Question

我有一个矩阵M1，每行都是一个与时间有关的信号。

我有另一个相同维度的矩阵M2，其中每行也是一个时间相关的信号，用作“模板”，用于识别第一个矩阵中的信号形状。

我想要一个列向量v，其中v [i]是M1的第i行和M2的第i行之间的相关性。

我查看了numpy的corrcoef函数并尝试了以下代码：

import numpy as np

M1 = np.array ([
    [1, 2, 3, 4],
    [2, 3, 1, 4]
])

M2 = np.array ([
    [10, 20, 30, 40],
    [20, 30, 10, 40]
])

print (np.corrcoef (M1, M2))

打印：

[[ 1.   0.4  1.   0.4]
 [ 0.4  1.   0.4  1. ]
 [ 1.   0.4  1.   0.4]
 [ 0.4  1.   0.4  1. ]]

我一直在阅读文档，但我仍然感到困惑的是，这个矩阵的哪些条目我必须选择作为我的向量v的条目。

有人可以帮忙吗？

（我已经研究了类似问题的几个S.O.答案，但还没有看到光......）

代码背景：

有256行（信号），我在'主信号'上运行了200个样本的滑动窗口，其中包含10k个样本的长度。因此M1和M2都是256行×200列。抱歉错误的10k样品。这是总信号长度。通过使用与滑动模板的相关性，我尝试找到模板匹配最佳的偏移量。实际上我正在寻找256通道侵入性心电图中的QRS复合波（或者更确切地说，电图，就像医生所说的那样）。

    lg.info ('Processor: {}, time: {}, markers: {}'.format (self.key, dt.datetime.now ().time (), len (self.data.markers)))

    # Compute average signal shape over preexisting markers and uses that as a template to find the others.
    # All generated markers will have the width of the widest preexisting one.

    template = np.zeros ((self.data.samples.shape [0], self.bufferWidthSteps))

    # Add intervals that were marked in advance
    nrOfTerms = 0
    maxWidthSteps = 0
    newMarkers = []
    for marker in self.data.markers:
        if marker.key == self.markerKey:

            # Find start and stop sample index    
            startIndex = marker.tSteps - marker.stampWidthSteps // 2
            stopIndex = marker.tSteps + marker.stampWidthSteps // 2

            # Extract relevant slice from samples and add it to template
            template += np.hstack ((self.data.samples [ : , startIndex : stopIndex], np.zeros ((self.data.samples.shape [0], self.bufferWidthSteps - marker.stampWidthSteps))))

            # Adapt nr of added terms to facilitate averaging
            nrOfTerms += 1

            # Remember maximum width of previously marked QRS complexes
            maxWidthSteps = max (maxWidthSteps, marker.stampWidthSteps)
        else:
            # Preexisting markers with non-matching keys are just copied to the new marker list
            # Preexisting markers with a matching key are omitted from the new marker list
            newMarkers.append (marker)

    # Compute average of intervals that were marked in advance
    template = template [ : , 0 : maxWidthSteps] / nrOfTerms
    halfWidthSteps = maxWidthSteps // 2

    # Append markers of intervals that yield an above threshold correlation with the averaged marked intervals
    firstIndex = 0
    stopIndex = self.data.samples.shape [1] - maxWidthSteps
    while firstIndex < stopIndex:
        corr = np.corrcoef (
            template,
            self.data.samples [ : , firstIndex : firstIndex + maxWidthSteps]
        )

        diag = np.diagonal (
            corr,
            template.shape [0]
        )

        meanCorr = np.mean (diag)

        if meanCorr > self.correlationThreshold:
            newMarkers.append ([self.markerFactories [self.markerKey] .make (firstIndex + halfWidthSteps, maxWidthSteps)])

            # Prevent overlapping markers
            firstIndex += maxWidthSteps
        else:
            firstIndex += 5

    self.data.markers = newMarkers

    lg.info ('Processor: {}, time: {}, markers: {}'.format (self.key, dt.datetime.now ().time (), len (self.data.markers)))

Answer 1

基于this solution找到两个2D数组之间的相关矩阵，我们可以找到一个类似的数据，用于查找计算两个数组中相应行之间相关性的相关向量。实现看起来像这样 -

def corr2_coeff_rowwise(A,B):
    # Rowwise mean of input arrays & subtract from input arrays themeselves
    A_mA = A - A.mean(1)[:,None]
    B_mB = B - B.mean(1)[:,None]

    # Sum of squares across rows
    ssA = (A_mA**2).sum(1);
    ssB = (B_mB**2).sum(1);

    # Finally get corr coeff
    return np.einsum('ij,ij->i',A_mA,B_mB)/np.sqrt(ssA*ssB)

我们可以通过在那里引入ssA魔法来进一步优化部件以获得ssB和einsum！

def corr2_coeff_rowwise2(A,B):
    A_mA = A - A.mean(1)[:,None]
    B_mB = B - B.mean(1)[:,None]
    ssA = np.einsum('ij,ij->i',A_mA,A_mA)
    ssB = np.einsum('ij,ij->i',B_mB,B_mB)
    return np.einsum('ij,ij->i',A_mA,B_mB)/np.sqrt(ssA*ssB)

示例运行 -

In [164]: M1 = np.array ([
     ...:     [1, 2, 3, 4],
     ...:     [2, 3, 1, 4.5]
     ...: ])
     ...: 
     ...: M2 = np.array ([
     ...:     [10, 20, 33, 40],
     ...:     [20, 35, 15, 40]
     ...: ])
     ...: 

In [165]: corr2_coeff_rowwise(M1, M2)
Out[165]: array([ 0.99411402,  0.96131896])

In [166]: corr2_coeff_rowwise2(M1, M2)
Out[166]: array([ 0.99411402,  0.96131896])

运行时测试 -

In [97]: M1 = np.random.rand(256,200)
    ...: M2 = np.random.rand(256,200)
    ...: 

In [98]: out1 = np.diagonal (np.corrcoef (M1, M2), M1.shape [0])
    ...: out2 = corr2_coeff_rowwise(M1, M2)
    ...: out3 = corr2_coeff_rowwise2(M1, M2)
    ...: 

In [99]: np.allclose(out1, out2)
Out[99]: True

In [100]: np.allclose(out1, out3)
Out[100]: True

In [101]: %timeit np.diagonal (np.corrcoef (M1, M2), M1.shape [0])
     ...: %timeit corr2_coeff_rowwise(M1, M2)
     ...: %timeit corr2_coeff_rowwise2(M1, M2)
     ...: 
100 loops, best of 3: 9.5 ms per loop
1000 loops, best of 3: 554 µs per loop
1000 loops, best of 3: 430 µs per loop

20x+ 加速内置einsum的{{1}}！

Answer 2

我认为是这样的：（如果错误请更正！）

import numpy as np

M1 = np.array ([
    [1, 2, 3, 4],
    [2, 3, 1, 4.5]
])

M2 = np.array ([
    [10, 20, 33, 40],
    [20, 35, 15, 40]
])

v = np.diagonal (np.corrcoef (M1, M2), M1.shape [0])

print (v)

打印哪些：

[ 0.99411402  0.96131896]

由于它只有一个维度，我可以将其视为列向量...

Answer 3

不太了解numpy数组魔法，我只是挑出行，将每一对分别送到corrcoeff

[np.corrcoef(i,j)[0][1] for i,j in zip(a,b)]

表示np.array列输出

c, c.shape = np.array([np.corrcoef(i,j)[0][1] for i,j in zip(a,b)]), (a.shape[0], 1)

我确信使用numpy广播/索引功能会更好

2个时间相关的多维信号（信号向量）的相关性

3 个答案: