这篇文章与我之前的问题相关:image processing in matlab 因为我已经在那里上传了我的算法。 我的想法是我正在尝试更改代码的过滤部分。 在 matlab filter.m 函数中可以接受过滤器(B,A,我的像素随时间的演变)并返回过滤后的值。 但目前我必须将整个时间序列一起传递。
但问题是,现在我想以某种方式更改代码,而不是将整个时间序列传递到过滤器中,我想一次传递一个值,但我希望过滤器函数将该值视为第 n 个值不是第一个值。 我创建了一个 sudo 代码,因为我将一张图片注入(inject)到代码中,但我不知道如何更改过滤部分。任何人都有任何想法吗?
clear all
j=1;
for i=0:3000
str = num2str(i);
str1 = strcat(str,'.mat');
load(str1);
D{j}=A(20:200,130:230);
j=j+1;
end
N=5;
w = [0.00000002 0.05;0.05 0.1;0.1 0.15;0.15 0.20;
0.20 0.25;0.25 0.30;0.30 0.35;0.35 0.40;
0.40 0.45;0.45 0.50;0.50 0.55;0.55 0.60;
0.60 0.65;0.65 0.70;0.70 0.75;0.75 0.80;
0.80 0.85;0.85 0.90;0.90 0.95;0.95 0.99999999];
for ind=1:20
wn = w(ind,:);
[b,a] = butter(N,wn);
bCoeff(ind,:)=b;
aCoeff(ind,:)=a;
end
ts=[];
sumLastImages=[];
for k=1:10 %number of images
for bands=1:20 %number of bands
for i=1:10 %image h
for j=1:10 %image w
pixelValue = D{k}(i,j);
% reflectivity elimination
% for the current pixel, have the summation of the same position from before
% images and create a mean value base on the temporal values
sumLastImages(i,j)=pixelValue+sumLastImages(i,j);
meanValue = sumLastImages(i,j)/k;
if(meanValue==0)
filteredimages{bands}(i,j)=0;
continue;
else
pixel_calculated_meanvalue = pixelValue/meanValue;
end
% filter part that need to be changed, and because we are adding
% one value then the reutrn of the filter is one too
ts_f = filter(bCoeff(bands,:), aCoeff(bands,:), ...
pixel_calculated_meanvalue);
filteredimages{bands}(i,j)=ts_f;
end
end
finalImagesSummation{bands}(:,:) = ...
(filteredimages{bands}(:,:)^2)+finalImagesSummation{bands}(:,:);
finalImages{bands}(:,:)=finalImagesSummation/k;
end
end
编辑
我设法像这样更改代码,现在我加载了前 200 张图像,之后我能够将图像一张一张地注入(inject)到过滤器中,但现在的问题是我得到了 "Out of内存。键入 HELP MEMORY 作为您的选项。"
错误为大
图片。
这是我的代码任何提高代码效率的想法:
%%
cd('D:\MatlabTest\06-06-Lentils');
clear all
%%
N=5;
W = [0.0 0.10;0.10 0.20;0.20 0.30;0.30 0.40;
0.40 0.50;0.50 0.60 ;0.60 0.70;0.70 0.80 ;
0.80 0.90;0.90 1.0];
[bCoeff{1},aCoeff{1}] = butter(N,0.1,'Low');
for ind=2:9
wn = W(ind,:);
[b,a] = butter(N,wn);
bCoeff{ind}=b;
aCoeff{ind}=a;
end
[bCoeff{10},aCoeff{10}] = butter(N,0.9,'high');
%%
j=1;
D = zeros(200,380,320);
T = 200;
K = 0:399;
l = T+1;
Yout = cell(1,10);
figure;
for i = 1:length(K)-200
disp(i)
if i == 1
for j = 1:T
str = int2str(K(1)+j-1);
str1 = strcat(str,'.mat');
load(str1);
D(j,1:380,1:320) = A;
end
else
str = int2str(l);
str1 = strcat(str,'.mat');
load(str1);
temp = D(2:200,1:380,1:320) ;
temp(200,1:380,1:320) = A;
D = temp;
clear temp
l = l +1;
end
for p = 1:380
for q = 1:320
x = D(:,p,q) - mean(D(:,p,q));
for k = 1:10
temp = filter(bCoeff{k},aCoeff{k},x);
if i == 1
Yout{k}(p,q) = mean(temp);
else
Yout{k}(p,q) = (Yout{k}(p,q) + mean(temp))/2;
end
end
end
end
for k = 1:10
subplot(5,2,k)
subimage(Yout{k}*1000,[0 255]);
color bar
colormap jet
end
pause(0.01);
end
disp('Done Loading...')
最佳答案
无需重写过滤函数,有一个简单的解决方案!
如果你想喂filter
一次处理一个样本,您还需要传递状态参数,以便根据其前一个样本处理每个输入样本。过滤后,新状态实际上作为第二个参数返回,因此 MATLAB 已经为您完成了大部分工作。 这是个好消息!
为了可读性,请允许我暂时用简单的字母替换您的变量名称:
a = aCoeff(bands, :);
b = bCoeff(bands, :);
x = pixel_calculated_meanvalue;
ts_f
由 y
表示。
所以,这个:
y = filter(b, a, x);
实际上相当于:
N = numel(x);
y = zeros(N, 1); %# Preallocate memory for output
z = zeros(max(length(a), length(b)) - 1, 1); %# This is the initial filter state
for i = 1:N
[y(i), z] = filter(b, a, x(i), z);
end
你可以自己检查一下结果是一样的!
对于您的示例,代码为:
N = numel(pixel_calculated_meanvalue);
ts_f = zeros(N, 1);
z = zeros(max(length(aCoeff(bands, :)), length(bCoeff(bands, :))) - 1, 1);
for i = 1:N
[ts_f(i), z] = filter(bCoeff(bands, :), aCoeff(bands, :), ...
pixel_calculated_meanvalue(i), z);
end
使用此方法,您可以一次处理一个输入样本,只需确保在每次 filter
调用后存储最后的过滤器状态即可。如果您计划使用多个过滤器,则必须为每个过滤器存储一个状态向量!
关于matlab - 在 matlab 中更改过滤器(B,A,X)并出现内存不足错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11583523/