假设我有以下数组:
array([2, 0, 0, 1, 0, 1, 0, 0])
如何获取出现值序列的索引:[0,0]
?因此,这种情况的预期输出为:[1,2,6,7]
。
编辑:
1) 请注意[0,0]
只是一个序列。它可以是 [0,0,0]
或 [4,6,8,9]
或 [5,2,0]
,什么都行。
2) 如果我的数组被修改为:array([2, 0, 0, 0, 0, 1, 0, 1, 0, 0])
,预期结果与[0,0]
的序列将是 [1,2,3,4,8,9]
。
我正在寻找一些 NumPy 快捷方式。
最佳答案
嗯,这基本上是一个 template-matching problem
这在图像处理中经常出现。这篇文章中列出了两种方法:基于纯 NumPy 和基于 OpenCV (cv2)。
方法 #1:使用 NumPy,可以创建一个 2D
滑动索引数组,横跨输入数组的整个长度。因此,每一行都是元素的滑动窗口。接下来,将每一行与输入序列匹配,这将带来 broadcasting
用于矢量化解决方案。我们寻找所有 True
行,表明这些行是完美匹配的,因此将是匹配的起始索引。最后,使用这些索引,创建一个延伸到序列长度的索引范围,为我们提供所需的输出。实现将是 -
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),我们有一个用于模板匹配
的内置函数:cv2.matchTemplate
.使用这个,我们将有起始匹配索引。其余步骤与之前的方法相同。下面是 cv2
的实现:
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
样本运行
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
似乎基于 Pure NumPy 的是最安全和最快的!
关于python - 在 NumPy 数组中搜索序列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36522220/