Multiple insert columns if not exist pandas

Question

I have the following df

list_columns = ['A', 'B', 'C']
list_data = [
    [1, '2', 3],
    [4, '4', 5],
    [1, '2', 3],
    [4, '4', 6]
    ]
df = pd.DataFrame(columns=list_columns, data=list_data)

I want to check if multiple columns exist, and if not to create them.

Example: If B,C,D do not exist, create them(For the above df it will create only D column) I know how to do this with one column:

if 'D' not in df:
    df['D']=0

Is there a way to test if all my columns exist, and if not create the one that are missing? And not to make an if for each column

score 22 · Accepted Answer · answered Jun 18 '20 at 12:10

22

Here loop is not necessary - use DataFrame.reindex with Index.union:

cols = ['B','C','D']

df = df.reindex(df.columns.union(cols, sort=False), axis=1, fill_value=0)
print (df)
   A  B  C  D
0  1  2  3  0
1  4  4  5  0
2  1  2  3  0
3  4  4  6  0

answered Jun 18 '20 at 12:10

jezrael

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how would you fill with a different value for each newly added column..say u have a separate array or dictionary for defaults for each column? – mike01010 Mar 31 '23 at 18:08
1

@mike01010 - Then use `df.fillna({'A':5, 'D':4})` – jezrael Apr 03 '23 at 05:17

Umar.H · Answer 2 · 2020-06-18T12:53:57.243

2

Just to add, you can unpack the set diff between your columns and the list with an assign and ** unpacking.

import numpy as np
cols = ['B','C','D','E']

df.assign(**{col : 0 for col in np.setdiff1d(cols,df.columns.values)})

   A  B  C  D  E
0  1  2  3  0  0
1  4  4  5  0  0
2  1  2  3  0  0
3  4  4  6  0  0

edited Jun 18 '20 at 12:53

answered Jun 18 '20 at 12:44

Umar.H

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Multiple insert columns if not exist pandas

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