CODEWITHSUNDEEP
pythonscipysparse-matrix
user3473016
user3473016

Reputation: 65

Replace elements in sparse matrix created by Scipy (Python)

I have a huge sparse matrix in Scipy and I would like to replace numerous elements inside by a given value (let's say -1).

Is there a more efficient way to do it than using:

SM[[rows],[columns]]=-1

Here is an example:

Nr=seg.shape[0] #size ~=50000

Im1=sparse.csr_matrix(np.append(np.array([-1]),np.zeros([1,Nr-1])))
Im1=sparse.csr_matrix(sparse.vstack([Im1,sparse.eye(Nr)]))
Im1[prev[1::]-1,Num[1::]-1]=-1 # this line is very slow

Im2=sparse.vstack([sparse.csr_matrix(np.zeros([1,Nr])),sparse.eye(Nr)])

IM=sparse.hstack([Im1,Im2]) #final result

Upvotes: 1

Views: 2325

Answers (1)

hpaulj
hpaulj

Reputation: 231665

I've played around with your sparse arrays. I'd encourage you to do some timings on smaller sizes, to see how different methods and sparse types behave. I like to use timeit in Ipython.

Nr=10 # seg.shape[0] #size ~=50000
Im2=sparse.vstack([sparse.csr_matrix(np.zeros([1,Nr])),sparse.eye(Nr)])

Im2 has a zero first row, and offset diagonal on the rest. So it's simpler, though not much faster, to start with an empty sparse matrix:

X = sparse.vstack([sparse.csr_matrix((1,Nr)),sparse.eye(Nr)])

Or use diags to construct the offset diagonal directly:

X = sparse.diags([1],[-1],shape=(Nr+1, Nr))

Im1 is similar, except it has a -1 in the (0,0) slot. How about stacking 2 diagonal matrices?

X = sparse.vstack([sparse.diags([-1],[0],(1,Nr)),sparse.eye(Nr)])

Or make the offset diagonal (copy Im2?), and modify [0,0]. A csr matrix gives an efficiency warning, recommending the use of lil format. It does, though, take some time to convert tolil().

X = sparse.diags([1],[-1],shape=(Nr+1, Nr)).tolil()
X[0,0] = -1  # slow warning with csr

Let's try your larger insertions:

prev = np.arange(Nr-2)  # what are these like?
Num = np.arange(Nr-2)
Im1[prev[1::]-1,Num[1::]-1]=-1

With Nr=10, and various Im1 formats:

lil - 267 us
csr - 1.44 ms
coo - not supported
todense - 25 us

OK, I've picked prev and Num such that I end up modifying diagonals of Im1. In this case it would be faster to construct those diagonals right from the start.

X2=Im1.todia()
print X2.data
[[ 1.  1.  1.  1.  1.  1.  1.  1.  1.  1.]
 [-1. -1. -1. -1. -1. -1. -1.  0.  0.  0.]]
print X2.offsets
[-1  0]

You may have to learn how various sparse formats are stored. csr and csc are a bit complex, designed for fast linear algebra operations. lil, dia, coo are simpler to understand.

Upvotes: 1

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