我在诊断和修复此错误时遇到问题。我正在尝试编写 OaA 算法，描述 in this paper .
#!/usr/bin/env python
# -*- coding: utf-8 -*-
""" Quick implementation of several convolution algorithms to compare times
"""
import numpy as np
import _kernel
from tqdm import trange, tqdm
from PIL import Image
from scipy.misc import imsave
from time import time, sleep
class convolve(object):
""" contains methods to convolve two images """
def __init__(self, image_array, kernel, back_same_size=True):
self.array = image_array
self.kernel = kernel
# Store these values as they will be accessed a _lot_
self.__rangeX_ = self.array.shape[0]
self.__rangeY_ = self.array.shape[1]
self.__rangeKX_ = self.kernel.shape[0]
self.__rangeKY_ = self.kernel.shape[1]
# Ensure the kernel is suitable to convolve the image
if (self.__rangeKX_ >= self.__rangeX_ or \
self.__rangeKY_ >= self.__rangeY_):
raise ValueError('Must submit suitably-sized arrays')
if (back_same_size):
# pad array for convolution
self.__offsetX_ = self.__rangeKX_ // 2
self.__offsetY_ = self.__rangeKY_ // 2
self.array = np.lib.pad(self.array, \
[(self.__offsetY_, self.__offsetY_), \
(self.__offsetX_, self.__offsetX_)],\
mode='constant', constant_values=0)
# Update these
self.__rangeX_ = self.array.shape[0]
self.__rangeY_ = self.array.shape[1]
else:
self.__offsetX_ = 0
self.__offsetY_ = 0
# to be returned instead of the originals
self.__arr_ = np.zeros([self.__rangeX_, self.__rangeY_])\
def spaceConv(self):
""" normal convolution, O(N^2*n^2). This is usually too slow """
# this is the O(N^2) part of this algorithm
for i in trange(self.__rangeX_):
for j in xrange(self.__rangeY_):
# Now the O(n^2) portion
total = 0.0
for k in xrange(self.__rangeKX_):
for t in xrange(self.__rangeKY_):
total += \
self.kernel[k][t] * self.array[i+k][j+t]
# Update entry in self.__arr_, which is to be returned
# http://stackoverflow.com/a/38320467/3928184
self.__arr_[i][j] = total
return self.__arr_[self.__offsetX_\
:self.__rangeX_ - self.__offsetX_,\
self.__offsetY_\
:self.__rangeY_ - self.__offsetY_]
def spaceConvDot(self):
""" Exactly the same as the former method """
def dot(ind, jnd):
""" perform a simple 'dot product' between the 2
dimensional image subsets. """
total = 0.0
# This is the O(n^2) part of the algorithm
for k in xrange(self.__rangeKX_):
for t in xrange(self.__rangeKY_):
total += \
self.kernel[k][t] * self.array[k + ind, t + jnd]
return total
# this is the O(N^2) part of the algorithm
for i in trange(self.__rangeX_):
for j in xrange(self.__rangeY_):
self.__arr_[i][j] = dot(i, j)
return self.__arr_[self.__offsetX_\
:self.__rangeX_ - self.__offsetX_,\
self.__offsetY_\
:self.__rangeY_ - self.__offsetY_]
def OAconv(self):
""" faster convolution algorithm, O(N^2*log(n)). """
from numpy.fft import fft2 as FFT, ifft2 as iFFT
# solve for the total padding along each axis
diffX = (self.__rangeKX_ - self.__rangeX_ + \
self.__rangeKX_ * (self.__rangeX_ //\
self.__rangeKX_)) % self.__rangeKX_
diffY = (self.__rangeKY_ - self.__rangeY_ + \
self.__rangeKY_ * (self.__rangeY_ //\
self.__rangeKY_)) % self.__rangeKY_
# padding on each side, i.e. left, right, top and bottom;
# centered as well as possible
right = diffX // 2
left = diffX - right
bottom = diffY // 2
top = diffY - bottom
# pad the array
self.array = np.lib.pad(self.array, \
((left, right), (top, bottom)), \
mode='constant', constant_values=0)
divX = self.array.shape[0] / float(self.__rangeKX_)
divY = self.array.shape[1] / float(self.__rangeKY_)
# Let's just make sure...
if not (divX % 1.0 == 0.0 or divY % 1.0 == 0.0):
raise ValueError('Image not partitionable (?)')
else:
divX = int(divX)
divY = int(divY)
# a list of tuples to partition the array by
subsets = [(i*self.__rangeKX_, (i + 1)*self.__rangeKX_,\
j*self.__rangeKY_, (j + 1)*self.__rangeKY_)\
for i in xrange(divX) \
for j in xrange(divY)]
# padding for individual blocks in the subsets list
padX = self.__rangeKX_ // 2
padY = self.__rangeKY_ // 2
self.__arr_ = np.lib.pad(self.__arr_, \
((left + padX, right + padX), \
(top + padY, bottom + padY)), \
mode='constant', constant_values=0)
kernel = np.pad(self.kernel, \
[(padY, padY), (padX, padX)], \
mode='constant', constant_values=0)
# We only need to do this once
trans_kernel = FFT(kernel)
# transform each partition and OA on conv_image
for tup in tqdm(subsets):
# slice and pad the array subset
subset = self.array[tup[0]:tup[1], tup[2]:tup[3]]
subset = np.lib.pad(subset, \
[(padY, padY), (padX, padX)],\
mode='constant', constant_values=0)
trans_subset = FFT(subset)
# multiply the two arrays entrywise
subset = trans_kernel * trans_subset
space = iFFT(subset).real
# overlap with indices and add them together
self.__arr_[tup[0]:tup[1] + 2 * padX, \
tup[2]:tup[3] + 2 * padY] += space
# crop image and get it back, convolved
return self.__arr_[self.__offsetX_ + padX + left \
:padX + left + self.__rangeX_ \
- self.__offsetX_, \
self.__offsetY_ + padY + bottom \
:padY + bottom + self.__rangeY_ \
- self.__offsetY_]
def OSconv(self):
""" Convolve an image using OS """
from numpy.fft import fft2 as FFT, ifft2 as iFFT
pass
def builtin(self):
""" Convolves using SciPy's convolution function - extremely
fast """
from scipy.ndimage.filters import convolve
return convolve(self.array, self.kernel)
if __name__ == '__main__':
try:
import pyplot as plt
except ImportError:
import matplotlib.pyplot as plt
image = np.array(Image.open('spider.jpg'))
image = np.rot90(np.rot90(np.rot90(image.T[0])))
times = []
#for i in range(3, 21, 2):
kern = _kernel.Kernel()
kern = kern.Kg2(11, 11, sigma=2.5, muX=0.0, muY=0.0)
kern /= np.sum(kern) # normalize volume
conv = convolve(image, kern)
#
# # Time the result of increasing kernel size
# _start = time()
convolved = conv.OAconv()
#convolved = conv.builtin()
# _end = time()
# times.append(_end - _start)
#x = np.array(range(3, 21, 2))
#plt.plot(range(3, 21, 2), times)
#plt.title('Kernel Size vs. spaceConv time', fontsize=12)
#plt.xlabel('Kernel Size (px)', fontsize=12)
#plt.ylabel('Time (s)', fontsize=12)
#plt.xticks(x, x)
#plt.show()
#conv = convolve(image[:2*kern.shape[0],:5*kern.shape[1]], kern)
plt.imshow(convolved, interpolation='none', cmap='gray')
plt.show()
#imsave('spider2', convolved, format='png')
但是现在，当我调用它时，我在测试图像中得到如下黑条:
这是我正在使用的示例高斯内核。
[[ 0. 0.02390753 0.03476507 0.02390753 0. ]
[ 0.02390753 0.06241541 0.07990366 0.06241541 0.02390753]
[ 0.03476507 0.07990366 0.10040324 0.07990366 0.03476507]
[ 0.02390753 0.06241541 0.07990366 0.06241541 0.02390753]
[ 0. 0.02390753 0.03476507 0.02390753 0. ]]
我相信我已经将问题缩小到乘法和 IFFT
space = np.real(IFFT(transformed_kernel*transformed_subset))
我认为它与离散高斯核的IFFT有关(出于某种原因)。这很奇怪，因为如果我只是策划
space = np.real(IFFT(transformed_subset))
我得到以下信息(没有黑条，并且它很好地拼凑起来):
如果我绘制相反的图，即
space = np.real(IFFT(transformed_kernel))
我再次没有看到黑条，而且似乎将它们放在了正确的位置。
我错过了什么？我已经盯着这个看好几天了，编辑索引等等，但我无法摆脱这个镶嵌:(
最佳答案
你的问题是，你的内核似乎在中间:
但事实并非如此 - 还涉及到移位 (11, 11)，当您查看卷积结果时，这一点会变得很明显。你的内核的中心应该在 (0,0) 并且(因为模数)你的内核应该是这样的:
所以我稍微修改了你的代码(抱歉从来没有用过这么多 np，但我希望你 可以得到要点):
... your code ...
kernel = np.pad(self.kernel, \
[(padY, padY), (padX, padX)], \
mode='constant', constant_values=0)
#Move the kernel center to the origin:
new_kernel=np.full_like(kernel, 0)
X,Y=kernel.shape
X_2=X//2
Y_2=Y//2
for x in xrange(X):
for y in xrange(Y):
n_x=(x+X_2)%X
n_y=(y+Y_2)%Y
new_kernel[n_x,n_y]=kernel[x,y]
# We only need to do this once
trans_kernel = FFT(new_kernel)# take the transform of the shifted kernel
.... your code ......
瞧:
没有黑格!
关于python - NumPy IFFT 在 OaA 卷积算法中引入黑条，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37771182/