根据另一个参考数组从一个数组中选择密切匹配

我有一个数组A和一个参考数组B。尺寸为A至少和B. e.g.

A = [2,100,300,793,1300,1500,1810,2400]
B = [4,305,789,1234,1890]

B实际上是指定时间信号中峰值的位置，并且A包含稍后时间的峰值位置。但其中的一些元素A实际上不是我想要的峰值（可能是由于噪音等），我想找到“真实”的峰值A基于B。中的“真实”元素A应该接近那些在B，在上面给出的示例中，“真实”的A应该A'=[2,300,793,1300,1810]。在这个例子中应该很明显的是100,1500,2400不是我们想要的，因为它们距离 B 中的任何元素都很远。我如何在 python/matlab 中以最有效/准确的方式编码？

方法#1： With NumPy broadcasting https://docs.scipy.org/doc/numpy/user/basics.broadcasting.html，我们可以在输入数组之间查找绝对逐元素减法，并使用适当的阈值来过滤掉不需要的元素A。对于给定的样本输入，阈值似乎为90 works.

因此，我们会有一个实现，就像这样 -

thresh = 90
Aout = A[(np.abs(A[:,None] - B) < thresh).any(1)]

样本运行 -

In [69]: A
Out[69]: array([   2,  100,  300,  793, 1300, 1500, 1810, 2400])

In [70]: B
Out[70]: array([   4,  305,  789, 1234, 1890])

In [71]: A[(np.abs(A[:,None] - B) < 90).any(1)]
Out[71]: array([   2,  300,  793, 1300, 1810])

方法#2：基于this post https://stackoverflow.com/a/40716474/3293881，这是一种使用内存有效的方法np.searchsorted https://docs.scipy.org/doc/numpy/reference/generated/numpy.searchsorted.html，这对于大型阵列可能至关重要 -

def searchsorted_filter(a, b, thresh):
    choices = np.sort(b) # if b is already sorted, skip it
    lidx = np.searchsorted(choices, a, 'left').clip(max=choices.size-1)
    ridx = (np.searchsorted(choices, a, 'right')-1).clip(min=0)
    cl = np.take(choices,lidx) # Or choices[lidx]
    cr = np.take(choices,ridx) # Or choices[ridx]
    return a[np.minimum(np.abs(a - cl), np.abs(a - cr)) < thresh]

样本运行 -

In [95]: searchsorted_filter(A,B, thresh = 90)
Out[95]: array([   2,  300,  793, 1300, 1810])

运行时测试

In [104]: A = np.sort(np.random.randint(0,100000,(1000)))

In [105]: B = np.sort(np.random.randint(0,100000,(400)))

In [106]: out1 = A[(np.abs(A[:,None] - B) < 10).any(1)]

In [107]: out2 = searchsorted_filter(A,B, thresh = 10)

In [108]: np.allclose(out1, out2)  # Verify results
Out[108]: True

In [109]: %timeit A[(np.abs(A[:,None] - B) < 10).any(1)]
100 loops, best of 3: 2.74 ms per loop

In [110]: %timeit searchsorted_filter(A,B, thresh = 10)
10000 loops, best of 3: 85.3 µs per loop

2018 年 1 月更新，性能进一步提升

我们可以避免第二次使用np.searchsorted(..., 'right')通过利用获得的指数np.searchsorted(..., 'left')还有absolute计算，就像这样 -

def searchsorted_filter_v2(a, b, thresh):
    N = len(b)

    choices = np.sort(b) # if b is already sorted, skip it

    l = np.searchsorted(choices, a, 'left')
    l_invalid_mask = l==N
    l[l_invalid_mask] = N-1
    left_offset = choices[l]-a
    left_offset[l_invalid_mask] *= -1    

    r = (l - (left_offset!=0))
    r_invalid_mask = r<0
    r[r_invalid_mask] = 0
    r += l_invalid_mask
    right_offset = a-choices[r]
    right_offset[r_invalid_mask] *= -1

    out = a[(left_offset < thresh) | (right_offset < thresh)]
    return out

更新了时序以测试进一步的加速 -

In [388]: np.random.seed(0)
     ...: A = np.random.randint(0,1000000,(100000))
     ...: B = np.unique(np.random.randint(0,1000000,(40000)))
     ...: np.random.shuffle(B)
     ...: thresh = 10
     ...: 
     ...: out1 = searchsorted_filter(A, B, thresh)
     ...: out2 = searchsorted_filter_v2(A, B, thresh)
     ...: print np.allclose(out1, out2)
True

In [389]: %timeit searchsorted_filter(A, B, thresh)
10 loops, best of 3: 24.2 ms per loop

In [390]: %timeit searchsorted_filter_v2(A, B, thresh)
100 loops, best of 3: 13.9 ms per loop

深层发掘 -

In [396]: a = A; b = B

In [397]: N = len(b)
     ...: 
     ...: choices = np.sort(b) # if b is already sorted, skip it
     ...: 
     ...: l = np.searchsorted(choices, a, 'left')

In [398]: %timeit np.sort(B)
100 loops, best of 3: 2 ms per loop

In [399]: %timeit np.searchsorted(choices, a, 'left')
100 loops, best of 3: 10.3 ms per loop

似乎searchsorted and sort占用了几乎所有的运行时间，并且它们对于该方法来说似乎是必不可少的。因此，使用这种基于排序的方法似乎无法进一步改进。