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"
8 Answers
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
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.
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
Summary of alternative solutions:
columns_add = ['a', 'b', 'c']
for loop:
for newcol in columns_add: df[newcol]= Nonedict method:
df.assign(**dict([(_,None) for _ in columns_add]))tuple assignment:
df['a'], df['b'], df['c'] = None, None, None
Why not just use loop:
for newcol in ['B','C','D']:
df[newcol]=np.nan
You can make use of Pandas broadcasting:
df = pd.DataFrame({'A': [1, 1, 1]})
df[['B', 'C']] = 2, 3
# df[['B', 'C']] = [2, 3]
Result:
A B C
0 1 2 3
1 1 2 3
2 1 2 3
To add empty columns:
df[['B', 'C', 'D']] = 3 * [np.nan]
Result:
A B C D
0 1 NaN NaN NaN
1 1 NaN NaN NaN
2 1 NaN NaN NaN
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)]
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)))
