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.
|
TensorFlow 1 version
View source on GitHub