在numpy/scipy中为稀疏的矩阵添加一个非常重复的矩阵？

Question

我试图实现与NumPy/SciPy的一个函数来计算詹森-香农散单(培训)向量和大量其他(观察)向量之间.观察向量存储在非常大的(500,000x65536)Scipy稀疏矩阵中(密集矩阵不适合存储器).

作为算法的一部分,我需要计算T +ø _我对于每个观测矢量O,_我,其中T是训练向量.我无法使用NumPy的常规广播规则找到一种方法,因为稀疏矩阵似乎不支持那些(如果T保留为密集阵列,Scipy会尝试使稀疏矩阵首先密集,哪些运行内存不足;如果我将T变成稀疏矩阵,则T + O _i失败,因为形状不一致).

目前,我正在采取将训练向量平铺为500,000x65536稀疏矩阵的非常低效的步骤:

training = sp.csr_matrix(training.astype(np.float32))
tindptr = np.arange(0, len(training.indices)*observations.shape[0]+1, len(training.indices), dtype=np.int32)
tindices = np.tile(training.indices, observations.shape[0])
tdata = np.tile(training.data, observations.shape[0])
mtraining = sp.csr_matrix((tdata, tindices, tindptr), shape=observations.shape)

但是当它只存储~1500个"真实"元素时,它占用了大量的内存(大约6GB).构建起来也很慢.

我试图通过使用stride_tricks使CSR矩阵的indptr变得聪明,数据成员不会在重复数据上使用额外的内存.

training = sp.csr_matrix(training)
mtraining = sp.csr_matrix(observations.shape,dtype=np.int32)
tdata = training.data
vdata = np.lib.stride_tricks.as_strided(tdata, (mtraining.shape[0], tdata.size), (0, tdata.itemsize))
indices = training.indices
vindices = np.lib.stride_tricks.as_strided(indices, (mtraining.shape[0], indices.size), (0, indices.itemsize))
mtraining.indptr = np.arange(0, len(indices)*mtraining.shape[0]+1, len(indices), dtype=np.int32)
mtraining.data = vdata
mtraining.indices = vindices

但是,这并不工作,因为跨入意见mtraining.data和mtraining.indices是错误的形状(根据这个答案有没有办法让它正确的形状).尝试使用.flat迭代器使它们看起来平坦失败,因为它看起来不像数组(例如它没有dtype成员),并且使用flatten()方法最终制作副本.

有没有办法完成这项工作？

Answer 1

您的另一个选择（我什至没有考虑过）是自己以稀疏格式实现总和，以便您可以充分利用数组的周期性性质。如果你滥用 scipy 稀疏矩阵的这种特殊行为，这可以很容易做到：

>>> a = sps.csr_matrix([1,2,3,4])
>>> a.data
array([1, 2, 3, 4])
>>> a.indices
array([0, 1, 2, 3])
>>> a.indptr
array([0, 4])

>>> b = sps.csr_matrix((np.array([1, 2, 3, 4, 5]),
...                     np.array([0, 1, 2, 3, 0]),
...                     np.array([0, 5])), shape=(1, 4))
>>> b
<1x4 sparse matrix of type '<type 'numpy.int32'>'
    with 5 stored elements in Compressed Sparse Row format>
>>> b.todense()
matrix([[6, 2, 3, 4]])

因此，您甚至不必寻找训练向量与观察矩阵的每一行之间的重合来将它们相加：只需将所有数据与正确的指针填入其中，需要求和的内容将得到求和当数据被访问时。

编辑

鉴于第一个代码的速度很慢，您可以用内存换取速度，如下所示：

def csr_add_sparse_vec(sps_mat, sps_vec) :
    """Adds a sparse vector to every row of a sparse matrix"""
    # No checks done, but both arguments should be sparse matrices in CSR
    # format, both should have the same number of columns, and the vector
    # should be a vector and have only one row.

    rows, cols = sps_mat.shape
    nnz_vec = len(sps_vec.data)
    nnz_per_row = np.diff(sps_mat.indptr)
    longest_row = np.max(nnz_per_row)

    old_data = np.zeros((rows * longest_row,), dtype=sps_mat.data.dtype)
    old_cols = np.zeros((rows * longest_row,), dtype=sps_mat.indices.dtype)

    data_idx = np.arange(longest_row) < nnz_per_row[:, None]
    data_idx = data_idx.reshape(-1)
    old_data[data_idx] = sps_mat.data
    old_cols[data_idx] = sps_mat.indices
    old_data = old_data.reshape(rows, -1)
    old_cols = old_cols.reshape(rows, -1)

    new_data = np.zeros((rows, longest_row + nnz_vec,),
                        dtype=sps_mat.data.dtype)
    new_data[:, :longest_row] = old_data
    del old_data
    new_cols = np.zeros((rows, longest_row + nnz_vec,),
                        dtype=sps_mat.indices.dtype)
    new_cols[:, :longest_row] = old_cols
    del old_cols
    new_data[:, longest_row:] = sps_vec.data
    new_cols[:, longest_row:] = sps_vec.indices
    new_data = new_data.reshape(-1)
    new_cols = new_cols.reshape(-1)
    new_pointer = np.arange(0, (rows + 1) * (longest_row + nnz_vec),
                            longest_row + nnz_vec)

    ret = sps.csr_matrix((new_data, new_cols, new_pointer),
                         shape=sps_mat.shape)
    ret.eliminate_zeros()

    return ret

它不像以前那么快，但它可以在大约 1 秒内完成 10,000 行：

In [2]: a
Out[2]: 
<10000x65536 sparse matrix of type '<type 'numpy.float64'>'
    with 15000000 stored elements in Compressed Sparse Row format>

In [3]: b
Out[3]: 
<1x65536 sparse matrix of type '<type 'numpy.float64'>'
    with 1500 stored elements in Compressed Sparse Row format>

In [4]: csr_add_sparse_vec(a, b)
Out[4]: 
<10000x65536 sparse matrix of type '<type 'numpy.float64'>'
    with 30000000 stored elements in Compressed Sparse Row format>

In [5]: %timeit csr_add_sparse_vec(a, b)
1 loops, best of 3: 956 ms per loop

编辑此代码非常非常慢

def csr_add_sparse_vec(sps_mat, sps_vec) :
    """Adds a sparse vector to every row of a sparse matrix"""
    # No checks done, but both arguments should be sparse matrices in CSR
    # format, both should have the same number of columns, and the vector
    # should be a vector and have only one row.

    rows, cols = sps_mat.shape

    new_data = sps_mat.data
    new_pointer = sps_mat.indptr.copy()
    new_cols = sps_mat.indices

    aux_idx = np.arange(rows + 1)

    for value, col in itertools.izip(sps_vec.data, sps_vec.indices) :
        new_data = np.insert(new_data, new_pointer[1:], [value] * rows)
        new_cols = np.insert(new_cols, new_pointer[1:], [col] * rows)
        new_pointer += aux_idx

    return sps.csr_matrix((new_data, new_cols, new_pointer),
                          shape=sps_mat.shape)