PyTorch 的 conv1d 和 SciPy 的 convolve 的不同结果

Question

I'm building a PyTorch model to estimate Impuse Responses . Currently I am calculating the loss from the real and estimated impulse response. I would like to convolve both the estimated and real impulse response with a signal and then calculate the loss from those.

The pyroomaccoustics package uses SciPy's fftconvolve to convolve the impulse response with a given signal. I cannot use this since it would break PyTorch's computation graph. PyTorch's conv1d uses cross-correlation. From this answer it seems that by flipping the filter conv1d can be used for convolution.

I am confused as to why the following code gives a different result for conv1d and convolve and what must be changed to get the outputs to be equal.

import torch
from scipy.signal import convolve

a = torch.tensor([.1, .2, .3, .4, .5])
b = torch.tensor([.0, .1, .0])

a1 = a.view(1, 1, -1)
b1 = torch.flip(b, (0,)).view(1, 1, -1)


print(torch.nn.functional.conv1d(a1, b1).view(-1))
# >>> tensor([0.0200, 0.0300, 0.0400])

print(convolve(a, b))
# >>> [0.   0.01 0.02 0.03 0.04 0.05 0.  ]

Answer 1

Take a look at the mode parameter of scipy.signal.convolve . Use mode='valid' to match PyTorch's conv1d :

In [20]: from scipy.signal import convolve                                      

In [21]: a = np.array([.1, .2, .3, .4, .5])                                     

In [22]: b = np.array([.0, .1, .0])                                             

In [23]: convolve(a, b, mode='valid')                                           
Out[23]: array([0.02, 0.03, 0.04])

To modify the call of PyTorch's conv1d to give the same output as the default behavior of scipy.signal.convolve (ie to match mode='full' ) for this example, set padding=2 in the call to conv1d . More generally, for a given convolution kernel b , set the padding to len(b) - 1 .

Answer 2

Additional example pictures for full,same,valid in convolve https://www.twblogs.net/a/5c8373f9bd9eee35fc13cb76

select [valid] will match with torch.conv1d