Hi all ,

I have a set of 3d points that rapresent atoms in a macro molecule , the atoms mass is the same for each atoms in the molecule, there is a way in paraview to found the center of a certain number of point that rapresent my data in a VTK polydata file ? there is no information about mass , I need just to find out the geometric center of the set of 3d points

How can I do that ?

there is not a tools … like filter to compute that points ?

What about `CellCenters`

filter?

@mwestphal I believe the user is asking for the center point of the dataset, not the center point of individual cells.

I believe the filter to do this doesn’t exist. However, you can easily build one. The tools for this are found in the tips and tricks post here: Pointing the camera at the center of a dataset using Python. What you are asking for is the center of the bounds.

yes I need to find the center of the points not the center of cell ! I need to know the coordinate of the center not just to center the camera is it possible ?

You will need to implement that as a Programmable Filter.

https://www.paraview.org/Wiki/Python_Programmable_Filter

I’m have not idea of how to implement … may give me an help ?

The following steps should get you the center of mass (being the average of a set of equally weighted points) without having to write a programmable filter. (It will give you the coordinates but not draw them in the 3D view.)

- Add the
`Calculator`

filter.

1.a. Set the`Result Array Name`

to`position`

1.b. Set the expression (the text box right above the calculator buttons) to`coords`

.

1.c Hit`Apply`

. (This will create a field named`position`

containing the coordinates of each point.) - Run the
`Descriptive Statistics`

filter.`Apply`

.

Once run, the `Descriptive Statistics`

filter will open up a spreadsheet view showing statistics like mean, min, max, etc. The mean for `position_0`

(X), `position_1`

(Y), and `position_2`

(Z) should give the center of gravity of your molecule. (That is, if I understand your problem correctly.)