import numpy as np
import tensorflow as tf
def concrete_shape(shape):
if isinstance(shape, tuple):
return shape
elif isinstance(shape, tf.TensorShape):
return tuple([d.value for d in shape])
else:
raise Exception("don't know how to interpret %s as a tensor shape" % shape)
def extract_shape(t):
if t.get_shape():
pass
shape = tuple([d.value for d in t.get_shape()])
if len(shape)==1 and shape[0] is None:
shape = ()
return shape
def shapes_equal(shape1, shape2):
ns1 = np.asarray(shape1)
ns2 = np.asarray(shape2)
try:
return (ns1 == ns2).all()
except:
return False
def shape_is_scalar(shape):
ns = np.asarray(shape)
return (ns == (1,)).all()
def logsumexp(x1, x2):
shift = tf.maximum(x1, x2)
return tf.log(tf.exp(x1 - shift) + tf.exp(x2-shift)) + shift
def reduce_logsumexp(x, **kwargs):
shift = tf.reduce_max(x, **kwargs)
return tf.log(tf.reduce_sum(tf.exp(x - shift), **kwargs)) + shift
def triangular_inv(L):
eye = tf.diag(tf.ones_like(tf.diag_part(L)))
invL = tf.matrix_triangular_solve(L, eye)
return invL
def broadcast_shape(**shapes):
result = None
xs = [np.empty(shape) for shape in shapes.values()]
return np.broadcast(*xs).shape
def differentiable_sq_singular_vals(A):
# the standard tensorflow SVD repr isn't differentiable,
# but if we fix the rotation we can get an approximate
# derivative
d, u, v = tf.svd(A)
vv = tf.stop_gradient(v)
ud = tf.matmul(A, vv)
dd = tf.reduce_sum(tf.square(ud), 0)
return dd
