Pandas: Apply a function to single or selected columns or rows in Dataframe
In this article we will discuss different ways to apply a given function to selected columns or rows.
Suppose we have a dataframe object i.e.
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# List of Tuples matrix = [(22, 34, 23), (33, 31, 11), (44, 16, 21), (55, 32, 22), (66, 33, 27), (77, 35, 11) ] # Create a DataFrame object dfObj = pd.DataFrame(matrix, columns=list('xyz'), index=list('abcdef')) |
Contents of the dataframe object dfObj are,
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Original Dataframe x y z a 22 34 23 b 33 31 11 c 44 16 21 d 55 32 22 e 66 33 27 f 77 35 11 |
Now if we want to call / apply a function on all the elements of a single or multiple columns or rows ? For example,
- Multiply all the values in column ‘x’ by 2
- Multiply all the values in row ‘c’ by 10
- Add 10 in all the values in column ‘y’ & ‘z’
Let’s see how to do that using different techniques,
Apply a function to a single column in Dataframe
Suppose we want to square all the values in column ‘z’ for above created dataframe object dfObj. We can do that using different methods i.e.
Method 1 : Using Dataframe.apply()
Apply a lambda function to all the columns in dataframe using Dataframe.apply() and inside this lambda function check if column name is ‘z’ then square all the values in it i.e.
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# Apply function numpy.square() to square the value one column only i.e. with column name 'z' modDfObj = dfObj.apply(lambda x: np.square(x) if x.name == 'z' else x) print("Modified Dataframe : Squared the values in column 'z'", modDfObj, sep='\n') |
Output:
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Modified Dataframe : Squared the values in column 'z' x y z a 22 34 529 b 33 31 121 c 44 16 441 d 55 32 484 e 66 33 729 f 77 35 121 |
There are 2 other ways to achieve the same effect i.e.
Method 2 : Using [] Operator
Select the column from dataframe as series using [] operator and apply numpy.square() method on it. Then assign it back to column i.e.
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# Apply a function to one column and assign it back to the column in dataframe dfObj['z'] = dfObj['z'].apply(np.square) |
It will basically square all the values in column ‘z’
Method 3 : Using numpy.square()
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# Method 3: # Apply a function to one column and assign it back to the column in dataframe dfObj['z'] = np.square(dfObj['z']) |
It will also square all the values in column ‘z’
Apply a function to a single row in Dataframe
Suppose we want to square all the values in row ‘b’ for above created dataframe object dfObj. We can do that using different methods i.e.
Method 1 : Using Dataframe.apply()
Apply a lambda function to all the rows in dataframe using Dataframe.apply() and inside this lambda function check if row index label is ‘b’ then square all the values in it i.e.
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# Apply function numpy.square() to square the values of one row only i.e. row with index name 'b' modDfObj = dfObj.apply(lambda x: np.square(x) if x.name == 'b' else x, axis=1) print("Modified Dataframe : Squared the values in row 'b'", modDfObj, sep='\n') |
Output:
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Modified Dataframe : Squared the values in row 'b' x y z a 22 34 23 b 1089 961 121 c 44 16 21 d 55 32 22 e 66 33 27 f 77 35 11 |
There are 2 other ways to achieve the same effect i.e.
Method 2 : Using [] Operator
Select the row from dataframe as series using dataframe.loc[] operator and apply numpy.square() method on it. Then assign it back to row i.e.
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# Apply a function to one row and assign it back to the row in dataframe dfObj.loc['b'] = dfObj.loc['b'].apply(np.square) |
It will basically square all the values in row ‘b’
Method 3 : Using numpy.square()
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# Apply a function to one row and assign it back to the column in dataframe dfObj.loc['b'] = np.square(dfObj.loc['b']) |
It will also square all the values in row ‘b’.
Apply a function to a certain columns in Dataframe
We can apply a given function to only specified columns too. For example square the values in column ‘x’ & ‘y’ i.e.
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# Apply function numpy.square() to square the value 2 column only i.e. with column names 'x' and 'y' only modDfObj = dfObj.apply(lambda x: np.square(x) if x.name in ['x', 'y'] else x) print("Modified Dataframe : Squared the values in column x & y :", modDfObj, sep='\n') |
Output:
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Modified Dataframe : Squared the values in column x & y : x y z a 484 1156 23 b 1089 961 11 c 1936 256 21 d 3025 1024 22 e 4356 1089 27 f 5929 1225 11 |
Basically we just modified the if condition in lambda function and squared the values in columns with name x & y.
Apply a function to a certain rows in Dataframe
We can apply a given function to only specified rows too. For example square the values in column ‘b’ & ‘c’ i.e.
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# Apply function numpy.square() to square the values of 2 rows only i.e. with row index name 'b' and 'c' only modDfObj = dfObj.apply(lambda x: np.square(x) if x.name in ['b', 'c'] else x, axis=1) print("Modified Dataframe : Squared the values in row b & c :", modDfObj, sep='\n') |
Output:
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Modified Dataframe : Squared the values in row b & c : x y z a 22 34 23 b 1089 961 121 c 1936 256 441 d 55 32 22 e 66 33 27 f 77 35 11 |
Basically we just modified the if condition in lambda function and squared the values in rows with name b & c.
Complete example is as follows :
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import pandas as pd import numpy as np def main(): # List of Tuples matrix = [(22, 34, 23), (33, 31, 11), (44, 16, 21), (55, 32, 22), (66, 33, 27), (77, 35, 11) ] # Create a DataFrame object dfObj = pd.DataFrame(matrix, columns=list('xyz'), index=list('abcdef')) print("Original Dataframe", dfObj, sep='\n') print('********* Apply a function to a single row or column in DataFrame ********') print('*** Apply a function to a single column *** ') # Method 1: # Apply function numpy.square() to square the value one column only i.e. with column name 'z' modDfObj = dfObj.apply(lambda x: np.square(x) if x.name == 'z' else x) print("Modified Dataframe : Squared the values in column 'z'", modDfObj, sep='\n') # Method 2: # Apply a function to one column and assign it back to the column in dataframe dfObj['z'] = dfObj['z'].apply(np.square) # Method 3: # Apply a function to one column and assign it back to the column in dataframe dfObj['z'] = np.square(dfObj['z']) print('*** Apply a function to a single row *** ') dfObj = pd.DataFrame(matrix, columns=list('xyz'), index=list('abcdef')) # Method 1: # Apply function numpy.square() to square the values of one row only i.e. row with index name 'b' modDfObj = dfObj.apply(lambda x: np.square(x) if x.name == 'b' else x, axis=1) print("Modified Dataframe : Squared the values in row 'b'", modDfObj, sep='\n') # Method 2: # Apply a function to one row and assign it back to the row in dataframe dfObj.loc['b'] = dfObj.loc['b'].apply(np.square) # Method 3: # Apply a function to one row and assign it back to the column in dataframe dfObj.loc['b'] = np.square(dfObj.loc['b']) print('********* Apply a function to certains row or column in DataFrame ********') dfObj = pd.DataFrame(matrix, columns=list('xyz'), index=list('abcdef')) print('Apply a function to certain columns only') # Apply function numpy.square() to square the value 2 column only i.e. with column names 'x' and 'y' only modDfObj = dfObj.apply(lambda x: np.square(x) if x.name in ['x', 'y'] else x) print("Modified Dataframe : Squared the values in column x & y :", modDfObj, sep='\n') print('Apply a function to certain rows only') # Apply function numpy.square() to square the values of 2 rows only i.e. with row index name 'b' and 'c' only modDfObj = dfObj.apply(lambda x: np.square(x) if x.name in ['b', 'c'] else x, axis=1) print("Modified Dataframe : Squared the values in row b & c :", modDfObj, sep='\n') if __name__ == '__main__': main() |
Output:
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Original Dataframe x y z a 22 34 23 b 33 31 11 c 44 16 21 d 55 32 22 e 66 33 27 f 77 35 11 ********* Apply a function to a single row or column in DataFrame ******** *** Apply a function to a single column *** Modified Dataframe : Squared the values in column 'z' x y z a 22 34 529 b 33 31 121 c 44 16 441 d 55 32 484 e 66 33 729 f 77 35 121 *** Apply a function to a single row *** Modified Dataframe : Squared the values in row 'b' x y z a 22 34 23 b 1089 961 121 c 44 16 21 d 55 32 22 e 66 33 27 f 77 35 11 ********* Apply a function to certains row or column in DataFrame ******** Apply a function to certain columns only Modified Dataframe : Squared the values in column x & y : x y z a 484 1156 23 b 1089 961 11 c 1936 256 21 d 3025 1024 22 e 4356 1089 27 f 5929 1225 11 Apply a function to certain rows only Modified Dataframe : Squared the values in row b & c : x y z a 22 34 23 b 1089 961 121 c 1936 256 441 d 55 32 22 e 66 33 27 f 77 35 11 |
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