work?

Thanks!

CodePudding user response：

Dictionaries/lists have a special meaning when assigned using loc. Pandas will try to expand them to Series.

You need to cheat a bit and use at:

s = df.name=='L2'
idx = s[s].index[0]
df.at[idx, 'DEF'] = dict1

updated dataframe:

  name             DEF
0   L1            None
1   L2  {'DEF120': 50}

CodePudding user response：

You can use pandas.at:

>>> df.at[1, 'DEF'] = dict1
>>> df
  name             DEF
0   L1            None
1   L2  {'DEF120': 50}

This way you can get the value 50 by doing:

>>> df.loc[df.name=='L2', 'DEF'].str['DEF120']
1    50
Name: DEF, dtype: int64

CodePudding user response：

You can try to pass dictionary as list or use np.where or Series.mask.

df.loc[df.name=='L2', 'DEF'] = [dict1]
# or
df['DEF'] = df['DEF'].mask(df.name=='L2', dict1)
# or
df['DEF'] = np.where(df.name=='L2', dict1, df['DEF'])

print(df)

  name             DEF
0   L1            None
1   L2  {'DEF120': 50}

CodePudding user response：

Basically you can do this like this:

import pandas as pd
dic = {"1st_entry":"hallo","2nd_entry":"good by"}
data = [1,2,3,dic]
df=pd.DataFrame(data)
print(df)

will result in

                                                0
0                                               a
1                                               b
2                                               c
3  {'1st_entry': 'hallo', '2nd_entry': 'good by'}

You have to find the right method to enter the data