有没有办法使用 numpy 删除循环？

Question

我有一个 3D numpy 数组 input_data (q x m x n)，我用它来构建直方图数据以最终绘制，它存储在 plot_data (m x n x 2) 中。这一步是我流程中的一个不错的瓶颈，我想知道是否有更快、更“numpy”的方式来做到这一点。

num_bins = 3
for i in range(m):

    for j in range(n):

        data = input_data[:, i, j]

        hist, bins = np.histogram(data, bins=num_bins)

        # Create the (x, y) pairs to plot
        plot_data[i][j] = np.stack((bins[:-1], hist), axis=1)

Answer 1

这是一种用于一般 bin 数量的矢量化方法 -

def vectorized_app(input_data, num_bins):
    s0 = input_data.min(0)
    s1 = input_data.max(0)

    m,n,r = input_data.shape
    ids = (num_bins*((input_data - s0)/(s1-s0))).astype(int).clip(max=num_bins-1)
    offset = num_bins*(r*np.arange(n)[:,None] + np.arange(r))
    ids3D = ids + offset
    count3D = np.bincount(ids3D.ravel(), minlength=n*r*num_bins).reshape(n,r,-1)
    bins3D = create_ranges_nd(s0, s1, num_bins+1)[...,:-1]

    out = np.empty((n,r,num_bins,2))
    out[...,0] = bins3D
    out[...,1] = count3D
    return out

辅助函数 -

# https://stackoverflow.com/a/46694364/ @Divakar
def create_ranges_nd(start, stop, N, endpoint=True):
    if endpoint==1:
        divisor = N-1
    else:
        divisor = N
    steps = (1.0/divisor) * (stop - start)
    return start[...,None] + steps[...,None]*np.arange(N)

运行时测试

原始方法-

def org_app(input_data, num_bins):
    q,m,n = input_data.shape
    plot_data = np.zeros((m,n,num_bins,2))
    for i in range(m):
        for j in range(n):
            data = input_data[:, i, j]
            hist, bins = np.histogram(data, bins=num_bins)
            plot_data[i][j] = np.stack((bins[:-1], hist), axis=1)
    return plot_data

时间和验证 -

让我们测试一个形状为(100, 100, 100) 的大型数据数组，bin 的数量为10：

In [967]: # Setup input
     ...: num_bins = 10
     ...: m = 100
     ...: n = 100
     ...: q = 100
     ...: input_data = np.random.rand(q,m,n)
     ...: 
     ...: out1 = org_app(input_data, num_bins)
     ...: out2 = vectorized_app(input_data, num_bins)
     ...: print np.allclose(out1, out2)
     ...: 
True

In [968]: %timeit org_app(input_data, num_bins)
1 loop, best of 3: 748 ms per loop

In [969]: %timeit vectorized_app(input_data, num_bins)
100 loops, best of 3: 12.7 ms per loop

In [970]: 748/12.7 # speedup with vectorized one over original
Out[970]: 58.89763779527559

Answer 2

我认为您的样本相似，因此直方图相似。在这种情况下，您可以简化比较并以更加矢量化的方式进行：

a=np.random.rand(100000,10,10)


def f():  # roughly your approach.
    plotdata=np.zeros((10,10,3),np.int32)
    for i in range(10):
        for j in range(10):
            bins,hist=np.histogram(a[:,i,j],3)
            plotdata[i,j]=bins 
    return plotdata

def g(): #vectored comparisons 
    u=(a < 1/3).sum(axis=0)
    w=(a > 2/3).sum(axis=0)
    v=len(a)-u-w
    return np.dstack((u,v,w))

对于 8 倍的改进：

In [213]: %timeit f()
548 ms ± 15.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [214]: %timeit g()
77.7 ms ± 5.46 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)