Estimate extrema iteratively

In this example, we compute extrema iteratively.

import openturns as ot
import openturns.viewer as otv

We first create a one-dimensional Uniform random variable to generate data.

dim = 1
distNormal = ot.Uniform()

The IterativeExtrema class needs the dimension of the sample (here 1):

iterExtrema = ot.IterativeExtrema(dim)

We can now perform the simulations. In our case most of the data should be in the [-3,3] interval. Consequently with few samples the expected minimum should be around -3 and the expected maximum should be around 3. We first increment the object with one Point at a time. At any given step the current minimum is obtained thanks to the getMin() method, the current maximum with the getMax() method and the current number of iterations is given by the getIterationNumber() method.

size = 2000
minEvolution = ot.Sample()
maxEvolution = ot.Sample()
for i in range(size):
    point = distNormal.getRealization()
    iterExtrema.increment(point)
    minEvolution.add(iterExtrema.getMin())
    maxEvolution.add(iterExtrema.getMax())

We display the evolution of the minimum (in blue) and the maximum (orange).

iterationSample = ot.Sample.BuildFromPoint(range(1, size + 1))
#
curveMin = ot.Curve(iterationSample, minEvolution)
curveMin.setLegend("min.")
#
curveMax = ot.Curve(iterationSample, maxEvolution)
curveMax.setLegend("max.")
#
graph = ot.Graph("Evolution of the min/max", "iteration nb", "min/max", True)
graph.add(curveMin)
graph.add(curveMax)
graph.setLegendPosition("upper left")
graph.setLogScale(ot.GraphImplementation.LOGX)
view = otv.View(graph)

We can also increment with a Sample.

sample = distNormal.getSample(size)
iterExtrema.increment(sample)

We print the total number of iterations and the extrema.

print("Total number of iterations: " + str(iterExtrema.getIterationNumber()))
print("Minimum: ", iterExtrema.getMin())
print("Maximum: ", iterExtrema.getMax())

otv.View.ShowAll()

Total number of iterations: 4000
Minimum:  [-0.999998]
Maximum:  [0.999377]