Delete columns of matrix of CSR format in Python

Question

I have a sparse matrix (22000x97482) in csr format and i want to delete some columns (indices of columns numbers are stored in a list)

Answer 1

If you have a very large number of columns then generating the full set of column indices can become rather costly. One slightly faster alternative would be to temporarily convert to COO format:

import numpy as np
from scipy import sparse

def dropcols_fancy(M, idx_to_drop):
    idx_to_drop = np.unique(idx_to_drop)
    keep = ~np.in1d(np.arange(M.shape[1]), idx_to_drop, assume_unique=True)
    return M[:, np.where(keep)[0]]

def dropcols_coo(M, idx_to_drop):
    idx_to_drop = np.unique(idx_to_drop)
    C = M.tocoo()
    keep = ~np.in1d(C.col, idx_to_drop)
    C.data, C.row, C.col = C.data[keep], C.row[keep], C.col[keep]
    C.col -= idx_to_drop.searchsorted(C.col)    # decrement column indices
    C._shape = (C.shape[0], C.shape[1] - len(idx_to_drop))
    return C.tocsr()

Check equivalence:

m, n, d = 1000, 2000, 20

M = sparse.rand(m, n, format='csr')
idx_to_drop = np.random.randint(0, n, d)

M_drop1 = dropcols_fancy(M, idx_to_drop)
M_drop2 = dropcols_coo(M, idx_to_drop)

print(np.all(M_drop1.A == M_drop2.A))
# True

Benchmark:

In [1]: m, n = 1000, 1000000

In [2]: %%timeit M = sparse.rand(m, n, format='csr')
   ...: dropcols_fancy(M, idx_to_drop)
   ...: 
1 loops, best of 3: 1.11 s per loop

In [3]: %%timeit M = sparse.rand(m, n, format='csr')
   ...: dropcols_coo(M, idx_to_drop)
   ...: 
1 loops, best of 3: 365 ms per loop

Answer 2

You can use fancy indexing to obtain a new csr_matrix with the columns that you have in your list:

all_cols = np.arange(old_m.shape[1])
cols_to_keep = np.where(np.logical_not(np.in1d(all_cols, cols_to_delete)))[0]
m = old_m[:, cols_to_keep]