convolution

Convolving image with kernel in Fourier domain

拜拜、爱过 提交于 2019-11-28 14:09:52
I'm using zero padding around my image and convolution kernel, converting them to the Fourier domain, and inverting them back to get the convolved image, see code below. The result, however, is wrong. I was expecting a blurred image, but the output is four shifted quarters. Why is the output wrong, and how can I fix the code? Input image: Result of convolution: from PIL import Image,ImageDraw,ImageOps,ImageFilter import numpy as np from scipy import fftpack from copy import deepcopy import imageio ## STEP 1 ## im1=Image.open("pika.jpeg") im1=ImageOps.grayscale(im1) im1.show() print("s",im1

Tensorflow + Keras + Convolution2d: ValueError: Filter must not be larger than the input: Filter: (5, 5) Input: (3, 350)

假如想象 提交于 2019-11-28 13:50:00
I have been trying to run the code below which I got from here and even though I have changed almost nothing other than the image size (350,350 instead of 150, 150) is still cannot get it to work. I am getting the above filter error (in title) which I do comprehend but I am not doing it wrong so I don't understand this. It basically says that I cannot have more nodes than inputs, correct? I was able to eventually hack my way to a solution by changing this line: model.add(Convolution2D(32, 5, 5, border_mode='valid', input_shape=(3, IMG_WIDTH, IMG_HEIGHT))) with this: model.add(Convolution2D(32,

Convolution of NumPy arrays of arbitrary dimension for Cauchy product of multivariate power series

戏子无情 提交于 2019-11-28 12:32:14
问题 I'm trying to implement the idea I have suggested here, for Cauchy product of multivariate finite power series (i.e. polynomials) represented as NumPy ndarrays. numpy.convolve does the job for 1D arrays, respectively. But to my best knowledge there is no implementations of convolution for arbitrary dimensional arrays. In the above link, I have suggested the equation: for convolution of two n dimensional arrays Phi of shape P=[p1,...,pn] and Psi of the shape Q=[q1,...,qn] , where: omega s are

Matlab's Conv2 equivalent in OpenCV

[亡魂溺海] 提交于 2019-11-28 09:37:13
问题 I have been trying to do Convolution of a 2D Matrix using OpenCV. I actually went through this code http://blog.timmlinder.com/2011/07/opencv-equivalent-to-matlabs-conv2-function/#respond but it yields correct answers only in positive cases. Is there a simple function like conv2 in Matlab for OpenCV or C++? Here is an example: A= [ 1 -2 3 4 ] I want to convolve it with [-0.707 0.707] And the result as by conv2 from Matlab is -0.7071 2.1213 -1.4142 -2.1213 -0.7071 2.8284 Some function to

How to write a convolution multiplication in Android Renderscript?

人盡茶涼 提交于 2019-11-28 08:49:47
I am new to Android Renderscript. I need to write a convolution multiplication in RenderScript since the final application is going to run on Android. Data stream is going to be an image. More specifically, I am not able to write the core logic using forEach functionality, though I can do it in Java, but speed it too slow! Please help! Steve During the rsForEach call (or other Renderscript function), you can access the neighbouring pixels of the original image (or whatever type of data you are using) by binding the original image allocation to a pointer within the Renderscript where it can

Improving Numpy Performance

China☆狼群 提交于 2019-11-28 05:58:21
I'd like to improve the performance of convolution using python, and was hoping for some insight on how to best go about improving performance. I am currently using scipy to perform the convolution, using code somewhat like the snippet below: import numpy import scipy import scipy.signal import timeit a=numpy.array ( [ range(1000000) ] ) a.reshape(1000,1000) filt=numpy.array( [ [ 1, 1, 1 ], [1, -8, 1], [1,1,1] ] ) def convolve(): global a, filt scipy.signal.convolve2d ( a, filt, mode="same" ) t=timeit.Timer("convolve()", "from __main__ import convolve") print "%.2f sec/pass" % (10 * t.timeit

Understanding behaviour of MATLAB's convn

。_饼干妹妹 提交于 2019-11-28 05:29:54
问题 I'm doing convolution of some tensors. Here is small test in MATLAB: ker= rand(3,4,2); a= rand(5,7,2); c=convn(a,ker,'valid'); c11=sum(sum(a(1:3,1:4,1).*ker(:,:,1)))+sum(sum(a(1:3,1:4,2).*ker(:,:,2))); c(1,1)-c11 % not equal! The third line performs a N-D convolution with convn , and I want to compare the result of the first row, first column of convn with computing the value manually. However, my computation in comparison to convn is not equal. So what is behind MATLAB's convn ? Is my

scipy.ndimage.filters.convolve and multiplying Fourier Transforms give different results

不打扰是莪最后的温柔 提交于 2019-11-28 04:50:16
问题 Here's my code: from scipy.ndimage import filters import numpy a = numpy.array([[2,43,42,123,461],[453,12,111,123,55] ,[123,112,233,12,255]]) b = numpy.array([[0,2,2,3,0],[0,15,12,100,0],[0,45,32,22,0]]) ab = filters.convolve(a,b, mode='constant', cval=0) af = numpy.fft.fftn(a) bf = numpy.fft.fftn(b) abf = af*bf abif = numpy.fft.ifftn(abf) print numpy.around(ab) print numpy.around(abif) The results are: [[ 1599 2951 7153 13280 18311] [ 8085 51478 13028 40239 30964] [18192 32484 23527 36122

Compute mean squared, absolute deviation and custom similarity measure - Python/NumPy

家住魔仙堡 提交于 2019-11-28 04:10:14
问题 I have a large image as an 2D array (let's assume that it is a 500 by 1000 pixels gray scale image). And I have one small image (let's say is it 15 by 15 pixels). I would like to slide the small image over the large one and for a given position of the small image I would like to calculate a measure of similarity between the small image and the underling part of the big image. I would like to be flexible in choosing a measure of similarity. For example I might want to calculate mean squared

scipy convolve2d outputs wrong values

我的未来我决定 提交于 2019-11-28 01:45:40
问题 Here is my code which I used for checking the correctness of convolve2d import numpy as np from scipy.signal import convolve2d X = np.random.randint(5, size=(10,10)) K = np.random.randint(5, size=(3,3)) print "Input's top-left corner:" print X[:3,:3] print 'Kernel:' print K print 'Hardcording the calculation of a valid convolution (top-left)' print (X[:3,:3]*K) print 'Sums to' print (X[:3,:3]*K).sum() print 'However the top-left value of the convolve2d result' Y = convolve2d(X, K, 'valid')