# 3. Data analysis¶

## 3.1. Object size in memory¶

medcoupling provides information on memory occupied by every object: mesh, field, array etc.:

print(mesh.getHeapMemorySizeStr())
print(field.getHeapMemorySizeStr())
print(array.getHeapMemorySizeStr())


## 3.2. Extract data¶

### 3.2.1. Extract for meshes¶

If m is a mesh (MEDCouplingUMesh) and Ids a list of cell ids, you can extract the mesh ids by simply doing :

part=m[Ids] m (to the left) and part extracted by calling m[1,2,4,5,7,8] (to the right)

Note

in medcoupling ids count from zero unlike SMESH where they count from one.

part is also a MEDCouplingUMesh with same coordinates than m. Reason is that medcoupling tries to reduce memory effort.

But it’s highly likely that some nodes in part will be not fetched by part.

It can be interesting to locate the fetched nodes.

subNodeIds=part.computeFetchedNodeIds() part.computeFetchedNodeIds() returns [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]. Ids 0 and 17 are not fetched

To extract coordinates, simply invoke

m.getCoords()[subNodeIds]


It can be interesting to reduce set of points part is lying on. Simply by doing.

part.zipCoords()


Or it can be interesting for further data handling to have both reduction and array.

o2n=part.zipCoordsTraducer()


Extraction in meshes often leads to locate cells/nodes regarding their neighborhood.

Let’s consider m2 3D mesh. To locate nodes on boundaries simply invoke :

bn = m2.findBoundaryNodes()


And now to extract cells lying on boundary nodes simply call :

bc = m2.getCellIdsLyingOnNodes(bn,False)


False means if a cell has at least one node in bn, take it. True means if all nodes of cell are in bn, take it.

If a mesh consists of several contiguous zones of cells, it is possible to retrieve cell ids of each zone: A mesh with two zones

# make a structured mesh 1x5
coords=DataArrayDouble(list(range(6)))
cmesh=MEDCouplingCMesh("cmesh")
cmesh.setCoords(coords,coords[:2])

# make a mesh with two zones
zmesh=cmesh.buildUnstructured()[0,1,3,4]

# get cells ids of zones
print([ ids.getValues() for ids in zoneArrays])


[[0, 1], [2, 3]]


### 3.2.2. Extract for arrays¶

Arrays are the common entry point to selection. If arr is a 2 component DataArrayDouble you can locate tuple ids by finding those whose first component is in [a,b):

tupleIds = arr[:,0].findIdsInRange(a,b)


Or you can find tuples whose magnitude is in [c,d):

tupleIds1 = arr.magnitude().findIdsInRange(c,d)


To find which of tupleIds are missing from tupleIds1, call

tupleIds2 = DataArrayInt.buildSubstraction(tupleIds,tupleIds1)


### 3.2.3. Extract for fields¶

If field4 is a MEDCouplingFieldDouble, you can extract a sub-part of field4 on a specified cell ids ids4 by doing

subField = field4[ids4]


Note

It works whatever the spatial discretization of field4 A field on nodes (to the left) and its sub-field on a half of nodes (to the right)

You can extract a field on plane by cutting field5 like this:

origin=[0,0,2]
normvec=[-1,-1,6]
slice5=field5.extractSlice3D(origin,normvec,1e-10)


The plane is defined by its origin and its normal vector normvec. The last argument is a half-thickness of the plane.

Note

It works for fields on cells only A field on cells (to the left) and a sub-field on a plane (to the right)

## 3.3. Geometric handling of unstructured meshes¶

Consider m2 as a 3D MEDCouplingUMesh instance. You can translate it by simply

m2.translate([1.,2.,3.])


Which is equivalent to

m2.getCoords()[:]+=DataArrayDouble([1.,2.,3.],1,3)


Samely you can simply rotate it around the point [1,2,1] along Y axis with an angle of pi/3 by doing

m2.rotate([1,2,1],[0,1,0],math.pi/3)


Which is equivalent to

MEDCouplingPointSet.Rotate3DAlg([1,2,1],[0,1,0],math.pi/3,m2.getCoords())


To scale m2 relative to point [1,2,4] by a factor of 6, call

m2.scale( [1,2,4], 6. )


It can also interesting to retrieve volume of cells in m2 (resp area, length in 2D, 1D):

volPerCell=m2.getMeasureField(True)


volPerCell is a field on cell (MEDCouplingFieldDouble). True means I don’t care of cell orientation. False tells I care of cell orientation using signed values. Area field of a cartesian mesh

You can compute total volume covered by mesh by doing

volPerCell.getArray().accumulate()


You also can locate cells (using cellIds) having volume greater than a threshold t1:

part=volPerCell.getArray().findIdsGreaterOrEqualTo(t1)


In this case it is easy to build a sub-mesh containing cells having a volume higher than t1:

m2[part]


There are other common geometric methods on meshes:

centers=m2.computeCellCenterOfMass()


centers will be a DataArrayDouble giving for each cell of m2 its center of mass.

It’s possible to compute a DataArrayDouble giving the center of mass of m2 simply by doing

(centers*volPerCell.getArray()).accumulate()/DataArrayDouble(volPerCell.accumulate())


Iso barycenter of nodes constituting each cell can be computed by calling

ibc=m2.computeIsoBarycenterOfNodesPerCell()


ibc will be a DataArrayDouble.

You can retrieve a field (MEDCouplingFieldDouble) of unitary vectors normal to cells:

ortho_field=m.buildOrthogonalField() A skin mesh with a normal field on it computed by buildOrthogonalField method

You also have a set of methods to caracterize mesh quality: getEdgeRatioField, getAspectRatioField, getWarpField, getSkewField, computeDiameterField.

medcoupling provides methods to intersect 2D meshes in 2D space. MEDCouplingUMesh.Intersect2DMeshWith1DLine partitions a 2D and a 1D mesh:

m2d,m1d,a2d,a1d=MEDCouplingUMesh.Intersect2DMeshWith1DLine( mesh2d, mesh1d, 1e-12 )


The last argument is a precision used to perform intersections and localization operations. 2D and 1D meshes before and after intersection

Intersect2DMeshWith1DLine returns new 2D and 1D meshes and two arrays. a2d gives for each cell in m2d the id in mesh2d it comes from. a1d is an array of pair that gives for each cell id i in m1d the cell in md2 on the left for the 1st component and the cell in m2d on the right for the 2nd component. -1 means no cell.

For the example in the picture above a2d is:

[0, 4, 1, 1, 2, 2, 3, 3]


and a1d is:

[-1, -1, -1, -1, 2, 3, 4, 5, 4, 5, 6, 7, 6, 7, -1, -1, -1, -1]


There also a method to partition a 2D mesh by another 2D mesh:

m,a1,a2=MEDCouplingUMesh.Intersect2DMeshes( mesh1, mesh2, 1e-12 ) Two 2D meshes before partitioning (to the left) and a result mesh after partitioning (to the right)

Intersect2DMeshes returns a new 2D mesh and two arrays. a1 gives for each result cell an id of the cell of mesh1 it comes from. a2 for each result cell gives an id of the cell of mesh2 it comes from.

You can compute distance from a set of points to cells of a mesh by calling

dist,cells=mesh.distanceToPoints(points)


This method returns distance and a closest cell id for each of given points. points is a DataArrayDouble with 3 components. Returned dist is a DataArrayDouble and cells is a DataArrayInt.

## 3.4. Mesh comparison¶

It is a common question. You have two meshes m1 and m2 coming from 2 different sources (2 files) and expected to be more or less equivalent.

medcoupling proposes some methods to help to caracterize equivalence of these 2 meshes.

The first, the strongest, is informatical equality:

m1.isEqual(m2,eps)


eps is the tolerance in coordinates.

If true is returned, you are lucky.

Sometimes only names (and or component names or units) are not the same:

m1.isEqualWithoutConsideringStr(m2,eps)


But sometime the last call also returns False. It may mean that there is a permutation of nodes and or cells.

If you know by construction that m1 and m2 share the same coords object:

a,_=m1.checkGeoEquivalWith(m2,20,1e-12)
assert(m1[a].isEqualWithoutConsideringStr(m2,1e-12))


checkGeoEquivalWith returns 2 elements. The first one is relative to cells and the second one is relative to nodes.

If the mapping between m1 and m2 is impossible regarding the specified code an exception is thrown. Code meaning:

• 20=2*10+0. 2 tells I know that coords are the same. 0 tells two cells are equal if and only if their connectivity is exactly the same.
• 21=2*10+1. 2 tells I know that coords are the same. 1 tells two cells are equal if and only if their connectivity is equal within a circular permutation.
• 22=2*10+2 . 2 tells I know that coords are the same. 2 tells two cells are equal if and only if nodes set is the same independently from order.

If you expect that two meshes are geometrically the same without knowing if there is any cell/node permutation use code 12:

a,b=m1.checkGeoEquivalWith(m2,12,1e-12)
m2.renumberNodes(b,len(b))
assert(m1[a].isEqualWithoutConsideringStr(m2,1e-12))


Code meaning: 12=1*10+2. 1 tells coords can be different. 2 tells two cells are equal if and only if nodes set is the same independently from order.

Remark

a and/or b may be None. It’s not a bug it only means that renumbering is equal to identity, meaning that no associated permutation is needed.

## 3.5. Common handling mesh¶

field1 is a node field containing non simplex cells. simplexize on field1.getMesh() can help to overpass this limitation.

assert(field1.getTypeOfField()==ON_NODES)
field1.getMesh().simplexize(PLANAR_FACE_5)
field1.getValueOnMulti(pts) Initial mesh (to the left) and its simplexization (to the right)

Remark

The mesh has been modified by simplexize. This method of simplexization is fast but leads to non conform mesh that can be a problem in some context

tetrahedrize method is dedicated to simplexization of 3D meshes only. It can create a conform mesh. Unlike simplexize method, tetrahedrize method can add new points to the result mesh.

mtet,n2ocells,np=m3.tetrahedrize(PLANAR_FACE_5)


The argument specifies how to split hexahedral cells. it must be in (PLANAR_FACE_5, PLANAR_FACE_6, GENERAL_24, GENERAL_48). n2ocells is a DataArrayInt holding, for each new cell, an id of old cell producing it. np is a number of new points.

Using medcoupling you can create a 3D extruded mesh. To do that you need a 2D mesh and a 1D mesh, which defines the vector of extrusion and the number of steps. The both meshes must be in 3D space. To extrude a 2D mesh m2 along a 1D mesh m1, call

m3=m2.buildExtrudedMesh(m1,0);


The last argument is a policy defining the type of extrusion:

• 0 means “translation only”: the cells of the 1D mesh represent the vectors along which the 2D mesh will be repeated to build each level.
• 1 means “translation and rotation”: the translation is done as above. For each level, an arc of circle is fitted on the 3 preceding points of the 1D mesh. The center of the arc is the center of rotation for each level, the rotation is done along an axis normal to the plane containing the arc, and finally the angle of rotation is defined by the first two points on the arc. A 2D mesh and an 1D mesh (to the left), an extrusion mesh built with policy=0 (in the middle) and with policy=1 (to the right)

In order to aggregate several meshes of the same dimension into one mesh, call

m5=MEDCouplingUMesh.MergeUMeshes([m3,m4])


To transform a linear mesh into a quadratic one, call

m1.convertLinearCellsToQuadratic(0)


A parameter of convertLinearCellsToQuadratic provides type of conversion:

• 0 creates cells of simple quadratic types, e.g. NORM_TRI6 and NORM_QUAD8
• 1 creates cells of complex quadratic types, e.g. NORM_TRI7 and NORM_QUAD9 Result quadratic 2D meshes converted with typeOfConversion=0 (to the left) and typeOfConversion=1 (to the right)

It’s common to deduce skin of a mesh m1:

skin = m1.computeSkin()


skin and m1 share the same coordinate array. A 2D mesh (to the left) and its skin (to the right)

In order to get a 1D mesh from a given 2D or 3D mesh, call

mesh1d,d,di,r,ri=mesh3d.explodeIntoEdges()


In addition to mesh1d, explodeIntoEdges method returns four arrays describing descending connectivity and reverse descending connectivity in indirect-indexing format. d and di describe the descending connectivity, i.e. enumerate cells of mesh1d bounding each cell of mesh3d. r and ri describe the reverse descending connectivity, i.e. enumerate cells of mesh3d bounded by each cell of mesh1d. A 2D mesh (to the left) and a 1D mesh returned by explodeIntoEdges (to the right)

There is also a method similar to explodeIntoEdges which returns a mesh of one less dimensions than a given mesh, i.e. 3D->2D or 2D->1D:

mesh2d,d,di,r,ri=mesh3d.buildDescendingConnectivity()


If a mesh is non-conformal, medcoupling can make it conformal. conformize2D method is to conformize a 2D mesh in 2D space, conformize3D is to conformize a 3D mesh in 3D space:

changedCells=mesh2d.conformize2D(eps)


changedCells is an array of ids of changed cells. The changed cells become polygons in 2D and polyhedrons in 3D. eps is the relative error to detect merged edges.

You can duplicate some nodes in a mesh by calling

mesh2d.duplicateNodes([3,4])


This will create new nodes at locations of nodes #3 and #4, the new nodes will replace nodes #3 and #4 within cells so that the nodes #3 and #4 will become orphan.

Inversly it is possible to merges nodes equal within a given precision:

m.mergeNodes(1e-12)


If your 2D mesh in 3D space includes incorrectly oriented cells, you can fix their orientation by calling:

vec=[0,0,-1]
skin.orientCorrectly2DCells(vec,False)


The last argument if True, only polygons are checked, else, all cells are checked. A mesh before applying orientCorrectly2DCells (to the left) and after (to the right)

If your mesh includes incorrectly oriented polyhedra, the following method can help to fix your mesh:

m3.orientCorrectlyPolyhedrons()


If m1d is a 1D line mesh, you can ensure consecutive order of its segments by calling

m1d.renumberCells(m1d.orderConsecutiveCells1D().invertArrayN2O2O2N(m1d.getNumberOfCells())) 1D mesh before and after renumbering

orderConsecutiveCells1D method returns a permutation map in new-to-old mode but renumberCells method requires the map in old-to-new mode, hence we use invertArrayN2O2O2N method to fit to that requirement.

To arrange cells to comply with MED format requirement you can call either of the following methods:

m3.renumberCells(m3.rearrange2ConsecutiveCellTypes())
m3.sortCellsInMEDFileFrmt()


It is also possible to rearrange, and even remove, nodes by calling renumberNodes:

m.renumberNodes([2,1,0,-1],3)


The code above rearranges a mesh with 4 nodes so that old node #0 becomes #2, old node #1 remains #1, old node #2 becomes #0, old node #3 is removed. The last argument 3 means that number of nodes becomes 3. The mesh before renumberNodes (to the left) and after (to the right). Shown Ids are a unit more than Ids in medcoupling

## 3.6. Operations on fields¶

Integral of a field can be computed by calling

field.integral(True)
field.integral(0,True)


The first call returns a list of integrals of all components. The second, returns integral of 0-th component. True means that abs is applied to support size used for computing the integral.

deviator method returns the stress deviator tensor field of a stress tensor field (with 6 components):

assert(field.getNumberOfComponents()==6)
diviatorfield=field.deviator()


To get value of a field at certain points call

points=DataArrayDouble([(0.,0.),(1,1)])
values=field.getValueOnMulti(points)