Python tf.linalg.LinearOperatorIdentity用法及代码示例

LinearOperator 的作用类似于 [batch] 方单位矩阵。

继承自：LinearOperator，Module

用法

tf.linalg.LinearOperatorIdentity(
    num_rows, batch_shape=None, dtype=None, is_non_singular=True,
    is_self_adjoint=True, is_positive_definite=True, is_square=True,
    assert_proper_shapes=False, name='LinearOperatorIdentity'
)

该运算符的作用类似于 [batch] 单位矩阵 A ，对于某些 b >= 0 ，其形状为 [B1,...,Bb, N, N] 。第一个 b 索引索引批处理成员。对于每个批次索引 (i1,...,ib) , A[i1,...,ib,::] 是一个 N x N 矩阵。此矩阵A 未具体化，但为了广播此形状将是相关的。

LinearOperatorIdentity 使用 num_rows 以及可选的 batch_shape 和 dtype 参数进行初始化。如果 batch_shape 是 None ，则此运算符有效地传递所有参数。如果提供了batch_shape，则可能会发生广播，这将需要制作副本。

# Create a 2 x 2 identity matrix.
operator = LinearOperatorIdentity(num_rows=2, dtype=tf.float32)

operator.to_dense()
==> [[1., 0.]
     [0., 1.]]

operator.shape
==> [2, 2]

operator.log_abs_determinant()
==> 0.

x = ... Shape [2, 4] Tensor
operator.matmul(x)
==> Shape [2, 4] Tensor, same as x.

y = tf.random.normal(shape=[3, 2, 4])
# Note that y.shape is compatible with operator.shape because operator.shape
# is broadcast to [3, 2, 2].
# This broadcast does NOT require copying data, since we can infer that y
# will be passed through without changing shape.  We are always able to infer
# this if the operator has no batch_shape.
x = operator.solve(y)
==> Shape [3, 2, 4] Tensor, same as y.

# Create a 2-batch of 2x2 identity matrices
operator = LinearOperatorIdentity(num_rows=2, batch_shape=[2])
operator.to_dense()
==> [[[1., 0.]
      [0., 1.]],
     [[1., 0.]
      [0., 1.]]]

# Here, even though the operator has a batch shape, the input is the same as
# the output, so x can be passed through without a copy.  The operator is able
# to detect that no broadcast is necessary because both x and the operator
# have statically defined shape.
x = ... Shape [2, 2, 3]
operator.matmul(x)
==> Shape [2, 2, 3] Tensor, same as x

# Here the operator and x have different batch_shape, and are broadcast.
# This requires a copy, since the output is different size than the input.
x = ... Shape [1, 2, 3]
operator.matmul(x)
==> Shape [2, 2, 3] Tensor, equal to [x, x]

形状兼容性

该运算符作用于具有兼容形状的 [batch] 矩阵。 x 是与 matmul 和 solve 的形状兼容的批处理矩阵，如果

operator.shape = [B1,...,Bb] + [N, N],  with b >= 0
x.shape =   [C1,...,Cc] + [N, R],
and [C1,...,Cc] broadcasts with [B1,...,Bb] to [D1,...,Dd]

性能

如果 batch_shape 初始化参数是 None ：

• operator.matmul(x) 是 O(1)
• operator.solve(x) 是 O(1)
• operator.determinant() 是 O(1)

如果提供了batch_shape 初始化arg，并且静态检查不能排除广播的需要：

• operator.matmul(x) 是 O(D1*...*Dd*N*R)
• operator.solve(x) 是 O(D1*...*Dd*N*R)
• operator.determinant() 是 O(B1*...*Bb)

矩阵属性提示

此 LinearOperator 使用 is_X 形式的布尔标志初始化，用于 X = non_singular, self_adjoint, positive_definite, square 。它们具有以下含义：

• 如果 is_X == True ，调用者应该期望操作符具有属性 X 。这是一个应该实现的承诺，但不是运行时断言。例如，有限的浮点精度可能会导致违反这些承诺。
• 如果 is_X == False ，调用者应该期望操作符没有 X 。
• 如果is_X == None(默认)，调用者应该没有任何期望。