Mean Squared Error in Numpy?

Question

Is there a method in numpy for calculating the Mean Squared Error between two matrices?

I've tried searching but found none. Is it under a different name?

If there isn't, how do you overcome this? Do you write it yourself or use a different lib?

Ulbricht

((A - B) ** 2).mean(axis=ax), where ax=0 is per-column, ax=1 is per-row and ax=None gives a grand total. – Norty

If you formulate that as an answer I will accept it. – Ulbricht

This answer is not correct because when you square a numpy matrix, it will perform a matrix multiplication rathar square each element individualy. Check my comment in Saullo Castro's answer. (PS: I've tested it using Python 2.7.5 and Numpy 1.7.1) – Visit

Also just as a note for anyone looking at this in the context of neural networks, you should sum the error, not average. Averaging the error will give you incorrect gradient values if you try to do grad checking (unless you account in backprop for the average, which is more work than it's worth) – Mimesis

Answer 1

You can use:

mse = ((A - B)**2).mean(axis=ax)

Or

mse = (np.square(A - B)).mean(axis=ax)

• with ax=0 the average is performed along the row, for each column, returning an array
• with ax=1 the average is performed along the column, for each row, returning an array
• with omitting the ax parameter (or setting it to ax=None) the average is performed element-wise along the array, returning a scalar value

Weissmann

Correct if I'm wrong, but I think if you do (MatrixA - MatrixB) ** 2 it will try to perform a matrix multiplication, which is different than square each element individually. If you try to use the following formula with a non-square matrix, it will raise a ValueError. – Visit

@Visit in a Numpy array this formula will be applied element-wise so that no matrix multiplication is performed – Weissmann

@Saulo Castro, I've just tested and I must insist that the result will not be element-wise. I'm using Python 2.7.5 and Numpy 1.7.1. I created the matrix "a" and squared it usign the following commands: a = numpy.matrix([[5, 5], [5, 5]]) and then a ** 2. The result is the numpy matrix matrix([[50, 50], [50, 50]]), which shows that numpy matrix multiplication will not be element-wise. – Visit

@Visit maybe you misunderstood me, using a np.ndarray will do an element-wise multiplication for a**2, but using a np.matrixlib.defmatrix.matrix will do a matrix multiplication for a**2... – Weissmann

Sorry, I misunderstood you. I thought you were using numpy.matrix. – Visit

Bear in mind that if you're comparing 2 uint matricies, this will not work because the difference will have negative numbers. You'll need to make int copies before hand (Acmp = np.array(A, dtype=int)) – Merline

np.nanmean(((A - B) ** 2)) if missing values – Mameluke

The first solution looks like numpy array syntax too, but it is not (see comment by Saullo Castro), so this very specific starting assumption (using a np.matrix) needs to be part of the answer, otherwise it is really misleading. – Bavaria

Is it possible to have an MSE between two 2D arrays that is equal to 263.562. – Miter

Answer 2

This isn't part of numpy, but it will work with numpy.ndarray objects. A numpy.matrix can be converted to a numpy.ndarray and a numpy.ndarray can be converted to a numpy.matrix.

from sklearn.metrics import mean_squared_error
mse = mean_squared_error(A, B)

See Scikit Learn mean_squared_error for documentation on how to control axis.

Rivulet

Answer 3

Even more numpy

np.square(np.subtract(A, B)).mean()

Florencia

Btw this way is equivalent to the Sci-kitLearn function, nice! – Vicar

Answer 4

Just for kicks

mse = (np.linalg.norm(A-B)**2)/len(A)

Hardening

Answer 5

Another alternative to the accepted answer that avoids any issues with matrix multiplication:

 def MSE(Y, YH):
     return np.square(Y - YH).mean()

From the documents for np.square:

Return the element-wise square of the input.

Despumate

Answer 6

The standard numpy methods for calculation mean squared error (variance) and its square root (standard deviation) are numpy.var() and numpy.std(), see here and here. They apply to matrices and have the same syntax as numpy.mean().

I suppose that the question and the preceding answers might have been posted before these functions became available.

Remarks on statistics
To answer the comment made by @Drew :
This answer is equivalent to the top answers in this thread. Technically, MSE differs from variance in that it uses "true" value of the parameter, rather than its estimate, see What's the difference between the variance and the mean squared error? and What is the Difference between Variance and MSE?. The two quantities then differ by the bias of our estimate of the central parameter. However, when calculating sample variance, as is done in the OP, we cannot really know the value of this parameter. I believe the OP uses term MSE in a loose sense.

Furthermore, the numpy functions proposed above allow for parameter ddof (the number of degrees of freedom), which allows to obtain unbiased variance estimates (contrary to what is claimed in some superficial comparisons between python and R.)

Villenage

MSE and variance are not the same unless the mean is zero (i.e., unless A and B have the same mean so that A-B has mean zero in the calculations above). – Sadiesadira

Answer 7

What about this to keep with the np.operation style?

mse = np.mean(np.square(A - B))

Just keep in mind that np.mean() with no axis keyword argument specified will output a scalar, just like np.sum().

Daytime