Fill matrix with transposed version

Question

I have a pairwise matrix:

>>> m
     a    b    c   d
a  1.0  NaN  NaN NaN
b  0.5  1.0  NaN NaN
c  0.6  0.0  1.0 NaN
d  0.5  0.4  0.3 1.0

I want to replace the NaN in the the top right with the same values as in the bottom left:

>>> m2
     a    b    c    d
a  1.0  0.5  0.6  0.5
b  0.5  1.0  0.0  0.4
c  0.6  0.0  1.0  0.3
d  0.5  0.4  0.3  1.0

I can do it by swapping columns and indexes:

cols = m.columns
idxs = m.index

for c in cols:
    for i in idxs:
        m[i][c] = m[c][i]

But that's slow with my actual data, and I'm sure there's a way to do it in one step. I know I can generate the upper right version with "m.T" but I don't know how to replace NaN with non-NaN values to get the complete matrix. There's probably a single-step way to do this in numpy, but I don't know from matrix algebra.

Answer 1

How about (docs):

>>> df.combine_first(df.T)
     a    b    c    d
a  1.0  0.5  0.6  0.5
b  0.5  1.0  0.0  0.4
c  0.6  0.0  1.0  0.3
d  0.5  0.4  0.3  1.0

Answer 2

Here is one alternative way:

>>> m[np.triu_indices_from(m, k=1)] = m.T[np.triu_indices_from(m, k=1)]
>>> m
array([[ 1. ,  0.5,  0.6,  0.5],
       [ 0.5,  1. ,  0. ,  0.4],
       [ 0.6,  0. ,  1. ,  0.3],
       [ 0.5,  0.4,  0.3,  1. ]])

m[np.triu_indices_from(m, k=1)] returns the values above the diagonal of m and assigns them to the values values above the diagonal of the transpose of m.

Answer 3

With numpy.isnan():

>>> m[np.isnan(m)] = m.T[np.isnan(m)]
>>> m
     a    b    c    d
a  1.0  0.5  0.6  0.5
b  0.5  1.0  0.0  0.4
c  0.6  0.0  1.0  0.3
d  0.5  0.4  0.3  1.0

or better, with panda.isnull():

>>> m[pd.isnull(m)] = m.T[pd.isnull(m)]
>>> m
     a    b    c    d
a  1.0  0.5  0.6  0.5
b  0.5  1.0  0.0  0.4
c  0.6  0.0  1.0  0.3
d  0.5  0.4  0.3  1.0

which is finally equivalent to @DSM 's solution!