numpy.convolve()

卷积函数:

numpy.convolve(avmode='full')

Parameters:

a : (N,) array_like

First one-dimensional input array.

v : (M,) array_like

Second one-dimensional input array.

mode : {‘full’, ‘valid’, ‘same’}, optional

‘full’:

By default, mode is ‘full’. This returns the convolution at each point of overlap, with an output shape of (N+M-1,). At the end-points of the convolution, the signals do not overlap completely, and boundary effects may be seen.

‘same’:

Mode ‘same’ returns output of length max(M, N). Boundary effects are still visible.

‘valid’:

Mode ‘valid’ returns output of length max(M, N) min(M, N) 1. The convolution product is only given for points where the signals overlap completely. Values outside the signal boundary have no effect.

Returns:

out : ndarray

Discrete, linear convolution of a and v.

The discrete convolution operation is defined as

(a * v)[n] = \sum_{m = -\infty}^{\infty} a[m] v[n - m]

It can be shown that a convolution x(t) * y(t) in time/space is equivalent to the multiplication X(f) Y(f)in the Fourier domain, after appropriate padding (padding is necessary to prevent circular convolution). Since multiplication is more efficient (faster) than convolution, the function scipy.signal.fftconvolveexploits the FFT to calculate the convolution of large data-sets.

Note how the convolution operator flips the second array before “sliding” the two across one another:

>>>
>>> np.convolve([1, 2, 3], [0, 1, 0.5])
array([ 0. ,  1. ,  2.5,  4. ,  1.5])

Only return the middle values of the convolution. Contains boundary effects, where zeros are taken into account:

>>>
>>> np.convolve([1,2,3],[0,1,0.5], 'same')
array([ 1. ,  2.5,  4. ])

The two arrays are of the same length, so there is only one position where they completely overlap:

>>>
>>> np.convolve([1,2,3],[0,1,0.5], 'valid')
array([ 2.5])
posted @ 2016-12-05 14:28  早起的鸟儿有  阅读(3824)  评论(0编辑  收藏  举报