Fill matrix with transposed version

2024/9/20 21:16:56

I have a pairwise matrix:

>>> ma    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:

>>> m2a    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.indexfor 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

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
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