tf.math.is_strictly_increasing

View source on GitHub

Returns True if x is strictly increasing.

View aliases

Compat aliases for migration

See Migration guide for more details.

tf.compat.v1.debugging.is_strictly_increasing, tf.compat.v1.is_strictly_increasing, tf.compat.v1.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

x1 = tf.constant([1.0, 2.0, 3.0])
tf.math.is_strictly_increasing(x1)
<tf.Tensor: shape=(), dtype=bool, numpy=True>
x2 = tf.constant([3.0, 1.0, 2.0])
tf.math.is_strictly_increasing(x2)
<tf.Tensor: shape=(), dtype=bool, numpy=False>

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.