Standard deviation from center of mass along Numpy array axis

Question

I am trying to find a well-performing way to calculate the standard deviation from the center of mass/gravity along an axis of a Numpy array.

In formula this is (sorry for the misalignment):

The best I could come up with is this:

def weighted_com(A, axis, weights):
    average = np.average(A, axis=axis, weights=weights)
    return average * weights.sum() / A.sum(axis=axis).astype(float)

def weighted_std(A, axis):
    weights = np.arange(A.shape[axis])
    w1com2 = weighted_com(A, axis, weights)**2
    w2com1 = weighted_com(A, axis, weights**2)
    return np.sqrt(w2com1 - w1com2)

In weighted_com, I need to correct the normalization from sum of weights to sum of values (which is an ugly workaround, I guess). weighted_std is probably fine.

To avoid the XY problem, I still ask for what I actually want, (a better weighted_std) instead of a better version of my weighted_com.

The .astype(float) is a safety measure as I'll apply this to histograms containing ints, which caused problems due to integer division when not in Python 3 or when from __future__ import division is not active.

Answer 1

You want to take the mean, variance and standard deviation of the vector [1, 2, 3, ..., n] — where n is the dimension of the input matrix A along the axis of interest —, with weights given by the matrix A itself.

For concreteness, say you want to consider these center-of-mass statistics along the vertical axis (axis=0) — this is what corresponds to the formulas you wrote. For a fixed column j, you would do

n = A.shape[0]
r = np.arange(1, n+1)
mu = np.average(r, weights=A[:,j])
var = np.average(r**2, weights=A[:,j]) - mu**2
std = np.sqrt(var)

In order to put all of the computations for the different columns together, you have to stack together a bunch of copies of r (one per column) to form a matrix (that I have called R in the code below). With a bit of care, you can make things work for both axis=0 and axis=1.

import numpy as np

def com_stats(A, axis=0):
    A = A.astype(float)    # if you are worried about int vs. float
    n = A.shape[axis]
    m = A.shape[(axis-1)%2]
    r = np.arange(1, n+1)
    R = np.vstack([r] * m)
    if axis == 0:
        R = R.T

    mu = np.average(R, axis=axis, weights=A)
    var = np.average(R**2, axis=axis, weights=A) - mu**2
    std = np.sqrt(var)
    return mu, var, std

For example,

A = np.array([[1, 1, 0], [1, 2, 1], [1, 1, 1]])
print(A)

# [[1 1 0]
#  [1 2 1]
#  [1 1 1]]

print(com_stats(A))

# (array([ 2. ,  2. ,  2.5]),                   # centre-of-mass mean by column
#  array([ 0.66666667,  0.5       ,  0.25  ]),  # centre-of-mass variance by column
#  array([ 0.81649658,  0.70710678,  0.5   ]))  # centre-of-mass std by column

EDIT:

One can avoid creating in-memory copies of r to build R by using numpy.lib.stride_tricks: swap the line

R = np.vstack([r] * m)

above with

from numpy.lib.stride_tricks import as_strided
R = as_strided(r, strides=(0, r.itemsize), shape=(m, n))

The resulting R is a (strided) ndarray whose underlying array is the same as r's — absolutely no copying of any values occurs.

from numpy.lib.stride_tricks import as_strided

FMT = '''\
Shape: {}
Strides: {}
Position in memory: {}
Size in memory (bytes): {}
'''

def find_base_nbytes(obj):
    if obj.base is not None:
        return find_base_nbytes(obj.base)
    return obj.nbytes

def stats(obj):
    return FMT.format(obj.shape,
                      obj.strides,
                      obj.__array_interface__['data'][0],
                      find_base_nbytes(obj))

n=10
m=1000
r = np.arange(1, n+1)
R = np.vstack([r] * m)
S = as_strided(r, strides=(0, r.itemsize), shape=(m, n))

print(stats(r))
print(stats(R))
print(stats(S))

Output:

Shape: (10,)
Strides: (8,)
Position in memory: 4299744576
Size in memory (bytes): 80

Shape: (1000, 10)
Strides: (80, 8)
Position in memory: 4304464384
Size in memory (bytes): 80000

Shape: (1000, 10)
Strides: (0, 8)
Position in memory: 4299744576
Size in memory (bytes): 80

Credit to this SO answer and this one for explanations on how to get the memory address and size of the underlying array of a strided ndarray.