Jack - 3 months ago 16

Python Question

Lets take a simplistic example where I have the data array

`A = np.asarray([[1,3], [2,4]])`

And this data is to be transformed into another form following a simple transformation:

`Q = np.asarray([[-0.5,1], [1,0.5]])`

B = np.dot(Q,np.dot(A,Q.T))

print B

Now assume that I have a set of data that takes the form of a 2d array for several time steps. For simplicity again assume that this data is just

`A`

`(2,2,N)`

`N =3`

`# Create the 3d data array`

AA = np.tile(A,(3,1,1)) # shape (3,2,2)

BB = np.dot(Q,np.dot(AA,Q.T))

print np.all( BB[:,0,:] == B ) # Returns true

So with this method I don't have to recast the

`Q`

`AA`

`AA`

`BB`

Edit:

Since

`dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m])`

Answer

```
In [91]: A=np.array([[1,3],[2,4]])
In [92]: Q=np.array([[-.5,1],[1,.5]])
In [93]: B=np.dot(Q,np.dot(A,Q.T))
In [94]: B
Out[94]:
array([[ 1.75, 2.75],
[ 4. , 4.5 ]])
```

The same calculation with `einsum`

:

```
In [95]: np.einsum('ij,jk,kl',Q,A,Q)
Out[95]:
array([[ 1.75, 2.75],
[ 4. , 4.5 ]])
```

If I make several copies of `A`

- on a new 1st dimension:

```
In [96]: AA = np.array([A,A,A])
In [97]: AA.shape
Out[97]: (3, 2, 2)
...
In [99]: BB=np.einsum('ij,pjk,kl->pil',Q,AA,Q)
In [100]: BB
Out[100]:
array([[[ 1.75, 2.75],
[ 4. , 4.5 ]],
[[ 1.75, 2.75],
[ 4. , 4.5 ]],
[[ 1.75, 2.75],
[ 4. , 4.5 ]]])
```

`BB`

has a (3,2,2) shape.

The newish `matmul`

(@ operator) lets me do the same thing

```
In [102]: Q@A@Q.T
Out[102]:
array([[ 1.75, 2.75],
[ 4. , 4.5 ]])
In [103]: Q@AA@Q.T
Out[103]:
array([[[ 1.75, 2.75],
[ 4. , 4.5 ]],
[[ 1.75, 2.75],
[ 4. , 4.5 ]],
[[ 1.75, 2.75],
[ 4. , 4.5 ]]])
```

With `einsum`

it is just as easy to work with the last dimension:

```
In [104]: AA3=np.stack([A,A,A],-1) # newish np.stack
In [105]: AA3.shape
Out[105]: (2, 2, 3)
In [106]: np.einsum('ij,jkp,kl->ilp',Q,AA3,Q)
Out[106]:
array([[[ 1.75, 1.75, 1.75],
[ 2.75, 2.75, 2.75]],
[[ 4. , 4. , 4. ],
[ 4.5 , 4.5 , 4.5 ]]])
In [107]: _.shape
Out[107]: (2, 2, 3)
```