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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/r2.3/api_docs/python/tf/math/is_strictly_increasing