Numpy Element-wise Standard Deviation

Question

Say I have n 2-dimensional matricies M1, M2, M3, ... all with the same dimensions.

Is there an efficient way to produce an output matrix MR where each element in MR corresponds to the standard deviation of the elements in that position in M1, M2, M3, ...

An example of the operation follows:

       1 4 5   8 2 3   -1 8 2      4.73  3.06  1.53
stdev( 3 9 2,  2 1 0,   0 3 1 ) =  1.53  4.16  1.00
       7 1 2   8 3 1    9 5 8      1.00  2.00  3.79

To clarify: the top left element of the resultant matrix is calculated as follows:

stdev(1,8,-1) = 4.7258

Whereas the bottom left element is calculated as:

stdev(7,8,9) = 1.00

If this is not a way to do this with built-in operators in one go is there an efficient alternative?

Here are the test matrices:

a=numpy.array( [[1,4,5],[3,9,2],[7,1,2]])
b=numpy.array( [[8,2,3],[2,1,0],[8,3,1]])
c=numpy.array([[-1,8,2],[0,3,1],[9,5,8]])

You could please be a bit more specific, which numbers you use to calculate your standard deviation. E.g, it is not clear to me, which numbers you use to obtain a std. dev. of 1.00. — Dietrich
– Dietrich, Commented Jun 3, 2014 at 8:26

lejlot · Accepted Answer · 2014-06-03 17:38:09Z

6

numpy is your friend

import numpy as np
print np.std((a,b,c), axis=0, ddof=1)

for provided matrices it gives

array([[ 4.72581563,  3.05505046,  1.52752523],
       [ 1.52752523,  4.163332  ,  1.        ],
       [ 1.        ,  2.        ,  3.7859389 ]])

as expected

edited Jun 3, 2014 at 17:38

answered Jun 3, 2014 at 8:26

lejlot

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2 Comments

Richard Over a year ago

Your suggestion doesn't produce the desired output.

Richard Over a year ago

Yep, just caught that.

tillsten · Accepted Answer · 2014-06-03 17:37:30Z

1

First put your data into one array:

d = np.dstack((a,b,c))

Than take the std along the thrid axies:

np.std(d, 2, ddof=1)

answered Jun 3, 2014 at 17:37

tillsten

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