Sort a sample

In this example we present useful methods of the Sample object such as marginals extraction and various sorting strategies.

import openturns as ot

ot.Log.Show(ot.Log.NONE)
ot.RandomGenerator.SetSeed(0)

We start by defining the distribution of a regular non-biased die.

die_distribution = ot.UserDefined([[i] for i in range(1, 7)])

We consider now an experiment with two independent dice and build the corresponding random vector :

two_dice_distribution = ot.ComposedDistribution([die_distribution] * 2)

We now build a sample of size from this distribution :

n = 5
sample = two_dice_distribution.getSample(n)
print(sample)

    [ v0 X0 ]
0 : [ 1  3  ]
1 : [ 3  2  ]
2 : [ 4  3  ]
3 : [ 4  4  ]
4 : [ 5  5  ]

Useful methods

We have access to the marginals by providing a list of the wanted indices :

# the first marginal
sample_die1 = sample.getMarginal([0])

# the second marginal
sample_die2 = sample.getMarginal([1])

Suppose we are interested in the sum of the two dice. This can be done by summing the two samples die1 and die2. Provided the dimensions are the same we can add samples with the + operator and produce a new sample :

experiment = sample_die1 + sample_die2

Note that the += operator is defined as well.

We can concatenate two samples having the same size with the stack method :

sample_die1.stack(sample_die2)
print(sample_die1)

    [ v0 X0 ]
0 : [ 1  3  ]
1 : [ 3  2  ]
2 : [ 4  3  ]
3 : [ 4  4  ]
4 : [ 5  5  ]

We can split a sample in two by giving an index (here 2).

remaining = sample_die1.split(2)
print(sample_die1)
print(remaining)

    [ v0 X0 ]
0 : [ 1  3  ]
1 : [ 3  2  ]
    [ v0 X0 ]
0 : [ 4  3  ]
1 : [ 4  4  ]
2 : [ 5  5  ]

Sorting samples

We can extract any marginal and sort it by ascending order by specifying the index :

sorted_marginal = sample.sort(1)
print(sorted_marginal)

0 : [ 2 ]
1 : [ 3 ]
2 : [ 3 ]
3 : [ 4 ]
4 : [ 5 ]

We can sort the sample in place, that is without creating a new sample, as well with sortInPlace. When the dimension is greater than one the sort is made according to the first marginal.

sample.sortInPlace()
print(sample)

    [ v0 X0 ]
0 : [ 1  3  ]
1 : [ 3  2  ]
2 : [ 4  3  ]
3 : [ 4  4  ]
4 : [ 5  5  ]

We can sort the rows according to the second marginal with the sortAccordingToAComponent :

another_sample = sample.sortAccordingToAComponent(1)
print(another_sample)

    [ v0 X0 ]
0 : [ 3  2  ]
1 : [ 1  3  ]
2 : [ 4  3  ]
3 : [ 4  4  ]
4 : [ 5  5  ]

There is also a sortAccordingToAComponentInPlace method that does the same without creating a new sample.

We can sort and remove the duplicates at the same time

print(sample_die2)
print(sample_die2.sortUnique())

    [ X0 ]
0 : [ 3  ]
1 : [ 2  ]
2 : [ 3  ]
3 : [ 4  ]
4 : [ 5  ]
    [ X0 ]
0 : [ 2  ]
1 : [ 3  ]
2 : [ 4  ]
3 : [ 5  ]

We note that the sample is smaller as expected. Sorting in place is also possible :

sample_die2.sortUniqueInPlace()
print(sample_die2)

    [ X0 ]
0 : [ 2  ]
1 : [ 3  ]
2 : [ 4  ]
3 : [ 5  ]

Let’s try with the sample in dimension 2 :

sampleUnique = sample.sortUnique()
print(sampleUnique)

    [ v0 X0 ]
0 : [ 1  3  ]
1 : [ 3  2  ]
2 : [ 4  3  ]
3 : [ 4  4  ]
4 : [ 5  5  ]

Nothing happens here because pairs are already unique !