numpy中的Python简单卷积

Question

我用numpy编写了这个简单的卷积函数。但是最终的数组值仍然是0。请帮我改正这个功能。

def convolve(a_prev, w, b):
pad = 0
stride = 1
s1 = a_prev.shape
s2 = w.shape
f = s2[1]
m = s1[0]
n_c = s2[0]
n_h = int((s1[1] - f + 2 * pad) / stride) + 1
n_w = int((s1[2] - f + 2 * pad) / stride) + 1
a = np.zeros((m,n_h,n_w,n_c), dtype=np.float32)
for n in range(m):
for z in range(n_c):
y = 0
x = 0
while ((y+f) <= n_h):
# Edit: forget to inialize the x = 0
while ((x+f) <= n_w):  
#a[n,y,x,z] = np.sum(a_prev[n,y:y+f,x:x+f]*w[z]) + b[z,0]
a[n,y,x,z] = np.sum(np.multiply(a_prev[n,y:y+f,x:x+f],w[z])) + b[z,0]
x += stride
y += stride

print(a[0,85,:,3])
return a

a_prev的形状为[num_examples,height, width, 3]， w为[num_filters,3,3,3]

我找到了原因，为什么它不工作，我犯了一个编程错误，忘记在while循环之前初始化x = 0．它现在工作得很好。下面是正确的函数:

def convolve(a_prev, kernel, b, pad = 0, stride = 1):
m = a_prev.shape[0]
prev_h = a_prev.shape[1]
prev_w = a_prev.shape[2]
f = kernel.shape[1]
n_c = kernel.shape[0]
new_h = int((prev_h - f + 2 * pad) / stride) + 1
new_w = int((prev_w - f + 2 * pad) / stride) + 1
az = np.zeros((m,new_h,new_w,n_c), dtype=np.float32)
for n in range(m):
for z in range(n_c):
y = 0
while (y+f) <= prev_h:
x = 0
while (x+f) <= prev_w:
az[n,y,x,z] = np.sum(a_prev[n,y:y+f,x:x+f]*kernel[z]) + b[z,0]
x += stride
y += stride

return az

Answer 1

有一种更简单的方法来进行卷积，而不需要使用前面的输入作为函数输入。

import numpy as np
def convolution(x, h):
# x and h are numpy arrays
M, N = np.size(x), np.size(h)
y = np.zeros(M+N-1)
# Initialise y with the length of the output signal
for m in np.arange(M):
for n in np.arange(N):
y[m+n] += x[m]*h[n]
return y

该函数使用离散信号卷积的基本定义