# Given that A is a sparse matrix, norm(A(i,:)) takes a very long time. Why and can one do better ?

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Cem Gormezano on 9 Aug 2020
Edited: Walter Roberson on 11 Aug 2020
I am computing norm(A(i,:)) within a loop, where A is a sparse matrix. This seems to be a bottleneck, is there anyway to perform this faster ? Thank you.

#### 1 Comment

Walter Roberson on 10 Aug 2020
Can you recode with the matrix transposed? Accessing sparse by columns is faster than by row.

John D'Errico on 10 Aug 2020
You really don't want to use loops to do something like this.
>> A = sprand(10000,10000,0.001);
>> timeit(@() norm(A(1,:)))
ans =
0.000121650132
>> timeit(@() sqrt(sum(A.^2,2)))
ans =
0.000442972132
So, the time to compute the norm of just one of the rows was a significant fraction of the total time to compute the norm of all rows at once.

James Tursa on 9 Aug 2020
Edited: James Tursa on 9 Aug 2020
Will you eventually need all of the rows? E.g., do this once at the beginning outside the loop
n = sqrt(sum(A.^2,2))
And then just pick off the element you want each iteration.
Alternatively, you could do this calculation one row at a time in a mex routine in order to avoid the data copy associated with forming the A(i,:) row explicitly.

Walter Roberson on 11 Aug 2020
Edited: Walter Roberson on 11 Aug 2020
If you were going to do this repeatedly, then
%one time cost
At = A.';
%then
norm(At(:,1))
is over 100 times faster than
norm(A(1,:))
For John's suggestion of A = sprand(10000,10000,0.001); then the cost of transposing A is roughly the same as the cost of calculating 11 row norms of A.
... And of course, it might happen to be practical to work with the transposed version instead of the original, so in some cases there would be no computation cost, just the cost of changing the code.