How to subtract each row of a pandas DataFrame from a number?

Question

Say I have a DataFrame:

ds = pd.DataFrame(np.abs(randn(3, 4)), index=[1,2,3], columns=['A','B','C','Average'])
ds
      A         B         C      Average
1  1.099679  0.042043  0.083903  0.410128
2  0.268205  0.718933  1.459374  0.758887
3  0.680566  0.538655  0.038236  1.169403

How do I subtract (and replace with the result) A, B and C in row one with the average in row 1?

Answer 1

One relatively simple way is to use the sub method (I'm assuming that Average is always the last column):

ds[ds.columns[:-1]].sub(ds.Average, axis=0)

This does the following:

• ds[ds.columns[:-1]] is a DataFrame containing all but the last column ( Average ) of ds .

• .sub(ds.Average, axis=0) subtracts the row-values in the Average column from the corresponding rows in the DataFrame.

To alter your original ds , make sure to rebind the relevant columns of ds to the new DataFrame of values:

ds[ds.columns[:-1]] = ds[ds.columns[:-1]].sub(ds.Average, axis=0)

Answer 2

How about

ds['A'] = ds['A'] - ds['Average']
ds['B'] = ds['B'] - ds['Average']
ds['C'] = ds['C'] - ds['Average']

Pandas is easy like that!

Oh, that does it for the entire DF. You only want it for the firs row is that right?

ds.loc[1, 'A'] = ds.loc[1, 'A'] - ds.loc[1, 'Average']
ds.loc[1, 'B'] = ds.loc[1, 'B'] - ds.loc[1, 'Average']
ds.loc[1, 'C'] = ds.loc[1, 'C'] - ds.loc[1, 'Average']

or in a loop:

for col in ['A', 'B', 'C']:
    ds.loc[1, col] = df.loc[1, col] - ds.loc[1, 'Average']

and so on...

if you have thousands of columns then simply do:

for col in ds.columns:
    ds[col] = ds[col] - ds['Average']