我正在尝试为信号实现一维卷积。
它应该具有与以下内容相同的输出:
ary1 = np.array([1, 1, 2, 2, 1])
ary2 = np.array([1, 1, 1, 3])
conv_ary = np.convolve(ary2, ary1, 'full')
>>>> [1 2 4 8 8 9 7 3]
我想出了这个方法:
def convolve_1d(signal, kernel):
n_sig = signal.size
n_ker = kernel.size
n_conv = n_sig - n_ker + 1
# by a factor of 3.
rev_kernel = kernel[::-1].copy()
result = np.zeros(n_conv, dtype=np.double)
for i in range(n_conv):
result[i] = np.dot(signal[i: i + n_ker], rev_kernel)
return result
但我的结果是
[8,8]
我可能必须对数组进行零填充并更改其索引。
有没有更顺利的方法来达到预期的结果?
这是一个可能的解决方案:
def convolve_1d(signal, kernel):
kernel = kernel[::-1]
return [
np.dot(
signal[max(0,i):min(i+len(kernel),len(signal))],
kernel[max(-i,0):len(signal)-i*(len(signal)-len(kernel)<i)],
)
for i in range(1-len(kernel),len(signal))
]
这是一个例子:
>>> convolve_1d([1, 1, 2, 2, 1], [1, 1, 1, 3])
[1, 2, 4, 8, 8, 9, 7, 3]
另一种解决方案,但效率不高:
import numpy as np
def convolve_1d(signal, kernel):
kernel = kernel[::-1]
k = len(kernel)
s = len(signal)
signal = [0]*(k-1)+signal+[0]*(k-1)
n = s+(k-1)
res = []
for i in range(s+k-1):
res.append(np.dot(signal[i:(i+k)], kernel))
return res
A = [1,1,2,2,1]
B = [1,1,1,3]
# result = [1,2,4,8,8,9,7,3]
print(convolve_1d(A, B))