如何点积(1，10^{13})(10^13，1)实数稀疏矩阵
问题描述
基本上就是标题所包含的内容。 这两个矩阵几乎都是零。第一个是1 x 9999999999999，第二个是9999999999999 x 1 当我尝试做点积时，我得到了这样的结果。
Unable to allocate 72.8 TiB for an array with shape (10000000000000,) and data type int64
Full traceback </br>
MemoryError: Unable to allocate 72.8 TiB for an array with shape (10000000000000,) and data type int64
In [31]: imputed.dot(s)
---------------------------------------------------------------------------
MemoryError Traceback (most recent call last)
<ipython-input-31-670cfc69d4cf> in <module>
----> 1 imputed.dot(s)
~/.local/lib/python3.8/site-packages/scipy/sparse/base.py in dot(self, other)
357
358 """
--> 359 return self * other
360
361 def power(self, n, dtype=None):
~/.local/lib/python3.8/site-packages/scipy/sparse/base.py in __mul__(self, other)
478 if self.shape[1] != other.shape[0]:
479 raise ValueError('dimension mismatch')
--> 480 return self._mul_sparse_matrix(other)
481
482 # If it's a list or whatever, treat it like a matrix
~/.local/lib/python3.8/site-packages/scipy/sparse/compressed.py in _mul_sparse_matrix(self, other)
499
500 major_axis = self._swap((M, N))[0]
--> 501 other = self.__class__(other) # convert to this format
502
503 idx_dtype = get_index_dtype((self.indptr, self.indices,
~/.local/lib/python3.8/site-packages/scipy/sparse/compressed.py in __init__(self, arg1, shape, dtype, copy)
32 arg1 = arg1.copy()
33 else:
---> 34 arg1 = arg1.asformat(self.format)
35 self._set_self(arg1)
36
~/.local/lib/python3.8/site-packages/scipy/sparse/base.py in asformat(self, format, copy)
320 # Forward the copy kwarg, if it's accepted.
321 try:
--> 322 return convert_method(copy=copy)
323 except TypeError:
324 return convert_method()
~/.local/lib/python3.8/site-packages/scipy/sparse/csc.py in tocsr(self, copy)
135 idx_dtype = get_index_dtype((self.indptr, self.indices),
136 maxval=max(self.nnz, N))
--> 137 indptr = np.empty(M + 1, dtype=idx_dtype)
138 indices = np.empty(self.nnz, dtype=idx_dtype)
139 data = np.empty(self.nnz, dtype=upcast(self.dtype))
MemoryError: Unable to allocate 72.8 TiB for an array with shape (10000000000000,) and data type int64
似乎scipy正在尝试创建一个temp数组。 我使用的是Scipy提供的.ot方法。 我也对不重要的解决方案持开放态度。 谢谢！
解决方案
In [105]: from scipy import sparse
如果我制作一个(100，1)CSR矩阵：
In [106]: A = sparse.random(100,1,format='csr')
In [107]: A
Out[107]:
<100x1 sparse matrix of type '<class 'numpy.float64'>'
with 1 stored elements in Compressed Sparse Row format>
数据和指数为：
In [109]: A.data
Out[109]: array([0.19060481])
In [110]: A.indices
Out[110]: array([0], dtype=int32)
In [112]: A.indptr
Out[112]:
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
因此，即使只有1个非零项，一个数组也很大(101)。
另一方面，同一数组的csc格式的存储空间要小得多。但形状为(1,100)的csc将看起来像CSR。
In [113]: Ac = A.tocsc()
In [114]: Ac.indptr
Out[114]: array([0, 1], dtype=int32)
In [115]: Ac.indices
Out[115]: array([88], dtype=int32)
数学，特别是矩阵乘积是用csr/csc格式完成的。因此，可能很难避免这种80 TB的内存使用。
查看回溯，我发现它正在尝试将other转换为与self匹配的格式。
所以A.dot(B)和A是(1，N)csr，小的形状。B是(N，1)csc，也是小形状。但B.tocsr()需要较大的(N+1,)形状indptr。
让我们尝试dot的替代方案
前2个矩阵：
In [122]: A = sparse.random(1,100, .2,format='csr')
In [123]: B = sparse.random(100,1, .2,format='csc')
In [124]: A
Out[124]:
<1x100 sparse matrix of type '<class 'numpy.float64'>'
with 20 stored elements in Compressed Sparse Row format>
In [125]: B
Out[125]:
<100x1 sparse matrix of type '<class 'numpy.float64'>'
with 20 stored elements in Compressed Sparse Column format>
In [126]: A@B
Out[126]:
<1x1 sparse matrix of type '<class 'numpy.float64'>'
with 1 stored elements in Compressed Sparse Row format>
In [127]: _.A
Out[127]: array([[1.33661021]])
它们的非零元素索引。只有匹配的才重要。
In [128]: A.indices, B.indices
Out[128]:
(array([16, 20, 23, 28, 30, 37, 39, 40, 43, 49, 54, 59, 61, 63, 67, 70, 74,
91, 94, 99], dtype=int32),
array([ 5, 8, 15, 25, 34, 35, 40, 46, 47, 51, 53, 60, 68, 70, 75, 81, 87,
90, 91, 94], dtype=int32))
等式矩阵：
In [129]: mask = A.indices[:,None]==B.indices
In [132]: np.nonzero(mask.any(axis=0))
Out[132]: (array([ 6, 13, 18, 19]),)
In [133]: np.nonzero(mask.any(axis=1))
Out[133]: (array([ 7, 15, 17, 18]),)
匹配索引：
In [139]: A.indices[Out[133]]
Out[139]: array([40, 70, 91, 94], dtype=int32)
In [140]: B.indices[Out[132]]
Out[140]: array([40, 70, 91, 94], dtype=int32)
相应data值的总和与[127]
In [141]: (A.data[Out[133]]*B.data[Out[132]]).sum()
Out[141]: 1.3366102138511582
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