If we want to use the permutations multiple times, we should convert the iterator to a list: Note that the itertools.permutations method returns an iterator, so we can only iterate over the permutations once. We can iterate over the permutations using a for loop and print each permutation. We then use the itertools.permutations method to generate all possible permutations of the items. In the code above, we import the itertools module and create a list of items that we want to permute. Permutations = itertools.permutations(items) You can use itertools.permutations to generate permutations in Python in the following way: We will also discuss how to convert an iterator returned by this method into a list for multiple uses. In this blog post, we will look at how to use the itertools.permutations method in Python to do so. Using the itertools module, we can generate all possible permutations of a given list of items. Best Corporate Meeting Management Software Tools.Best Online Meeting Management Software Tools.Best Board Meeting Management Software Tools.Best Online Communication Software Tools.With 22 stored elements in Compressed Sparse Row format> This permutation works with the example matrix in In : B = np.matrix('1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 1 0 1 1 0 1 0 0 0 0 0 0 The MATLAB docs say that with this permutation, the eigenvalues remain the same. In numpy terms, that means: In : I,J=np.ix_(permutation_array,permutation_array)Īnd indeed there are more 0 bands in the 2 off diagonal corners.Īnd using the bandwidth calculation on the MATLAB page, In : i,j=A.nonzero() This is a permutation r such that S(r,r) tends to have its nonzero elements closer to the diagonal. R = symrcm(S) returns the symmetric reverse Cuthill-McKee ordering of S. I don't see any improvement in clustering of 1s in res. You on the other hand appear to be doing: In : res = sparse.lil_matrix(A.shape,dtype=A.dtype) I see some clustering maybe the best we can expect from a random matrix. My first inclination is to use such an array to simply index rows of the original matrix: In : A.A (oops, used the wrong matrix here): In : permutation_array = csgraph.reverse_cuthill_mckee(A, False) In : A = sparse.csr_matrix((np.ones().astype(np.uint8),ran.indices, ran.indptr)) Make a random matrix (float) and convert it to a uint8 (beware, csr calculations might not work with this dtype): In : ran=sparse.random(10,10.3, format='csr') I wonder if you are applying the permutation array correctly. So this array is something like: (size = 200,000) The result of the reverse_cuthill_mckee call is an array which is like a tupel containing the indices for my permutation. Permutation_array = csgraph.reverse_cuthill_mckee(A, false) I tried a for-loop but my matrix has dimension like 200,000 x 150,000 and it takes too much time. Now is there any efficient solution for doing this permutation on my sparse csr_matrix in any other sparse matrix (csr, lil_matrix, etc)? The result of this method is a permutation array whichs gives me the indices of how to permutate the rows of my matrix as I understood. I used the Scipy Reverse Cuthill-McKee implementation ( _cuthill_mckee) for creating a band matrix using a (high-dimensional) sparse csr_matrix.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |