Python 为什么einsum在浮点数上比numpy.correlate快，但在复数上慢？

试图找出在某些复数上实现相关性的最快方法—数据形状是长度为~1e5的一维数组，特征核长度为~20。Inner1d不能接受复数，并且fft卷积在这种内核大小下是无效的，所以我测试的两种方法是np.correlate和einsum

对于浮动，einsum与使用as_Stread的滑动窗口相结合比np.correlate快一点

data = np.ones(1e5, dtype='float64')
kernel = np.ones(20, dtype='float64')

%timeit xc1 = np.correlate(data, kernel)
100 loops, best of 3: 2.13 ms per loop

%timeit xc2 = np.einsum("ij,j->i", as_strided(data, shape=(data.shape[0]-(kernel.shape[0]-1), kernel.shape[0]), strides=data.strides * 2), kernel)
1000 loops, best of 3: 1.35 ms per loop

%timeit xc3 = np.einsum("ij,j->i", as_strided(data_Re, shape=(data_Re.shape[0]-(kernel_Re.shape[0]-1), kernel_Re.shape[0]), strides=data_Re.strides * 2), kernel_Re) + np.einsum("ij,j->i", as_strided(data_Im, shape=(data_Im.shape[0]-(kernel_Im.shape[0]-1), kernel_Im.shape[0]), strides=data_Im.strides * 2), kernel_Im)
100 loops, best of 3: 4.21 ms per loop

但对于复数，einsum明显比np慢

data = np.ones(1e5, dtype='complex128')
kernel = np.ones(20, dtype='complex128')
data_conj = np.empty(data.shape[0], dtype='complex128')
data_conj = np.conj(data)

%timeit xc1 = np.correlate(data, kernel)
100 loops, best of 3: 2.21 ms per loop

%timeit xc2 = np.einsum("ij,j->i", as_strided(data_conj, shape=(data_conj.shape[0]-(kernel.shape[0]-1), kernel.shape[0]), strides=data_conj.strides * 2), kernel)
100 loops, best of 3: 5.78 ms per loop

我想我理解为什么einsum比以前花费的时间更长，但是np.correlate怎么会比浮动花费更多的时间呢？是否有一个复杂的数学技巧，我可能可以利用的einsum

就其价值而言，将einsum关联分解为实部和虚部的速度更快，但仍然比np.CONNECTOR慢得多

data = np.ones(1e5, dtype='float64')
kernel = np.ones(20, dtype='float64')

%timeit xc1 = np.correlate(data, kernel)
100 loops, best of 3: 2.13 ms per loop

%timeit xc2 = np.einsum("ij,j->i", as_strided(data, shape=(data.shape[0]-(kernel.shape[0]-1), kernel.shape[0]), strides=data.strides * 2), kernel)
1000 loops, best of 3: 1.35 ms per loop

%timeit xc3 = np.einsum("ij,j->i", as_strided(data_Re, shape=(data_Re.shape[0]-(kernel_Re.shape[0]-1), kernel_Re.shape[0]), strides=data_Re.strides * 2), kernel_Re) + np.einsum("ij,j->i", as_strided(data_Im, shape=(data_Im.shape[0]-(kernel_Im.shape[0]-1), kernel_Im.shape[0]), strides=data_Im.strides * 2), kernel_Im)
100 loops, best of 3: 4.21 ms per loop