What is the most efficient way to handle conversion from full to symmetric second order tensors using numpy?

Question

I am processing symmetric second order tensors (of stress) using numpy. In order to transform the tensors I have to generate a fully populated tensor, do the transformation and then recover the symmetric tensor in the rotated frame.

My input is a 2D numpy array of symmetric tensors (nx6). The code below works, but I’m pretty sure there must be a more efficient and/or elegant way to manipulate the arrays but I can’t seem to figure it out.

I anyone can anyone suggest an improvement I’d be very grateful? The sample input is just 2 symmetric tensors but in use this could be millions of tensors, hence the concernr with efficiency

Thanks,

Doug

# Sample symmetric input (S11, S22, S33, S12, S23, S13)
sym_tens_in=np.array([[0,9], [1,10], [2,11], [3,12], [4,13], [5,14]])
   
# Expand to full tensor
tens_full=np.array([[sym_tens_in[0], sym_tens_in[3], sym_tens_in[4]],
                    [sym_tens_in[3], sym_tens_in[1], sym_tens_in[5]],
                    [sym_tens_in[4], sym_tens_in[5], sym_tens_in[2]]])

# Transpose and reshape to n x 3 x 3 
tens_full=np.transpose(tens_full, axes=(2, 0, 1))

# This where the work on the full tensor will go....

# Reshape for extraction of the symmetric tensor
tens_full=np.reshape(tens_full, (2,9))

# Create an array for the test ouput symmetric tensor
sym_tens_out=np.empty((2,6), dtype=np.int32)

# Extract the symmetric components
sym_tens_out[:,0]=tens_full[:,0]
sym_tens_out[:,1]=tens_full[:,4]
sym_tens_out[:,2]=tens_full[:,8]
sym_tens_out[:,3]=tens_full[:,2]
sym_tens_out[:,4]=tens_full[:,3]
sym_tens_out[:,5]=tens_full[:,5]

# Transpose....
sym_tens_out=np.transpose(sym_tens_out)

Asked By: scotsman60

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Source

Answer 1

I guess that numpy.einsum should be faster and more memory-efficient than creating a new array for the full tensor, reshaping it, and then extracting the symmetric components

import numpy as np

sym_tens_in = np.array([[0, 9], [1, 10], [2, 11], [3, 12], [4, 13], [5, 14]])

# Compute the indices for the full tensor using einsum
full_idx = np.einsum('ij->ijkl', np.ones((2, 6)))

# Extract the full tensor using the computed indices
tens_full = sym_tens_in[full_idx]

# This where the work on the full tensor will go....

# Compute the indices for the symmetric tensor using einsum
sym_idx = np.einsum('ijkl->ij', full_idx * np.array([1, 0, 0, 1, 1, 0]))

# Extract the symmetric tensor using the computed indices
sym_tens_out = sym_tens_in[sym_idx]

Answered By: di.bezrukov

Answer 2

This won’t be any faster, but it’s more compact:

In [166]: idx=np.array([0,3,4,3,1,5,4,5,2]).reshape(3,3)
In [167]: sym_tens_in[idx].transpose(2,0,1)
Out[167]: 
array([[[ 0,  3,  4],
        [ 3,  1,  5],
        [ 4,  5,  2]],

       [[ 9, 12, 13],
        [12, 10, 14],
        [13, 14, 11]]])

The transpose could be done first:

sym_tens_in.T[:,idx]

Similarly the reverse mapping can be done with:

In [168]: idx1 = [0,4,8,1,2,5]
In [171]: tens_full.reshape(2,-1)[:,idx1]
Out[171]: 
array([[ 0,  1,  2,  3,  4,  5],
       [ 9, 10, 11, 12, 13, 14]])

with the optional transpose.

Answered By: hpaulj

Answer 3

OK – Based on the answers provided here I found a really cool solution. Now, I have to say that in my original question I omitted the actual reason that I was trying to get the full tensor into nx3x3 form. Basically, I’m implementing a function to rotate 2nd order stress tensors which requires solution of σ′=R⋅σ⋅R^T.
I was planning to use numpy.matmul for the matrix multiplication but to transform multiple stress tensors, matmul requires the 3×3 tensors to be in the last two indices of the nx3x3 matrix – hence the effort to get the data into nx3x3 from from the original 3x3xn form….

However, after I let go of numpy.matmul as my target solution and embraced numpy.einsum instead……. everything became much easier….

# Sample symmetric input (S11, S22, S33, S12, S23, S13)
sym_tens_in=np.array([[0,9], [1,10], [2,11], [3,12], [4,13], [5,14]])
idx=np.array([0,3,5,3,1,4,5,4,2]).reshape(3,3)
full=sym_tens_in[idx]
full_transformed=np.einsum('ij, jkn, lk->nil', rot_mat, full, rot_mat)

Thanks for the inspiration!!!!

Answered By: scotsman60

What is the most efficient way to handle conversion from full to symmetric second order tensors using numpy?

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