概述

作为算法的一部分,我需要为每个观察向量Oi计算T Oi,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 Oi失败,因为形状不一致).@H_419_3@

目前,我正在采取将训练向量平铺为500,000×65536稀疏矩阵的非常低效的步骤：@H_419_3@

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.@H_419_3@

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.@H_419_3@

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,indices.size),indices.itemsize))
mtraining.indptr = np.arange(0,len(indices)*mtraining.shape[0]+1,len(indices),dtype=np.int32)
mtraining.data = vdata
mtraining.indices = vindices

但是这不起作用,因为跨步视图mtraining.data和mtraining.indices是错误的形状(根据this answer,没有办法使它成为正确的形状).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.@H_419_3@

有没有办法完成这项工作？@H_419_3@

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

>>> b = sps.csr_matrix((np.array([1,4,5]),...                     np.array([0,0]),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([[6,4]])

因此,您甚至不必在训练向量和观察矩阵的每一行之间寻找巧合来添加它们：只需用正确的指针填充所有数据,并且需要求和的东西将得到求和何时访问数据.@H_419_3@

编辑@H_419_3@

鉴于第一个代码的速度很慢,您可以按如下方式将内存换成速度：@H_419_3@

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,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

它没有以前那么快,但它可以在大约1秒内完成10,000行：@H_419_3@

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@H_419_3@

def csr_add_sparse_vec(sps_mat,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,[col] * rows)
        new_pointer += aux_idx

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

总结

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