Python: Calculate Center Of Mass

I have a data set with 4 columns: x,y,z, and value, let's say: x y z value 0 0 0 0 0 1 0 0 0 2 0 0 1 0 0 0 1 1 0 1 1 2 0 1 2 0 0 0 2 1 0 0 2 2 0 0 I

Solution 1:

The simplest way I can think of is this: just find an average of the coordinates of mass components weighted by each component's contribution.

import numpy
masses = numpy.array([[0,  0,  0,  0],
[0,  1,  0,  0],
[0,  2,  0,  0],
[1,  0,  0,  0],
[1,  1,  0,  1],
[1,  2,  0,  1],
[2,  0,  0,  0],
[2,  1,  0,  0],
[2,  2,  0,  0]])

nonZeroMasses = masses[numpy.nonzero(masses[:,3])] # Not really necessary, can just use masses because 0 mass used as weight will work just fine.

CM = numpy.average(nonZeroMasses[:,:3], axis=0, weights=nonZeroMasses[:,3])

Solution 2:

Another option is to use the scipy center of mass:

from scipy import ndimage
import numpy

masses = numpy.array([[0,  0,  0,  0],
[0,  1,  0,  0],
[0,  2,  0,  0],
[1,  0,  0,  0],
[1,  1,  0,  1],
[1,  2,  0,  1],
[2,  0,  0,  0],
[2,  1,  0,  0],
[2,  2,  0,  0]])

ndimage.measurements.center_of_mass(masses)

Solution 3:

How about:

#                   x      y     z  value
table = np.array([[ 5. ,  1.3,  8.3,  9. ],
                  [ 6. ,  6.7,  1.6,  5.9],
                  [ 9.1,  0.2,  6.2,  3.7],
                  [ 2.2,  2. ,  6.7,  4.6],
                  [ 3.4,  5.6,  8.4,  7.3],
                  [ 4.8,  5.9,  5.7,  5.8],
                  [ 3.7,  1.1,  8.2,  2.2],
                  [ 0.3,  0.7,  7.3,  4.6],
                  [ 8.1,  1.9,  7. ,  5.3],
                  [ 9.1,  8.2,  3.3,  5.3]])

def com(xyz, mass):
    mass = mass.reshape((-1, 1))
    return (xyz * mass).mean(0)

print(com(table[:, :3], table[:, 3]))

Solution 4:

Why did ndimage.measurements.center_of_mass not give the expected result?

The key is in how the input data masses was represented by an array of 4-tuples (x, y, z, value)

# x   y   z   value
[[0,  0,  0,  0],
 [0,  1,  0,  0],
 [0,  2,  0,  0],
 [1,  0,  0,  0],
 [1,  1,  0,  1],
 [1,  2,  0,  1],
 [2,  0,  0,  0],
 [2,  1,  0,  0],
 [2,  2,  0,  0]]

The array masses here represents the 3-D position and weights of each mass. Note however that this python array structure is only a 2-D array. It's shape is (9, 4).

The input you need to pass to ndimage to get the expected result is a 3-D array containing zeros everywhere and the weight of each mass at the appropriate coordinates within the array, like this:

from scipy import ndimage
import numpy

masses = numpy.zeros((3, 3, 1))
#      x  y  z    value
masses[1, 1, 0] = 1
masses[1, 2, 0] = 1

CM = ndimage.measurements.center_of_mass(masses)
#  x    y    z# (1.0, 1.5, 0.0)

Which is exactly the expected output.

Note the limitation of this solution (and the ndimage library) is it requires non-negative integer coordinates. Also will not be efficient for large and/or sparse volumes because each "pixel" of the ndimage needs to be instantiated in memory.