tf.math.is_non_decreasing
Stay organized with collections
Save and categorize content based on your preferences.
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.
|
Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License, and code samples are licensed under the Apache 2.0 License. For details, see the Google Developers Site Policies. Java is a registered trademark of Oracle and/or its affiliates. Some content is licensed under the numpy license.
Last updated 2023-10-06 UTC.
[[["Easy to understand","easyToUnderstand","thumb-up"],["Solved my problem","solvedMyProblem","thumb-up"],["Other","otherUp","thumb-up"]],[["Missing the information I need","missingTheInformationINeed","thumb-down"],["Too complicated / too many steps","tooComplicatedTooManySteps","thumb-down"],["Out of date","outOfDate","thumb-down"],["Samples / code issue","samplesCodeIssue","thumb-down"],["Other","otherDown","thumb-down"]],["Last updated 2023-10-06 UTC."],[],[]]