tf.math.is_strictly_increasing

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.

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Licensed under the Creative Commons Attribution License 3.0.
Code samples licensed under the Apache 2.0 License.
https://www.tensorflow.org/versions/r2.4/api_docs/python/tf/math/is_strictly_increasing