Question

我正在计算两个大向量集（具有相同特征）之间的余弦相似度。每组向量均表示为稀疏的CSR稀疏矩阵A和B。我要计算A x B ^ T ，它不会稀疏。但是，我只需要跟踪超过某个阈值的值，例如0.8。我正在尝试使用带有普通RDD的Pyspark来实现此目标，并想到使用对Scipy CSR矩阵实施的快速矢量运算。

A和B的行被标准化，因此要计算余弦相似度，我只需要找到A的每一行与B的每一行的点积。 A的尺寸为5,000,000 x 5,000。 B的尺寸为2,000,000 x 5,000。

假设A和B太大而无法作为广播变量放入我的工作程序节点上的内存中。我应该如何以最佳方式并行处理A和B？

编辑发布解决方案后，我一直在探索其他更清晰，更优化的方法，尤其是为Spark MLlib IndexedRowMatrix对象实现的columnSimilarities（）函数。（Which pyspark abstraction is appropriate for my large matrix multiplication?）

Answer 1

我能够在此框架中实现解决方案。
欢迎了解为什么这种解决方案很慢-是自定义序列化吗？

def csr_mult_helper(pair):
    threshold=0.8
    A_row = pair[0][0]  # keep track of the row offset
    B_col = pair[1][0]   # offset for B (this will be a column index, after the transpose op)
    A = sparse.csr_matrix(pair[0][1], pair[0][2])  # non-zero entires, size data
    B = sparse.csr_matrix(pair[1][1], pair[1][2])

    C = A * B.T  # scipy sparse mat mul

    for row_idx, row in enumerate(C):  # I think it would be better to use a filter Transformation instead
        col_indices = row.indices      #  but I had trouble with the row and column index book keeping
        col_values = row.data

        for col_idx, val in zip(col_indices, col_values):
            if val > threshold:
                yield (A_row + row_idx, B_col + col_idx, val)  # source vector, target vector, cosine score            

def parallelize_sparse_csr(M, rows_per_chunk=1):
    [rows, cols] = M.shape
    i_row = 0
    submatrices = []
    while i_row < rows:
        current_chunk_size = min(rows_per_chunk, rows - i_row)
        submat = M[i_row:(i_row + current_chunk_size)]
        submatrices.append(   (i_row,                                #  offset
                              (submat.data, submat.indices, submat.indptr),  # sparse matrix data
                              (current_chunk_size, cols)) )      # sparse matrix shape
        i_row += current_chunk_size
    return sc.parallelize(submatrices)

########## generate test data ###########
K,L,M,N = 5,2000,3,2000  # matrix dimensions (toy example)
A_ = sparse.rand(K,L, density=0.1, format='csr')
B_ = sparse.rand(M,N, density=0.1, format='csr')
print("benchmark: {} \n".format((A_ * B_.T).todense()))  # benchmark solution for comparison

########## parallelize, multiply, and filter #########
t_start = time.time()
A = parallelize_sparse_csr(A_, rows_per_chunk=10)
B = parallelize_sparse_csr(B_, rows_per_chunk=10) # number of elements per partition, from B
            # warning: this code breaks if the B_ matrix rows_per_chunk parameter != 1
            # although I don't understand why yet

print("custom pyspark solution: ")
result = A.cartesian(B).flatMap(csr_mult_helper).collect()
print(results)

print("\n {} s elapsed".format(time.time() - t_start))

使用pyspark并行化Scipy CSR稀疏矩阵进行大矩阵乘法

1 个答案: