搜索NumPy数组中的序列

Question

假设我有以下数组:

 array([2, 0, 0, 1, 0, 1, 0, 0])

如何获得我出现值序列的索引:[0,0]？因此,这种情况的预期输出将是:[1,2,6,7].

编辑:

1)请注意,这[0,0]只是一个序列.它可能是[0,0,0]或者,[4,6,8,9]或者说[5,2,0],任何东西.

2)如果我的数组被修改为:array([2, 0, 0, 0, 0, 1, 0, 1, 0, 0]),那么具有相同序列的预期结果[0,0]将是[1,2,3,4,8,9].

我正在寻找一些NumPy快捷方式.

Answer 1

嗯,这基本上是template-matching problem图像处理中出现的很多.在这篇文章中列出了两种方法:基于Pure NumPy和基于OpenCV(cv2).

方法#1:使用NumPy,可以2D在输入数组的整个长度上创建一个滑动索引数组.因此,每行将是元素的滑动窗口.接下来,将每一行与输入序列匹配,这将引入broadcasting矢量化解决方案.我们寻找所有True行,表明那些是完美匹配的行,因此将是匹配的起始索引.最后,使用这些索引,创建一系列延伸到序列长度的索引,以便为我们提供所需的输出.实施将是 -

def search_sequence_numpy(arr,seq):
    """ Find sequence in an array using NumPy only.

    Parameters
    ----------    
    arr    : input 1D array
    seq    : input 1D array

    Output
    ------    
    Output : 1D Array of indices in the input array that satisfy the 
    matching of input sequence in the input array.
    In case of no match, an empty list is returned.
    """

    # Store sizes of input array and sequence
    Na, Nseq = arr.size, seq.size

    # Range of sequence
    r_seq = np.arange(Nseq)

    # Create a 2D array of sliding indices across the entire length of input array.
    # Match up with the input sequence & get the matching starting indices.
    M = (arr[np.arange(Na-Nseq+1)[:,None] + r_seq] == seq).all(1)

    # Get the range of those indices as final output
    if M.any() >0:
        return np.where(np.convolve(M,np.ones((Nseq),dtype=int))>0)[0]
    else:
        return []         # No match found

方法#2:使用OpenCV(cv2),我们有一个内置函数template-matching:cv2.matchTemplate.使用它,我们将有起始匹配索引.其余步骤与前一种方法相同.以下是实施cv2:

from cv2 import matchTemplate as cv2m

def search_sequence_cv2(arr,seq):
    """ Find sequence in an array using cv2.
    """

    # Run a template match with input sequence as the template across
    # the entire length of the input array and get scores.
    S = cv2m(arr.astype('uint8'),seq.astype('uint8'),cv2.TM_SQDIFF)

    # Now, with floating point array cases, the matching scores might not be 
    # exactly zeros, but would be very small numbers as compared to others.
    # So, for that use a very small to be used to threshold the scorees 
    # against and decide for matches.
    thresh = 1e-5 # Would depend on elements in seq. So, be careful setting this.

    # Find the matching indices
    idx = np.where(S.ravel() < thresh)[0]

    # Get the range of those indices as final output
    if len(idx)>0:
        return np.unique((idx[:,None] + np.arange(seq.size)).ravel())
    else:
        return []         # No match found

样品运行

In [512]: arr = np.array([2, 0, 0, 0, 0, 1, 0, 1, 0, 0])

In [513]: seq = np.array([0,0])

In [514]: search_sequence_numpy(arr,seq)
Out[514]: array([1, 2, 3, 4, 8, 9])

In [515]: search_sequence_cv2(arr,seq)
Out[515]: array([1, 2, 3, 4, 8, 9])

运行时测试

In [477]: arr = np.random.randint(0,9,(100000))
     ...: seq = np.array([3,6,8,4])
     ...: 

In [478]: np.allclose(search_sequence_numpy(arr,seq),search_sequence_cv2(arr,seq))
Out[478]: True

In [479]: %timeit search_sequence_numpy(arr,seq)
100 loops, best of 3: 11.8 ms per loop

In [480]: %timeit search_sequence_cv2(arr,seq)
10 loops, best of 3: 20.6 ms per loop

看起来像Pure NumPy一样是最安全和最快的!