我有一个 csr 格式的稀疏矩阵 (22000x97482),我想删除一些列(列号的索引存储在列表中)
最佳答案
如果您有非常多的列,那么生成完整的列索引集可能会变得相当昂贵。一种稍微快一点的替代方法是暂时转换为 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()
检查等价性:
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
基准:
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
关于python - 在Python中删除CSR格式矩阵的列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23966923/