Python:查找二维数组中给定数组的出现次数 [英] Python: Find number of occurrences of given array within two-dimensional array

问题描述

如何最好地在python(带numpy)中找到一组数组(二维数组)中给定数组的出现次数? 这是(简化)我需要在python代码中表达的内容:

How best to find the number of occurrences of a given array within a set of arrays (two-dimensional array) in python (with numpy)? This is (simplified) what I need expressed in python code:

patterns = numpy.array([[1, -1, 1, -1],
                   [1, 1, -1, 1],
                   [1, -1, 1, -1],
                   ...])
findInPatterns = numpy.array([1, -1, 1, -1])
numberOfOccurrences = findNumberOfOccurrences(needle=findInPatterns, haystack=patterns)
print(numberOfOccurrences) # should print e.g. 2

实际上，我需要找出可以在集合中找到每个数组的频率.但是上面代码中描述的功能已经对我有很大帮助.

In reality, I need to find out how often each array can be found within the set. But the functionality described in the code above would already help me a lot on my way.

现在，我知道我可以使用循环来执行此操作，但我想知道是否有更有效的方法来执行此操作?谷歌搜索只让我进入numpy.bincount，它完全满足我的需要，但不适合二维数组，仅适合整数.

Now, I know I could use loops to do that but was wondering if there was a more efficient way to do this? Googling only let me to numpy.bincount which does exactly what I need but not for two-dimensional arrays and only for integers.

推荐答案

使用1 s和-1 s数组，从性能上讲，使用np.dot并不能胜过一切:如果(且仅当)全部项匹配，则点积将总计一行中的项数.所以你可以做

With an array of 1s and -1s, performance wise nothing is going to beat using np.dot: if (and only if) all items match then the dot product will add up to the number of items in the row. So you can do

>>> haystack = np.array([[1, -1, 1, -1],
...                      [1, 1, -1, 1],
...                      [1, -1, 1, -1]])
>>> needle = np.array([1, -1, 1, -1])
>>> haystack.dot(needle)
array([ 4, -2,  4])
>>> np.sum(haystack.dot(needle) == len(needle))
2

这是基于卷积的图像匹配的一种特殊情况，您可以轻松地重写它以查找比整行短的图案，甚至可以使用FFT加快速度.

This is sort of a toy particular case of convolution based image matching, and you could rewrite it easily to look for patterns shorter than a full row, and even speed it up using FFTs.

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