tf.math.is_strictly_increasing

Returns True if x is strictly increasing.

View aliases

Main aliases

`tf.debugging.is_strictly_increasing`, `tf.is_strictly_increasing`

Compat aliases for migration

tf.compat.v1.debugging.is_strictly_increasing, tf.compat.v1.is_strictly_increasing, tf.compat.v1.math.is_strictly_increasing, `tf.compat.v2.math.is_strictly_increasing`

tf.math.is_strictly_increasing(
    x, name=None
)

Elements of x are compared in row-major order. The tensor [x[0],...] is strictly increasing if for every adjacent pair we have x[i] < x[i+1]. If x has less than two elements, it is trivially strictly increasing.

See also: is_non_decreasing

Args
`x`	Numeric `Tensor`.
`name`	A name for this operation (optional). Defaults to "is_strictly_increasing"

Returns
Boolean `Tensor`, equal to `True` iff `x` is strictly increasing.

Raises
`TypeError`	if `x` is not a numeric tensor.

© 2020 The TensorFlow Authors. All rights reserved.
Licensed under the Creative Commons Attribution License 3.0.
Code samples licensed under the Apache 2.0 License.
https://www.tensorflow.org/versions/r1.15/api_docs/python/tf/math/is_strictly_increasing