Eliminated redundant computation in repeated calls to leapfrog() function #414
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dustinvtran
reviewed
Jan 19, 2017
| for _ in range(self.n_steps): | ||
| new_sample, new_r_sample = leapfrog(new_sample, new_r_sample, | ||
| self.step_size, self._log_joint) | ||
| new_sample, new_r_sample = leapfrog(new_sample, new_r_sample, |
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cool! i think we can also remove the two lines above this now as well.
dustinvtran
reviewed
Jan 19, 2017
| grad_log_joint = tf.gradients(log_joint(z_new), list(six.itervalues(z_new))) | ||
| for i, key in enumerate(six.iterkeys(z_old)): | ||
| r_new[key] += 0.5 * step_size * grad_log_joint[i] | ||
| for n in range(n_steps): |
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since we're not using the looping variable, we should do for _ in ... for readability
dustinvtran
reviewed
Jan 19, 2017
| z_new = {} | ||
| r_new = {} | ||
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| for key in z_old: |
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i think these lines can be condensed with the the copy method in dict's, e.g., z_new = z_old.copy() and r_new = r_old.copy().
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thanks for the catch; this looks great. only minor comments above |
dustinvtran
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hi @matthewdhoffman: i made the minor changes from my comments above. let me know if you approve/have comments, and i'll merge. |
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LGTM, thanks for cleaning it up. (And for showing me a couple of better idioms!) |
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In the old code, each call to
leapfrog()computed the gradient twice with respect to latent variables that hadn't changed. This patch eliminates that duplicated computation, resulting in a near 2x speedup.It works by adding an
n_stepsparameter toleapfrog()and moving the loop into theleapfrog()function.