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# Compare unconditional and conditional histogramsΒΆ

In this example, we compare unconditional and conditional histograms for a simulation. We consider the flooding model. Let be a function which takes four inputs , , and and returns one output .

We first consider the (unconditional) distribution of the input .

Let be a given threshold on the output : we consider the event . Then we consider the conditional distribution of the input given that : .

If these two distributions are significantly different, we conclude that the input has an impact on the event .

In order to approximate the distribution of the output , we perform a Monte-Carlo simulation with size 500. The threshold is chosen as the 90% quantile of the empirical distribution of . In this example, the distribution is aproximated by its empirical histogram (but this could be done with another distribution approximation as well, such as kernel smoothing for example).

```
import numpy as np
from openturns.usecases import flood_model
import openturns as ot
import openturns.viewer as viewer
from matplotlib import pylab as plt
ot.Log.Show(ot.Log.NONE)
```

We use the FloodModel data class that contains all the case parameters.

```
fm = flood_model.FloodModel()
```

Create an input sample from the joint distribution defined in the data class. We build an output sample by taking the image by the model.

```
size = 500
inputSample = fm.distribution.getSample(size)
outputSample = fm.model(inputSample)
```

Merge the input and output samples into a single sample.

```
sample = ot.Sample(size, 5)
sample[:, 0:4] = inputSample
sample[:, 4] = outputSample
sample[0:5, :]
```

Extract the first column of inputSample into the sample of the flowrates .

```
sampleQ = inputSample[:, 0]
```

```
def computeConditionnedSample(
sample, alpha=0.9, criteriaComponent=None, selectedComponent=0
):
"""
Return values from the selectedComponent-th component of the sample.
Selects the values according to the alpha-level quantile of
the criteriaComponent-th component of the sample.
"""
dim = sample.getDimension()
if criteriaComponent is None:
criteriaComponent = dim - 1
sortedSample = sample.sortAccordingToAComponent(criteriaComponent)
quantiles = sortedSample.computeQuantilePerComponent(alpha)
quantileValue = quantiles[criteriaComponent]
sortedSampleCriteria = sortedSample[:, criteriaComponent]
indices = np.where(np.array(sortedSampleCriteria.asPoint()) > quantileValue)[0]
conditionnedSortedSample = sortedSample[int(indices[0]):, selectedComponent]
return conditionnedSortedSample
```

Create an histogram for the unconditional flowrates.

```
numberOfBins = 10
histogram = ot.HistogramFactory().buildAsHistogram(sampleQ, numberOfBins)
```

Extract the sub-sample of the input flowrates Q which leads to large values of the output H.

```
alpha = 0.9
criteriaComponent = 4
selectedComponent = 0
conditionnedSampleQ = computeConditionnedSample(
sample, alpha, criteriaComponent, selectedComponent
)
```

We could as well use:
```
`
conditionnedHistogram = ot.HistogramFactory().buildAsHistogram(conditionnedSampleQ)
`
```

but this creates an histogram with new classes, corresponding
to conditionnedSampleQ.
We want to use exactly the same classes as the full sample,
so that the two histograms match.

```
first = histogram.getFirst()
width = histogram.getWidth()
conditionnedHistogram = ot.HistogramFactory().buildAsHistogram(
conditionnedSampleQ, first, width
)
```

Then creates a graphics with the unconditional and the conditional histograms.

```
graph = histogram.drawPDF()
graph.setLegends(["Q"])
#
graphConditionnalQ = conditionnedHistogram.drawPDF()
graphConditionnalQ.setColors(["blue"])
graphConditionnalQ.setLegends([r"$Q|H>H_{%s}$" % (alpha)])
graph.add(graphConditionnalQ)
view = viewer.View(graph)
plt.show()
```

We see that the two histograms are very different. The high values of the input seem to often lead to a high value of the output .

We could explore this situation further by comparing the unconditional distribution of (which is known in this case) with the conditonal distribution of , estimated by kernel smoothing. This would have the advantage of accuracy, since the kernel smoothing is a more accurate approximation of a distribution than the histogram.