## Using autograd in nonlinear regression

Posted November 17, 2017 at 07:49 AM | categories: python, autograd, regression | tags: | View Comments

Table raw-data contains the energy as a function of volume for some solid material from a set of density functional theory calculations. Our goal is to fit the Murnaghan equation of state to this data. The model is moderately nonlinear. I have previously done this with the standard nonlinear regression functions in scipy, so today we will use autograd along with a builtin optimizer to minimize an objective function to achieve the same thing.

The basic idea is we define an objective function, in this case the summed squared errors between predicted values from the model and known values from our data. The objective function takes two arguments: the model parameters, and the "step". This function signature is a consequence of the built in optimizer we use; it expects that signature (it is useful for batch training, but we will not use that here). We use autograd to create a gradient of the objective function which the adam optimizer will use to vary the parameters with the goal of minimizing the objective function.

The adam optimizer function takes as one argument a callback function, which we call `summary`

to print out intermediate results during the convergence. We run the optimizer in a loop because the optimizer runs a fixed number of steps on each call. We check if the objective function is sufficiently small, and if it is we break out.

import autograd.numpy as np from autograd import grad from autograd.misc.optimizers import adam np.set_printoptions(precision=3, suppress=True) # input data Vinput = np.array([row[0] for row in data]) Eknown = np.array([row[1] for row in data]) def Murnaghan(pars, vol): ''' given a vector of parameters and volumes, return a vector of energies. equation From PRB 28,5480 (1983) ''' E0, B0, BP, V0 = pars E = E0 + B0 * vol / BP * (((V0 / vol)**BP) / (BP - 1.0) + 1.0) - V0 * B0 / (BP - 1.) return E def objective(pars, step): "This is what we want to minimize by varying the pars." predicted = Murnaghan(pars, Vinput) # Note Eknown is not defined in this function scope errors = Eknown - predicted return np.sum(errors**2) objective_grad = grad(objective) def summary(pars, step, gradient): # Note i, N are not defined in this function scope if step % N == 0: print('step {0:5d}: {1:1.3e}'.format(i * N + step, objective(pars, step))) pars = np.array([-400, 0.5, 2, 210]) # The initial guess N = 200 # num of steps to take on each optimization learning_rate = 0.001 for i in range(100): pars = adam(objective_grad, pars, step_size=learning_rate, num_iters=N, callback=summary) SSE = objective(pars, None) if SSE < 0.00002: print('Tolerance met.', SSE) break print(pars)

step 0: 3.127e+02 step 200: 1.138e+02 step 400: 2.011e+01 step 600: 1.384e+00 step 800: 1.753e-01 step 1000: 2.044e-03 step 1200: 1.640e-03 step 1400: 1.311e-03 step 1600: 1.024e-03 step 1800: 7.765e-04 step 2000: 5.698e-04 step 2200: 4.025e-04 step 2400: 2.724e-04 step 2600: 1.762e-04 step 2800: 1.095e-04 step 3000: 6.656e-05 step 3200: 3.871e-05 step 3400: 2.359e-05 ('Tolerance met.', 1.5768901008364176e-05) [-400.029 0.004 4.032 211.847]

There are some subtleties in the code above. One is the variables that are used kind of all over the place, which is noted in a few places. Those could get tricky to keep track of. Another is the variable I called learning_rate. I borrowed that terminology from the machine learning community. It is the `step_size`

in this implementation of the optimizer. If you make it too large, the objective function doesn't converge, but if you set it too small, it will take a long time to converge. Note that it took at about 3400 steps of "training". This is a lot more than is typically required by something like `pycse.nlinfit`

. This isn't the typical application for this approach to regression. More on that another day.

As with any fit, it is wise to check it out at least graphically. Here is the fit and data.

%matplotlib inline import matplotlib matplotlib.rc('axes.formatter', useoffset=False) import matplotlib.pyplot as plt plt.plot(Vinput, Eknown, 'ko', label='known') vinterp = np.linspace(Vinput.min(), Vinput.max(), 200) plt.plot(vinterp, Murnaghan(pars, vinterp), 'r-', label='predicted') plt.xlabel('Vol') plt.ylabel('E')

The fit looks pretty good.

volume | energy |
---|---|

324.85990899 | -399.9731688470 |

253.43999457 | -400.0172393178 |

234.03826687 | -400.0256270548 |

231.12159387 | -400.0265690700 |

228.40609504 | -400.0273551120 |

225.86490337 | -400.0280030862 |

223.47556626 | -400.0285313450 |

221.21992353 | -400.0289534593 |

219.08319566 | -400.0292800709 |

217.05369547 | -400.0295224970 |

215.12089909 | -400.0296863867 |

213.27525144 | -400.0297809256 |

211.51060823 | -400.0298110000 |

203.66743321 | -400.0291665573 |

197.07888649 | -400.0275017142 |

191.39717952 | -400.0250998136 |

186.40163591 | -400.0221371852 |

181.94435510 | -400.0187369863 |

177.92077043 | -400.0149820198 |

174.25380090 | -400.0109367042 |

170.88582166 | -400.0066495100 |

167.76711189 | -400.0021478258 |

164.87096104 | -399.9974753449 |

159.62553397 | -399.9876885136 |

154.97005460 | -399.9774175487 |

150.78475335 | -399.9667603369 |

146.97722201 | -399.9557686286 |

143.49380641 | -399.9445262604 |

Copyright (C) 2017 by John Kitchin. See the License for information about copying.

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