Norm along row in pandas

One line, using whatever function you desire (including lambda functions), e.g.

df.apply(np.linalg.norm, axis=1)

or

df.apply(lambda x: (x**2).sum()**.5, axis=1)

Numpy provides norm... Use:

np.linalg.norm(df[['X','Y','Z']].values,axis=1)

I found a faster solution than what @elyase suggested:

np.sqrt(np.square(df).sum(axis=1))