How to handle params of size 2D in OpenMDAO?

Question

I have a param that is a 2D array. It works fine with getting the correct output but when I try to do anything with the gradients such as optimization or check_total_derivatives I get a sizing error. I was wondering what the best way is to handle params that are of size 2D. Here is a sample code:

import numpy as np
from openmdao.api import Group, Problem, Component, IndepVarComp, ExecComp

class C1(Component):
    def __init__(self, n):
        super(C1, self).__init__()
        self.add_param('grid', val=np.zeros((n, n)))
        self.add_output('x', shape=1)
        self.n = n

def solve_nonlinear(self, params, unknowns, resids):
    x = 0
    for i in range(self.n):
        for j in range(self.n):
            x += params['grid'][i][j]
    unknowns['x'] = x

def linearize(self, params, unknowns, resids):
    J = {}
    J['x', 'grid'] = np.ones((self.n, self.n))
    return J

class Group1(Group):
    def __init__(self, n):
        super(Group1, self).__init__()
        self.add('grid', IndepVarComp('grid', np.zeros((n, n))), promotes=['*'])
        self.add('c1', C1(n), promotes=['*'])
        self.add('obj_cmp', ExecComp('obj = -x', x=1.0), promotes=['*'])
n = 3
p = Problem()
p.root = Group1(n)
p.setup(check=False)
p['grid'] = np.ones((n, n))
p.run()
p.check_total_derivatives()
print p['x']

I get the error:

ValueError: In component 'c1', the derivative of 'x' wrt 'grid' should have shape '(1, 3)' but has shape '(3, 3)' instead.

I feel like the derivative in this case should be of size (3, 3) because that is the size of the input param. How do you handle 2D params?

While there was a small mistake in the defined Jacobian here, the error msg was also incorrect. As a note, the error msg was fixed in this pull request: github.com/OpenMDAO/OpenMDAO/pull/444 — Justin Gray
– Justin Gray, Commented Jan 14, 2016 at 13:09

Kenneth Moore · Accepted Answer · 2016-01-13 17:13:46Z

5

You have a small mistake in the Jacobian; it should look like this:

def linearize(self, params, unknowns, resids):
    J = {}
    J['x', 'grid'] = np.ones((1, self.n*self.n))
    return J

The output x is length 1, while the param grid is n by n, so it is length n*n, so the resulting J should be 1 by 9. With that change, I get the right answer.

I did notice a mistake in the error message. It should say that the expected shape is (1, 9) instead of (1, 3). I will put in a fix for that.

answered Jan 13, 2016 at 17:13

Kenneth Moore

2,2571 gold badge11 silver badges13 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Rob Falck · Accepted Answer · 2016-01-13 17:08:13Z

2

When you have a 2D variable and need to construct the gradient, flatten it (in row-major order) and formulate the gradient based on the flattened version.

answered Jan 13, 2016 at 17:08

Rob Falck

2,72418 silver badges15 bronze badges

Collectives™ on Stack Overflow

How to handle params of size 2D in OpenMDAO?

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related