pandas.DataFrame.assign¶
-
DataFrame.
assign
(**kwargs)[source]¶ Assign new columns to a DataFrame, returning a new object (a copy) with the new columns added to the original ones. Existing columns that are re-assigned will be overwritten.
Parameters: kwargs : keyword, value pairs
keywords are the column names. If the values are callable, they are computed on the DataFrame and assigned to the new columns. The callable must not change input DataFrame (though pandas doesn’t check it). If the values are not callable, (e.g. a Series, scalar, or array), they are simply assigned.
Returns: df : DataFrame
A new DataFrame with the new columns in addition to all the existing columns.
Notes
Assigning multiple columns within the same
assign
is possible. For Python 3.6 and above, later items in ‘**kwargs’ may refer to newly created or modified columns in ‘df’; items are computed and assigned into ‘df’ in order. For Python 3.5 and below, the order of keyword arguments is not specified, you cannot refer to newly created or modified columns. All items are computed first, and then assigned in alphabetical order.Changed in version 0.23.0: Keyword argument order is maintained for Python 3.6 and later.
Examples
>>> df = pd.DataFrame({'A': range(1, 11), 'B': np.random.randn(10)})
Where the value is a callable, evaluated on df:
>>> df.assign(ln_A = lambda x: np.log(x.A)) A B ln_A 0 1 0.426905 0.000000 1 2 -0.780949 0.693147 2 3 -0.418711 1.098612 3 4 -0.269708 1.386294 4 5 -0.274002 1.609438 5 6 -0.500792 1.791759 6 7 1.649697 1.945910 7 8 -1.495604 2.079442 8 9 0.549296 2.197225 9 10 -0.758542 2.302585
Where the value already exists and is inserted:
>>> newcol = np.log(df['A']) >>> df.assign(ln_A=newcol) A B ln_A 0 1 0.426905 0.000000 1 2 -0.780949 0.693147 2 3 -0.418711 1.098612 3 4 -0.269708 1.386294 4 5 -0.274002 1.609438 5 6 -0.500792 1.791759 6 7 1.649697 1.945910 7 8 -1.495604 2.079442 8 9 0.549296 2.197225 9 10 -0.758542 2.302585
Where the keyword arguments depend on each other
>>> df = pd.DataFrame({'A': [1, 2, 3]})
>>> df.assign(B=df.A, C=lambda x:x['A']+ x['B']) A B C 0 1 1 2 1 2 2 4 2 3 3 6