tf.math.is_non_decreasing
Returns True if x is non-decreasing.
tf.math.is_non_decreasing(
x, name=None
)
Elements of x are compared in row-major order. The tensor [x[0],...] is non-decreasing if for every adjacent pair we have x[i] <= x[i+1]. If x has less than two elements, it is trivially non-decreasing.
See also: is_strictly_increasing
x1 = tf.constant([1.0, 1.0, 3.0])
tf.math.is_non_decreasing(x1)
<tf.Tensor: shape=(), dtype=bool, numpy=True>
x2 = tf.constant([3.0, 1.0, 2.0])
tf.math.is_non_decreasing(x2)
<tf.Tensor: shape=(), dtype=bool, numpy=False>
| Args | |
|---|---|
x |
Numeric Tensor. |
name |
A name for this operation (optional). Defaults to "is_non_decreasing" |
| Returns | |
|---|---|
Boolean Tensor, equal to True iff x is non-decreasing. |
| 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/r2.3/api_docs/python/tf/math/is_non_decreasing