I'm trying to figure out how to add multiple columns to pandas simultaneously with Pandas. I would like to do this in one step rather than multiple repeated steps.
import pandas as pd data = {'col_1': [0, 1, 2, 3], 'col_2': [4, 5, 6, 7]} df = pd.DataFrame(data) I thought this would work here...
df[['column_new_1', 'column_new_2', 'column_new_3']] = [np.nan, 'dogs', 3] 14 Answers
I would have expected your syntax to work too. The problem arises because when you create new columns with the column-list syntax (df[[new1, new2]] = ...), pandas requires that the right hand side be a DataFrame (note that it doesn't actually matter if the columns of the DataFrame have the same names as the columns you are creating).
Your syntax works fine for assigning scalar values to existing columns, and pandas is also happy to assign scalar values to a new column using the single-column syntax (df[new1] = ...). So the solution is either to convert this into several single-column assignments, or create a suitable DataFrame for the right-hand side.
Here are several approaches that will work:
import pandas as pd import numpy as np df = pd.DataFrame({ 'col_1': [0, 1, 2, 3], 'col_2': [4, 5, 6, 7] }) Then one of the following:
1) Three assignments in one, using list unpacking:
df['column_new_1'], df['column_new_2'], df['column_new_3'] = [np.nan, 'dogs', 3] 2) DataFrame conveniently expands a single row to match the index, so you can do this:
df[['column_new_1', 'column_new_2', 'column_new_3']] = pd.DataFrame([[np.nan, 'dogs', 3]], index=df.index) 3) Make a temporary DataFrame with new columns, then combine to the original DataFrame with .concat:
df = pd.concat( [ df, pd.DataFrame( [[np.nan, 'dogs', 3]], index=df.index, columns=['column_new_1', 'column_new_2', 'column_new_3'] ) ], axis=1 ) 4) Similar to 3, but using join instead of concat (may be less efficient):
df = df.join(pd.DataFrame( [[np.nan, 'dogs', 3]], index=df.index, columns=['column_new_1', 'column_new_2', 'column_new_3'] )) 5) Using a dict is a more "natural" way to create the new DataFrame than the previous two, but the new columns will be sorted alphabetically (at least before Python 3.6 or 3.7):
df = df.join(pd.DataFrame( { 'column_new_1': np.nan, 'column_new_2': 'dogs', 'column_new_3': 3 }, index=df.index )) 6) Use .assign() with multiple column arguments.
I like this variant on @zero's answer a lot, but like the previous one, the new columns will always be sorted alphabetically, at least with early versions of Python:
df = df.assign(column_new_1=np.nan, column_new_2='dogs', column_new_3=3) 7) This is interesting (based on this answer), but I don't know when it would be worth the trouble:
new_cols = ['column_new_1', 'column_new_2', 'column_new_3'] new_vals = [np.nan, 'dogs', 3] df = df.reindex(columns=df.columns.tolist() + new_cols) # add empty cols df[new_cols] = new_vals # multi-column assignment works for existing cols 8) In the end, it's hard to beat three separate assignments:
df['column_new_1'] = np.nan df['column_new_2'] = 'dogs' df['column_new_3'] = 3 Note: many of these options have already been covered in other questions:
- Add multiple columns to DataFrame and set them equal to an existing column
- Is it possible to add several columns at once to a pandas DataFrame?
- Add multiple empty columns to pandas DataFrame
You could use assign with a dict of column names and values.
In [1069]: df.assign(**{'col_new_1': np.nan, 'col2_new_2': 'dogs', 'col3_new_3': 3}) Out[1069]: col_1 col_2 col2_new_2 col3_new_3 col_new_1 0 0 4 dogs 3 NaN 1 1 5 dogs 3 NaN 2 2 6 dogs 3 NaN 3 3 7 dogs 3 NaN My goal when writing Pandas is to write efficient readable code that I can chain. I won't go into why I like chaining so much here, I expound on that in my book, Effective Pandas.
I often want to add new columns in a succinct manner that also allows me to chain. My general rule is that I update or create columns using the .assign method.
To answer your question, I would use the following code:
(df .assign(column_new_1=np.nan, column_new_2='dogs', column_new_3=3 ) ) To go a little further. I often have a dataframe that has new columns that I want to add to my dataframe. Let's assume it looks like say... a dataframe with the three columns you want:
df2 = pd.DataFrame({'column_new_1': np.nan, 'column_new_2': 'dogs', 'column_new_3': 3}, index=df.index ) In this case I would write the following code:
(df .assign(**df2) ) With the use of concat:
In [128]: df Out[128]: col_1 col_2 0 0 4 1 1 5 2 2 6 3 3 7 In [129]: pd.concat([df, pd.DataFrame(columns = [ 'column_new_1', 'column_new_2','column_new_3'])]) Out[129]: col_1 col_2 column_new_1 column_new_2 column_new_3 0 0.0 4.0 NaN NaN NaN 1 1.0 5.0 NaN NaN NaN 2 2.0 6.0 NaN NaN NaN 3 3.0 7.0 NaN NaN NaN Not very sure of what you wanted to do with [np.nan, 'dogs',3]. Maybe now set them as default values?
In [142]: df1 = pd.concat([df, pd.DataFrame(columns = [ 'column_new_1', 'column_new_2','column_new_3'])]) In [143]: df1[[ 'column_new_1', 'column_new_2','column_new_3']] = [np.nan, 'dogs', 3] In [144]: df1 Out[144]: col_1 col_2 column_new_1 column_new_2 column_new_3 0 0.0 4.0 NaN dogs 3 1 1.0 5.0 NaN dogs 3 2 2.0 6.0 NaN dogs 3 3 3.0 7.0 NaN dogs 3 Dictionary mapping with .assign():
This is the most readable and dynamic way to assign new column(s) with value(s) when working with many of them.
import pandas as pd import numpy as np new_cols = ["column_new_1", "column_new_2", "column_new_3"] new_vals = [np.nan, "dogs", 3] # Map new columns as keys and new values as values col_val_mapping = dict(zip(new_cols, new_vals)) # Unpack new column/new value pairs and assign them to the data frame df = df.assign(**col_val_mapping) If you're just trying to initialize the new column values to be empty as you either don't know what the values are going to be or you have many new columns.
import pandas as pd import numpy as np new_cols = ["column_new_1", "column_new_2", "column_new_3"] new_vals = [None for item in new_cols] # Map new columns as keys and new values as values col_val_mapping = dict(zip(new_cols, new_vals)) # Unpack new column/new value pairs and assign them to the data frame df = df.assign(**col_val_mapping) use of list comprehension, pd.DataFrame and pd.concat
pd.concat( [ df, pd.DataFrame( [[np.nan, 'dogs', 3] for _ in range(df.shape[0])], df.index, ['column_new_1', 'column_new_2','column_new_3'] ) ], axis=1) 
if adding a lot of missing columns (a, b, c ,....) with the same value, here 0, i did this:
new_cols = ["a", "b", "c" ] df[new_cols] = pd.DataFrame([[0] * len(new_cols)], index=df.index) It's based on the second variant of the accepted answer.
You can use tuple unpacking:
df = pd.DataFrame({'col1': [1, 2], 'col2': [3, 4]}) df['col3'], df['col4'] = 'a', 10 Result:
col1 col2 col3 col4 0 1 3 a 10 1 2 4 a 10 Just want to point out that option2 in @Matthias Fripp's answer
(2) I wouldn't necessarily expect DataFrame to work this way, but it does
df[['column_new_1', 'column_new_2', 'column_new_3']] = pd.DataFrame([[np.nan, 'dogs', 3]], index=df.index)
is already documented in pandas' own documentation
0You can pass a list of columns to [] to select columns in that order. If a column is not contained in the DataFrame, an exception will be raised. Multiple columns can also be set in this manner. You may find this useful for applying a transform (in-place) to a subset of the columns.
If you just want to add empty new columns, reindex will do the job
df col_1 col_2 0 0 4 1 1 5 2 2 6 3 3 7 df.reindex(list(df)+['column_new_1', 'column_new_2','column_new_3'], axis=1) col_1 col_2 column_new_1 column_new_2 column_new_3 0 0 4 NaN NaN NaN 1 1 5 NaN NaN NaN 2 2 6 NaN NaN NaN 3 3 7 NaN NaN NaN full code example
import numpy as np import pandas as pd df = {'col_1': [0, 1, 2, 3], 'col_2': [4, 5, 6, 7]} df = pd.DataFrame(df) print('df',df, sep='\n') print() df=df.reindex(list(df)+['column_new_1', 'column_new_2','column_new_3'], axis=1) print('''df.reindex(list(df)+['column_new_1', 'column_new_2','column_new_3'], axis=1)''',df, sep='\n') otherwise go for zeros answer with assign
I am not comfortable using "Index" and so on...could come up as below
df.columns Index(['A123', 'B123'], dtype='object') df=pd.concat([df,pd.DataFrame(columns=list('CDE'))]) df.rename(columns={ 'C':'C123', 'D':'D123', 'E':'E123' },inplace=True) df.columns Index(['A123', 'B123', 'C123', 'D123', 'E123'], dtype='object') You could instantiate the values from a dictionary if you wanted different values for each column & you don't mind making a dictionary on the line before.
>>> import pandas as pd >>> import numpy as np >>> df = pd.DataFrame({ 'col_1': [0, 1, 2, 3], 'col_2': [4, 5, 6, 7] }) >>> df col_1 col_2 0 0 4 1 1 5 2 2 6 3 3 7 >>> cols = { 'column_new_1':np.nan, 'column_new_2':'dogs', 'column_new_3': 3 } >>> df[list(cols)] = pd.DataFrame(data={k:[v]*len(df) for k,v in cols.items()}) >>> df col_1 col_2 column_new_1 column_new_2 column_new_3 0 0 4 NaN dogs 3 1 1 5 NaN dogs 3 2 2 6 NaN dogs 3 3 3 7 NaN dogs 3 Not necessarily better than the accepted answer, but it's another approach not yet listed.
import pandas as pd df = pd.DataFrame({ 'col_1': [0, 1, 2, 3], 'col_2': [4, 5, 6, 7] }) df['col_3'], df['col_4'] = [df.col_1]*2 >> df col_1 col_2 col_3 col_4 0 4 0 0 1 5 1 1 2 6 2 2 3 7 3 3 I tried your original approach (the one you said didn't work for you) and it worked fine for me, at least in my pandas version (1.5.2)
import pandas as pd import numpy as np data = {'col_1': [0, 1, 2, 3], 'col_2': [4, 5, 6, 7]} df = pd.DataFrame(data) df[['column_new_1', 'column_new_2', 'column_new_3']] = [np.nan, 'dogs', 3] print(pd.__version__) print(df) This is what I got:
1.5.2 col_1 col_2 column_new_1 column_new_2 column_new_3 0 0 4 NaN dogs 3 1 1 5 NaN dogs 3 2 2 6 NaN dogs 3 3 3 7 NaN dogs 3 But there's a cooler and more versatile approach
Since probably you'll want to use some logic when adding new columns, another way to add new columns* to a dataframe in one go is to apply a row-wise function with the logic you want. In your example:
def add_3_new_fields_to_each_row(row: pd.Series) -> pd.Series: """ Adding 3 new fields to each row of a dataframe is the same as adding 3 new columns to the dataframe """ row['column_new_1'] = np.nan row['column_new_2'] = 'dogs' row['column_new_3'] = 3 # the good thing of this approach is that you could even make the # values of "later" fields be dependent on the values of # "earlier" fields, all in one go return row # this row now has 3 more fields df = pd.DataFrame(data) df_new = df.apply(add_3_new_fields_to_each_row, axis='columns') By doing this, df is unchanged, but df_new is the dataframe you want:
col_1 col_2 column_new_1 column_new_2 column_new_3 0 0.0 4.0 NaN dogs 3 1 1.0 5.0 NaN dogs 3 2 2.0 6.0 NaN dogs 3 3 3.0 7.0 NaN dogs 3 * (actually, it returns a new dataframe with the new columns, and doesn't modify the original dataframe)
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