Saullo Castro - 1 month ago 10

Python Question

In this question it is explained how to access the

`lower`

`upper`

`m = np.matrix([[11, 12, 13],`

[21, 22, 23],

[31, 32, 33]])

Here I need to transform the matrix in a 1D array, which can be done doing:

`indices = np.triu_indices_from(m)`

a = np.asarray( m[indices] )[-1]

#array([11, 12, 13, 22, 23, 33])

After doing a lot of calculations with

`a`

`new = np.zeros(m.shape)`

for i,j in enumerate(zip(*indices)):

new[j]=a[i]

new[j[1],j[0]]=a[i]

Returning:

`array([[ 11., 12., 13.],`

[ 12., 22., 23.],

[ 13., 23., 33.]])

Is there a better way to accomplish this? More especifically, avoiding the Python loop to rebuild the 2D array?

Answer

Do you just want to form a symmetric array? You can skip the diagonal indices completely.

```
>>> m=np.array(m)
>>> inds = np.triu_indices_from(m,k=1)
>>> m[(inds[1], inds[0])] = m[inds]
>>> m
array([[11, 12, 13],
[12, 22, 23],
[13, 23, 33]])
```

Creating a symmetric array from a:

```
>>> new = np.zeros((3,3))
>>> vals = np.array([11, 12, 13, 22, 23, 33])
>>> inds = np.triu_indices_from(new)
>>> new[inds] = vals
>>> new[(inds[1], inds[0])] = vals
>>> new
array([[ 11., 12., 13.],
[ 12., 22., 23.],
[ 13., 23., 33.]])
```

Source (Stackoverflow)

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