tf.math.is_strictly_increasing

TensorFlow 1 version View source on GitHub

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.