I have a large scipy.sparse.csc_matrix and would like to normalize it. That is subtract the column mean from each element and divide by the column standard deviation (std)i.
scipy.sparse.csc_matrix has a .mean() but is there an efficient way to compute the variance or std?
You can calculate the variance yourself using the mean, with the following formula:
E[X^2] - (E[X])^2
E[X] stands for the mean. So to calculate E[X^2] you would have to square the csc_matrix and then use the mean function. To get (E[X])^2 you simply need to square the result of the mean function obtained using the normal input.
Sicco has the better answer.
However, another way is to convert the sparse matrix to a dense numpy array one column at a time (to keep the memory requirements lower compared to converting the whole matrix at once):
# mat is the sparse matrix
# Get the number of columns
cols = mat.shape[1]
arr = np.empty(shape=cols)
for i in range(cols):
arr[i] = np.var(mat[:, i].toarray())
The efficient way is actually to densify the entire matrix, then standardize it in the usual way with
X = X.toarray()
X -= X.mean()
X /= X.std()
As @Sebastian has noted in his comments, standardizing destroys the sparsity structure (introduces lots of non-zero elements) in the subtraction step, so there's no use keeping the matrix in a sparse format.
