Divide multiple columns by another column in pandas

I need to divide all but the first columns in a DataFrame by the first column.

Here's what I'm doing, but I wonder if this isn't the "right" pandas way:

df = pd.DataFrame(np.random.rand(10,3), columns=list('ABC'))

df[['B', 'C']] = (df.T.iloc[1:] / df.T.iloc[0]).T

Is there a way to do something like df[['B','C']] / df['A']? (That just gives a 10x12 dataframe of nan.)

Also, after reading some similar questions on SO, I tried df['A'].div(df[['B', 'C']]) but that gives a broadcast error.

I believe df[['B','C']].div(df.A, axis=0) and df.iloc[:,1:].div(df.A, axis=0) work.

do: df.iloc[:,1:] = df.iloc[:,1:].div(df.A, axis=0)

This will divide all columns other than the 1st column with the 'A' column used as divisor.

Results are 1st column + all columns after / 'divisor column'.

You are actually doing a matrix multiplication (Apparently numpy understands that "/" operator multiplies by the inverse), so you need the shapes to match (see here).

e.g.

df['A'].shape --> (10,)
df[['B','C']].shape --> (10,2)

You should make them match as (2,10)(10,):
df[['B','C']].T.shape, df['A'].shape -->((2, 10), (10,))

But then your resulting matrix is: ( df[['B','C']].T / df['A'] ).shape --> (2,10)

Therefore:

( df[['B','C']].T / df['A'] ).T

Shape is (10,2). It gives you the results that you wanted!