CODEWITHSUNDEEP
pythonnumpymatrix
RiskTech
RiskTech

Reputation: 1165

Average of consecutive columns in a numpy matrix

I m looking to return a matrix that is the average of the consecutive columns of another matrix, however the first column is the average between 1 and the first column of the input matrix.

I was able to achieve it by the following code but i believe this is not the most efficient way to deal with this problem.

import numpy as np

a = np.matrix([[ 0.6,1.2,1.8,2.4,3],[0.8,1.6,2.4,3.2,4]])

matrix_resultado = numpy.zeros(shape=a.shape) #Same shape resulting matrix

matrix_resultado[:,[0]] = ((1+a[:,0])/2) #First column average

#loop to calculate the average between the columns
for n in range(1,5):
    matrix_resultado[:,[n]] = ((a[:,(n-1)]+a[:,n])/2)

matrix_resultado
array([[ 0.8,  0.9,  1.5,  2.1,  2.7],
       [ 0.9,  1.2,  2. ,  2.8,  3.6]])

What is the most efficient way to achieve the same result ?

Upvotes: 0

Views: 162

Answers (1)

Divakar
Divakar

Reputation: 221614

Approach #1 : Here's one approach -

In [14]: col0_avg = (a[:,0,None] + 1)/2

In [15]: avg = (a[:,1:] + a[:,:-1])/2

In [16]: np.c_[col0_avg, avg]
Out[16]: 
matrix([[ 0.8,  0.9,  1.5,  2.1,  2.7],
        [ 0.9,  1.2,  2. ,  2.8,  3.6]])

Approach #1-S : To avoid the concatenation, we can start off with the input array and then perform those averaging steps in-place, like so -

out = a.copy()
out[:,1:] += a[:,:-1]
out[:,0] += 1
out /= 2

Approach #2 : Probably faster using uniform-filter -

In [35]: from scipy.ndimage.filters import uniform_filter1d as unif1d

In [150]: unif1d(a,size=2, axis=1, mode='constant', cval=1.0)
Out[150]: 
array([[ 0.8,  0.9,  1.5,  2.1,  2.7],
       [ 0.9,  1.2,  2. ,  2.8,  3.6]])

Runtime test -

In [133]: a = np.asmatrix(np.random.randn(200,10000))

In [134]: %%timeit
     ...: col0_avg = (a[:,0,None] + 1)/2
     ...: avg = (a[:,1:] + a[:,:-1])/2
     ...: out0 = np.c_[col0_avg, avg]
100 loops, best of 3: 7.3 ms per loop

In [135]: %%timeit
     ...: out = a.copy()
     ...: out[:,1:] += a[:,:-1]
     ...: out[:,0] += 1
     ...: out /= 2
100 loops, best of 3: 5.22 ms per loop

In [152]: %timeit unif1d(a,size=2, axis=1, mode='constant', cval=1.0)
100 loops, best of 3: 6.71 ms per loop

Upvotes: 3

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