How to divide one dataframe row by the value of the row using apply?-CodePudding

I want to do something like this, but it throws an error:

df.apply(lambda row: row / row[2] )

My input data is the following:

Index	Col1	Col2	Col3
8206731245	1829	8440.19	140
6523960772	12452	523065.36	138
2546262822	14889	144067.84	138

Expected output:

Index	Col1	Col2	Col3
8206731245	1829/140	8440.19/140	140/140
6523960772	12452/138	523065.36/138	138/138
2546262822	14889/138	144067.84/138	138/138

CodePudding user response：

Use something like:

pd.concat([df[["Index"]],df.drop(columns=["Index"]).apply(lambda row: row / row["Col3"] ,axis=1)], axis=1)

Output

Index       Col1        Col2        Col3
8206731245  13.064286   60.287071   1.0
6523960772  90.231884   3790.328696 1.0
2546262822  107.891304  1043.969855 1.0

CodePudding user response：

df.apply() is done over columns by default. Use axis=1 to apply over rows.

df.apply(lambda row: row / row[2], axis=1)

                  Col1         Col2  Col3
Index                                    
8206731245   13.064286    60.287071   1.0
6523960772   90.231884  3790.328696   1.0
2546262822  107.891304  1043.969855   1.0

(I'm assuming your "Index" column represents the index.)

CodePudding user response：

Instead of apply, you could use div on axis=0 to leverage vectorized division:

out = df[['Index']].join(df.drop(columns='Index').div(df['Col3'], axis=0))

Or, if "Index" is the index, not a column (and it seems likely that's the case since you index row[2]), then you can simply use:

out = df.div(df['Col3'], axis=0)

If you don't want to specify the column name, you could also use iloc to use the location:

out = df.div(df.iloc[:, 2], axis=0)

Output:

        Index        Col1         Col2  Col3
0  8206731245   13.064286    60.287071   1.0
1  6523960772   90.231884  3790.328696   1.0
2  2546262822  107.891304  1043.969855   1.0