84
votes

How do I add multiple empty columns to a DataFrame from a list?

I can do:

    df["B"] = None
    df["C"] = None
    df["D"] = None

But I can't do:

    df[["B", "C", "D"]] = None

KeyError: "['B' 'C' 'D'] not in index"

7
None is different to 0, but some answers are assuming it's equivalent. Also, assigning None will give a dtype of object, but assigning 0 will give a dtype of int. - smci
Also you can't do df[['B','C','D']] = None, None, None or [None, None, None] or pd.DataFrame([None, None, None]) - smci

7 Answers

76
votes

I'd concat using a DataFrame:

In [23]:
df = pd.DataFrame(columns=['A'])
df

Out[23]:
Empty DataFrame
Columns: [A]
Index: []

In [24]:    
pd.concat([df,pd.DataFrame(columns=list('BCD'))])

Out[24]:
Empty DataFrame
Columns: [A, B, C, D]
Index: []

So by passing a list containing your original df, and a new one with the columns you wish to add, this will return a new df with the additional columns.


Caveat: See the discussion of performance in the other answers and/or the comment discussions. reindex may be preferable where performance is critical.

86
votes

You could use df.reindex to add new columns:

In [18]: df = pd.DataFrame(np.random.randint(10, size=(5,1)), columns=['A'])

In [19]: df
Out[19]: 
   A
0  4
1  7
2  0
3  7
4  6

In [20]: df.reindex(columns=list('ABCD'))
Out[20]: 
   A   B   C   D
0  4 NaN NaN NaN
1  7 NaN NaN NaN
2  0 NaN NaN NaN
3  7 NaN NaN NaN
4  6 NaN NaN NaN

reindex will return a new DataFrame, with columns appearing in the order they are listed:

In [31]: df.reindex(columns=list('DCBA'))
Out[31]: 
    D   C   B  A
0 NaN NaN NaN  4
1 NaN NaN NaN  7
2 NaN NaN NaN  0
3 NaN NaN NaN  7
4 NaN NaN NaN  6

The reindex method as a fill_value parameter as well:

In [22]: df.reindex(columns=list('ABCD'), fill_value=0)
Out[22]: 
   A  B  C  D
0  4  0  0  0
1  7  0  0  0
2  0  0  0  0
3  7  0  0  0
4  6  0  0  0
37
votes

If you don't want to rewrite the name of the old columns, then you can use reindex:

df.reindex(columns=[*df.columns.tolist(), 'new_column1', 'new_column2'], fill_value=0)

Full example:

In [1]: df = pd.DataFrame(np.random.randint(10, size=(3,1)), columns=['A'])

In [1]: df
Out[1]: 
   A
0  4
1  7
2  0

In [2]: df.reindex(columns=[*df.columns.tolist(), 'col1', 'col2'], fill_value=0)
Out[2]: 

   A  col1  col2
0  1     0     0
1  2     0     0

And, if you already have a list with the column names, :

In [3]: my_cols_list=['col1','col2']

In [4]: df.reindex(columns=[*df.columns.tolist(), *my_cols_list], fill_value=0)
Out[4]: 
   A  col1  col2
0  1     0     0
1  2     0     0
6
votes

Why not just use loop:

for newcol in ['B','C','D']:
    df[newcol]=np.nan
4
votes

Summary of alternative solutions:

columns_add = ['a', 'b', 'c']
  1. for loop:

    for newcol in columns_add:
        df[newcol]= None
    
  2. dict method:

    df.assign(**dict([(_,None) for _ in columns_add]))
    
  3. tuple assignment:

    df['a'], df['b'], df['c'] = None, None, None
    
1
votes

I'd use

df["B"], df["C"], df["D"] = None, None, None

or

df["B"], df["C"], df["D"] = ["None" for a in range(3)]
0
votes

Just to add to the list of funny ways:

columns_add = ['a', 'b', 'c']
df = df.assign(**dict(zip(columns_add, [0] * len(columns_add)))