2d numpy数组的乘法/除法以生成3d数组 [英] Multiplication/Division of 2d numpy arrays to produce 3d array

问题描述

我正在寻找一种快速(即矢量化)的方法来替换以下循环.我有2个numpy数组，其尺寸为(20738,14)和(31,14).我需要将它们逐个元素相乘以获得(20738,31,14)的数组.我一直在尝试各种广播配置，但似乎无法获得理想的结果.

I am looking for a fast (ie vectorized) method to replace the following loop. I have 2 numpy arrays with dimensions: (20738,14) and (31,14). I need to multiply them element-wise to get an array of (20738,31,14). I've been experimenting with various configurations of broadcasting, but can't seem to get the desired result.

    v_mu_3d = np.zeros((v.shape[0], ALT_LEN, NEST_LEN))
    for k in range(NEST_LEN):
        v_mu_3d[:,:,k] = v * MU[:,k]
    v_mu_3d = np.exp(v_mu_3d)

以下是类似的操作:

p2_3d = np.zeros((v.shape[0], ALT_LEN, NEST_LEN))
    for j in range(ALT_LEN):
        num = v_mu_3d[:,:,:].sum(axis=1)
        temp_MU = MU[j,:]
        num = ne.evaluate('where(temp_MU >0, num / temp_MU, 0)')
        denom = num.sum(axis=1)
        denom = denom[:, np.newaxis]
        p2_3d[:, j, :] = num / denom

我可以将底部替换为:

p2_3d = v_mu_3d.sum(axis=1) / v_mu_3d.sum(axis=2).sum(axis=1)[:,None]

但似乎无法弄清楚:

num / MU

，它应采用(20738,14)num数组，然后将元素逐个除以(31,14)MU数组. np.repeat()可能有效，但是速度至关重要，因为包含函数在最小化例程中运行了数千次.

which should take the (20738,14) num array and divide element-wise by the (31,14) MU array. np.repeat() might work, but speed is critical because the containing function is run several thousand times in a minimization routine.

更新 根据Nils Werner的回答，这些循环可以简化为: 1)

UPDATE Based on the answer by Nils Werner, these loops can be reduced to: 1)

p1_3d = v_mu_3d / v_mu_3d.sum(axis=1)[:,None,:]

和2)

num = v_mu_3d.sum(axis=1)
num = np.where(MU>0, num[:,None,:] / MU, 0)
p2_3d = num / num.sum(axis=1)[:,None,:]

推荐答案

只需使用广播:

v.shape
# (20738, 14)
MU.shape
# (31, 14)

v_mu_3d = v[:, None, :] * MU
p2_3d = v[:, None, :] / MU

v_mu_3d.shape
# (20738, 31, 14)
p2_3d.shape
# (20738, 31, 14)

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