设置csr_matrix行

时间:2015-02-10 08:37:23

标签: python performance sparse-matrix

我有一个稀疏的csr_matrix,我想将单行的值更改为不同的值。但是我找不到简单有效的实现方法。这就是它必须做的事情:

A = csr_matrix([[0, 1, 0],
                [1, 0, 1],
                [0, 1, 0]])
new_row = np.array([-1, -1, -1])
print(set_row_csr(A, 2, new_row).todense())

>>> [[ 0,  1, 0],
     [ 1,  0, 1],
     [-1, -1, -1]]

这是我set_row_csr的当前实现:

def set_row_csr(A, row_idx, new_row):
    A[row_idx, :] = new_row
    return A

但这给了我一个SparseEfficiencyWarning。有没有办法在没有手动索引杂耍的情况下完成这项工作,或者这是我唯一的出路?

5 个答案:

答案 0 :(得分:3)

最后,我设法通过索引杂耍完成了这项工作。

def set_row_csr(A, row_idx, new_row):
    '''
    Replace a row in a CSR sparse matrix A.

    Parameters
    ----------
    A: csr_matrix
        Matrix to change
    row_idx: int
        index of the row to be changed
    new_row: np.array
        list of new values for the row of A

    Returns
    -------
    None (the matrix A is changed in place)

    Prerequisites
    -------------
    The row index shall be smaller than the number of rows in A
    The number of elements in new row must be equal to the number of columns in matrix A
    '''
    assert sparse.isspmatrix_csr(A), 'A shall be a csr_matrix'
    assert row_idx < A.shape[0], \
            'The row index ({0}) shall be smaller than the number of rows in A ({1})' \
            .format(row_idx, A.shape[0])
    try:
        N_elements_new_row = len(new_row)
    except TypeError:
        msg = 'Argument new_row shall be a list or numpy array, is now a {0}'\
        .format(type(new_row))
        raise AssertionError(msg)
    N_cols = A.shape[1]
    assert N_cols == N_elements_new_row, \
            'The number of elements in new row ({0}) must be equal to ' \
            'the number of columns in matrix A ({1})' \
            .format(N_elements_new_row, N_cols)

    idx_start_row = A.indptr[row_idx]
    idx_end_row = A.indptr[row_idx + 1]
    additional_nnz = N_cols - (idx_end_row - idx_start_row)

    A.data = np.r_[A.data[:idx_start_row], new_row, A.data[idx_end_row:]]
    A.indices = np.r_[A.indices[:idx_start_row], np.arange(N_cols), A.indices[idx_end_row:]]
    A.indptr = np.r_[A.indptr[:row_idx + 1], A.indptr[(row_idx + 1):] + additional_nnz]

答案 1 :(得分:2)

物理吸引力的答案确实明显更快。它比我的解决方案快得多,后者只是添加一个单独的矩阵与该单行集。虽然添加溶液比切片解决方案更快。

对我而言,在csr_matrix或csc_matrix中的列中设置行的最快方法是自己修改基础数据。

def time_copy(A, num_tries = 10000):
    start = time.time()
    for i in range(num_tries):
        B = A.copy()
    end = time.time()
    return end - start

def test_method(func, A, row_idx, new_row, num_tries = 10000):
    start = time.time()
    for i in range(num_tries):
        func(A.copy(), row_idx, new_row)
    end = time.time()
    copy_time = time_copy(A, num_tries)
    print("Duration {}".format((end - start) - copy_time))

def set_row_csr_slice(A, row_idx, new_row):
    A[row_idx,:] = new_row

def set_row_csr_addition(A, row_idx, new_row):
    indptr = np.zeros(A.shape[1] + 1)
    indptr[row_idx +1:] = A.shape[1]
    indices = np.arange(A.shape[1])
    A += csr_matrix((new_row, indices, indptr), shape=A.shape)

>>> A = csr_matrix((np.ones(1000), (np.random.randint(0,1000,1000), np.random.randint(0, 1000, 1000))))
>>> test_method(set_row_csr_slice, A, 200, np.ones(A.shape[1]), num_tries = 10000)
Duration 4.938395977020264

>>> test_method(set_row_csr_addition, A, 200, np.ones(A.shape[1]), num_tries = 10000)
Duration 2.4161765575408936

>>> test_method(set_row_csr, A, 200, np.ones(A.shape[1]), num_tries = 10000)
Duration 0.8432261943817139

切片解决方案也随着矩阵的大小和稀疏性而变得更加糟糕。

# Larger matrix, same fraction sparsity
>>> A = csr_matrix((np.ones(10000), (np.random.randint(0,10000,10000), np.random.randint(0, 10000, 10000))))
>>> test_method(set_row_csr_slice, A, 200, np.ones(A.shape[1]), num_tries = 10000)
Duration 18.335174798965454

>>> test_method(set_row_csr, A, 200, np.ones(A.shape[1]), num_tries = 10000)
Duration 1.1089558601379395

# Super sparse matrix
>>> A = csr_matrix((np.ones(100), (np.random.randint(0,10000,100), np.random.randint(0, 10000, 100))))
>>> test_method(set_row_csr_slice, A, 200, np.ones(A.shape[1]), num_tries = 10000)
Duration 13.371600151062012

>>> test_method(set_row_csr, A, 200, np.ones(A.shape[1]), num_tries = 10000)
Duration 1.0454308986663818

答案 2 :(得分:0)

这个set_row_csr出了点问题。是的,它很快,似乎适用于一些测试用例。但是,在我的测试用例中,它似乎会破坏csr稀疏矩阵的内部csr结构。之后尝试lil_matrix(A),您会看到错误消息。

答案 3 :(得分:0)

在physicalattraction的答案中,len(new_row)必须等于A.shape[1]在添加稀疏行时可能没什么兴趣。

所以,根据他的回答,我提出了一种在csr中设置行的方法,同时它保留了sparcity属性。另外,我添加了一种方法将密集数组转换为稀疏数组(数据,索引格式)

def to_sparse(dense_arr):
    sparse = [(data, index) for index, data in enumerate(dense_arr) if data != 0]

    # Convert list of tuples to lists
    sparse = list(map(list, zip(*sparse)))

    # Return data and indices
    return sparse[0], sparse[1]

def set_row_csr_unbounded(A, row_idx, new_row_data, new_row_indices):
    '''
    Replace a row in a CSR sparse matrix A.

    Parameters
    ----------
    A: csr_matrix
        Matrix to change
    row_idx: int
        index of the row to be changed
    new_row_data: np.array
        list of new values for the row of A
    new_row_indices: np.array
        list of indices for new row

    Returns
    -------
    None (the matrix A is changed in place)

    Prerequisites
    -------------
    The row index shall be smaller than the number of rows in A
    Row data and row indices must have the same size
    '''
    assert isspmatrix_csr(A), 'A shall be a csr_matrix'
    assert row_idx < A.shape[0], \
            'The row index ({0}) shall be smaller than the number of rows in A ({1})' \
            .format(row_idx, A.shape[0])

    try:
        N_elements_new_row = len(new_row_data)
    except TypeError:
        msg = 'Argument new_row_data shall be a list or numpy array, is now a {0}'\
        .format(type(new_row_data))
        raise AssertionError(msg)

    try:
        assert N_elements_new_row == len(new_row_indices), \
                'new_row_data and new_row_indices must have the same size'
    except TypeError:
        msg = 'Argument new_row_indices shall be a list or numpy array, is now a {0}'\
        .format(type(new_row_indices))
        raise AssertionError(msg)

    idx_start_row = A.indptr[row_idx]
    idx_end_row = A.indptr[row_idx + 1]

    A.data = np.r_[A.data[:idx_start_row], new_row_data, A.data[idx_end_row:]]
    A.indices = np.r_[A.indices[:idx_start_row], new_row_indices, A.indices[idx_end_row:]]
    A.indptr = np.r_[A.indptr[:row_idx + 1], A.indptr[(row_idx + 1):] + N_elements_new_row]

答案 4 :(得分:0)

这是我的方法:

A = A.tolil()
A[index, :] = new_row
A = A.tocsr()

只需转换为lil_matrix,更改行并转换回即可。