假設(shè)我有以下數(shù)組：

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

如何獲得我出現(xiàn)值序列的索引：[0,0]？因此,這種情況的預(yù)期輸出將是：[1,2,6,7].

編輯：

1)請(qǐng)注意[0,0]只是一個(gè)序列.它可能是[0,0,0]或[4,6,8,9]或[5,2,0],只是任何東西.

2)如果我的數(shù)組被修改為：array([2,0,0,0,0,1,0,1,0,0]),具有相同序列[0,0]的預(yù)期結(jié)果將是[ 1,2,3,4,8,9.

我正在尋找一些NumPy快捷方式.

解決方法:

嗯,這基本上是圖像處理中出現(xiàn)的template-matching problem.在這篇文章中列出了兩種方法：基于Pure NumPy和基于OpenCV(cv2).

方法#1：使用NumPy,可以在輸入數(shù)組的整個(gè)長(zhǎng)度上創(chuàng)建一個(gè)滑動(dòng)索引的2D數(shù)組.因此,每行將是元素的滑動(dòng)窗口.接下來(lái),將每一行與輸入序列匹配,這將為矢量化解決方案帶來(lái)broadcasting.我們尋找所有True行,表明那些是完美的匹配,因此將是匹配的起始索引.最后,使用這些索引,創(chuàng)建一系列延伸到序列長(zhǎng)度的索引,以便為我們提供所需的輸出.實(shí)施將是 –

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),我們有一個(gè)用于模板匹配的內(nèi)置函數(shù)：cv2.matchTemplate.使用這個(gè),我們將得到起始匹配索引.其余步驟與前一種方法相同.這是cv2的實(shí)現(xiàn)：

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

樣品運(yùn)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])

運(yùn)行時(shí)測(cè)試

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

看起來(lái)像Pure NumPy一樣是最安全和最快的！來(lái)源：https://www.icode9.com/content-1-480101.html

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