Returns True if x 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.
|
TensorFlow 1 version
View source on GitHub