python - 用 2d 矩阵填充 4D 矩阵

Question

我想知道是否有一个 numpy 函数可以使它更快。这是我正在尝试做的一个示例。

def _sparse_4D_rand_mat(self, x, y, density):
    _4D_mat = np.empty((x, y, x, y))
    for i in range(self.size):
        for j in range(self.size):
            _4D_mat[:,i,j,:] = self._rand_sparse(x, y, density)
    return _4D_mat
def _rand_sparse(self, m, n, density, format='csr'):
    nnz = max(min(int(m * n * density), m * n), 0)
    row = np.random.random_integers(low=0, high=m - 1, size=nnz)
    col = np.random.random_integers(low=0, high=n - 1, size=nnz)
    data = np.ones(nnz, dtype=float)
    data = np.random.dirichlet(data)
    return csr_matrix((data, (row, col)), shape=(m, n)).toarray()

感谢您的贡献。我是新人;)

Answer 1

由于密度保持不变，您可以调用一次_rand_sparse来生成一个大的稀疏二维数组，而不是多次调用_rand_sparse来生成许多小的稀疏二维数组，然后使用 reshape 方法将 2D 结果 reshape 为 4D 数组:

_4D_mat = _rand_sparse(x * y * x, y, density)
_4D_mat = _4D_mat.reshape((x, y, x, y))

例如，

import numpy as np
import scipy.sparse as sparse

def _rand_sparse(m, n, density, format='csr'):
    nnz = max(min(int(m * n * density), m * n), 0)
    # use randint since random_integer is deprecated in NumPy 1.11.0
    row = np.random.randint(low=0, high=m, size=nnz)
    col = np.random.randint(low=0, high=n, size=nnz)
    data = np.ones(nnz, dtype=float)
    data = np.random.dirichlet(data)
    return sparse.csr_matrix((data, (row, col)), shape=(m, n)).toarray()

def orig(x, y, density):
    _4D_mat = np.empty((x, y, x, y))
    for i in range(y):
        for j in range(x):
            _4D_mat[:, i, j, :] = _rand_sparse(x, y, density)
    return _4D_mat

def alt(x, y, density):
    _4D_mat = _rand_sparse(x * y * x, y, density)
    _4D_mat = _4D_mat.reshape((x, y, x, y))
    return _4D_mat

x, y, density = 2, 4, 0.5

由于 alt 消除了双 for 循环，因此该解决方案将比 orig 快得多，因为 x 和 的值>y 变大(即，随着 for 循环中的迭代次数增加)。事实上，即使对于上面使用的小值，alt 也已经比 orig 快(几乎 8 倍):

In [108]: %timeit orig(x, y, density)
100 loops, best of 3: 2.24 ms per loop

In [109]: %timeit alt(x, y, density)
1000 loops, best of 3: 281 µs per loop

<小时/>

I need the sum for each 2D array in the 4D array to be 1

要规范化适当的切片，您可以使用:

totals = np.nansum(_4D_mat, axis=0, keepdims=True)
totals = np.nansum(totals, axis=3, keepdims=True)
_4D_mat /= totals