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I'm writing a program that finds the indices of a matrix G where there is only a single 1 for either a column index or a row index and removes any found index if it has a 1 for both the column and row index. Then I want to take these indices and use them as indices in an array U, which is where the trouble comes. The indices do not seem to be stored as integers and I'm not sure what they are being stored as or why. I'm quite new to Matlab (but thats probably obvious) and so I don't really understand how types work for Matlab or how they're assigned. So I'm not sure why I',m getting the error message mentioned in the title and I'm not sure what to do about it. Any assistance you can provide would be greatly appreciated.
I forgot to mention this before but G is a matrix that only contains 1s or 0s and U is an array of strings (i think what would be called a cell?)
function A = ISClinks(U, G)
B = [];
[rownum,colnum] = size(G);
j = 1;
for i=1:colnum
s = sum(G(:,i));
if s == 1
B(j,:) = i;
j = j + 1;
end
end
for i=1:rownum
s = sum(G(i,:));
if s == 1
if ismember(i, B)
B(B == i) = [];
else
B(j,:) = i;
j = j+1;
end
end
end
A = [];
for i=1:size(B,1)
s = B(i,:);
A(i,:) = U(s,:);
end
end
This is the problem code, but I'm not sure what's wrong with it.
A = [];
for i=1:size(B,1)
s = B(i,:);
A(i,:) = U(s,:);
end
Your program seems to be structured as though it had been written in a language like C. In MATLAB, you can usually substitute specialized functions (e.g. any() ) for low-level loops in many cases. Your function could be written more efficiently as:
function A = ISClinks(U, G)
% Find columns and rows that are set in the input
active_columns=any(G,1);
active_rows=any(G,2).';
% (Optional) Prevent columns and rows with same index from being simultaneously set
%exclusive_active_columns = active_columns & ~active_rows; %not needed; this line is only for illustrative purposes
%exclusive_active_rows = active_rows & ~active_columns; %same as above
% Merge column state vector and row state vector by XORing them
active_indices=xor(active_columns,active_rows);
% Select appropriate rows of matrix U
A=U(active_indices,:);
end
This function does not cause errors with the example input matrices I tested. If U is a cell array (e.g. U={'Lorem','ipsum'; 'dolor','sit'; 'amet','consectetur'}), then return value A will also be a cell array.
Sorry for asking really basic question but I am confused in multiple loops.
I want to run loops all at same time and then break the loop and go to next loop for R, e and x combine.
Means for R =1 ,e=1 ,x=1 and then R=2 ,e=2 ,x=2 and so on.
Can someone tell me where I am at fault or what is missing to get my desired results?
Code:
threshold = [0.4:0.1:1.1];
limit_for_idx = [0.4:0.1:1.1];
limit = [0.4:0.1:1.1];
D=1;
E=1;
J=0;
for R = 1:numel(threshold);
for e = 1:numel(limit_for_idx);
for x = 1:numel(limit)
J = J+1 ;
% Perform Tasks and go to next loop for R ,e and x
break
end
break
end
end
The wrong in your code is that in the 2nd iteration and so on, e and x will get 1 again, and not 2 or which value that you want.
To solve that, just iterate one variable and equal them all to it:
threshold = [0.4:0.1:1.1];
limit_for_idx = [0.4:0.1:1.1];
limit = [0.4:0.1:1.1];
D=1;
E=1;
J=0;
for R=1:numel(threshold)
e=R;
x=R;
% do your stuff...
end
Hello again logical friends!
I’m aware this is quite an involved question so please bear with me! I think I’ve managed to get it down to two specifics:- I need two loops which I can’t seem to get working…
Firstly; The variable rollers(1).ink is a (12x1) vector containing ink values. This program shares the ink equally between rollers at each connection. I’m attempting to get rollers(1).ink to interact with rollers(2) only at specific timesteps. The ink should transfer into the system once for every full revolution i.e. nTimesSteps = each multiple of nBins_max. The ink should not transfer back to rollers(1).ink as the system rotates – it should only introduce ink to the system once per revolution and not take any back out. Currently I’ve set rollers(1).ink = ones but only for testing. I’m truly stuck here!
Secondly; The reason it needs to do this is because at the end of the sim I also wish to remove ink in the form of a printed image. The image should be a reflection of the ink on the last roller in my system and half of this value should be removed from the last roller and taken out of the system at each revolution. The ink remaining on the last roller should be recycled and ‘re-split’ in the system ready for the next rotation.
So…I think it’s around the loop beginning line86 where I need to do all this stuff. In pseudo, for the intermittent in-feed I’ve been trying something like:
For k = 1:nTimeSteps
While nTimesSteps = mod(nTimeSteps, nBins_max) == 0 % This should only output when nTimeSteps is a whole multiple of nBins_max i.e. one full revolution
‘Give me the ink on each segment at each time step in a matrix’
End
The output for averageAmountOfInk is the exact format I would like to return this data except I don’t really need the average, just the actual value at each moment in time. I keep getting errors for dimensional mismatches when I try to re-create this using something like:
For m = 1:nTimeSteps
For n = 1:N
Rollers(m,n) = rollers(n).ink’;
End
End
I’ll post the full code below if anyone is interested to see what it does currently. There’s a function at the end also which of course needs to be saved out to a separate file.
I’ve posted variations of this question a couple of times but I’m fully aware it’s quite a tricky one and I’m finding it difficult to get my intent across over the internets!
If anyone has any ideas/advice/general insults about my lack of programming skills then feel free to reply!
%% Simple roller train
% # Single forme roller
% # Ink film thickness = 1 micron
clc
clear all
clf
% # Initial state
C = [0,70; % # Roller centres (x, y)
10,70;
21,61;
11,48;
21,34;
27,16;
0,0
];
R = [5.6,4.42,9.8,6.65,10.59,8.4,23]; % # Roller radii (r)
% # Direction of rotation (clockwise = -1, anticlockwise = 1)
rotDir = [1,-1,1,-1,1,-1,1]';
N = numel(R); % # Amount of rollers
% # Find connected rollers
isconn = #(m, n)(sum(([1, -1] * C([m, n], :)).^2)...
-sum(R([m, n])).^2 < eps);
[Y, X] = meshgrid(1:N, 1:N);
conn = reshape(arrayfun(isconn, X(:), Y(:)), N, N) - eye(N);
% # Number of bins for biggest roller
nBins_max = 50;
nBins = round(nBins_max*R/max(R))';
% # Initialize roller struct
rollers = struct('position',{}','ink',{}','connections',{}',...
'rotDirection',{}');
% # Initialise matrices for roller properties
for ii = 1:N
rollers(ii).ink = zeros(1,nBins(ii));
rollers(ii).rotDirection = rotDir(ii);
rollers(ii).connections = zeros(1,nBins(ii));
rollers(ii).position = 1:nBins(ii);
end
for ii = 1:N
for jj = 1:N
if(ii~=jj)
if(conn(ii,jj) == 1)
connInd = getConnectionIndex(C,ii,jj,nBins(ii));
rollers(ii).connections(connInd) = jj;
end
end
end
end
% # Initialize averageAmountOfInk and calculate initial distribution
nTimeSteps = 1*nBins_max;
averageAmountOfInk = zeros(nTimeSteps,N);
inkPerSeg = zeros(nTimeSteps,N);
for ii = 1:N
averageAmountOfInk(1,ii) = mean(rollers(ii).ink);
end
% # Iterate through timesteps
for tt = 1:nTimeSteps
rollers(1).ink = ones(1,nBins(1));
% # Rotate all rollers
for ii = 1:N
rollers(ii).ink(:) = ...
circshift(rollers(ii).ink(:),rollers(ii).rotDirection);
end
% # Update all roller-connections
for ii = 1:N
for jj = 1:nBins(ii)
if(rollers(ii).connections(jj) ~= 0)
index1 = rollers(ii).connections(jj);
index2 = find(ii == rollers(index1).connections);
ink1 = rollers(ii).ink(jj);
ink2 = rollers(index1).ink(index2);
rollers(ii).ink(jj) = (ink1+ink2)/2;
rollers(index1).ink(index2) = (ink1+ink2)/2;
end
end
end
% # Calculate average amount of ink on each roller
for ii = 1:N
averageAmountOfInk(tt,ii) = sum(rollers(ii).ink);
end
end
image(5:20) = (rollers(7).ink(5:20))./2;
inkPerSeg1 = [rollers(1).ink]';
inkPerSeg2 = [rollers(2).ink]';
inkPerSeg3 = [rollers(3).ink]';
inkPerSeg4 = [rollers(4).ink]';
inkPerSeg5 = [rollers(5).ink]';
inkPerSeg6 = [rollers(6).ink]';
inkPerSeg7 = [rollers(7).ink]';
This is an extended comment rather than a proper answer, but the comment box is a bit too small ...
Your code overwhelms me, I can't see the wood for the trees. I suggest that you eliminate all the stuff we don't need to see to help you with your immediate problem (all those lines drawing figures for example) -- I think it will help you to debug your code yourself to put all that stuff into functions or scripts.
Your code snippet
For k = 1:nTimeSteps
While nTimesSteps = mod(nTimeSteps, nBins_max) == 0
‘Give me the ink on each segment at each time step in a matrix’
End
might be (I don't quite understand your use of the while statement, the word While is not a Matlab keyword, and as you have written it the value returned by the statement doesn't change from iteration to iteration) equivalent to
For k = 1:nBins_max:nTimeSteps
‘Give me the ink on each segment at each time step in a matrix’
End
You seem to have missed an essential feature of Matlab's colon operator ...
1:8 = [1 2 3 4 5 6 7 8]
but
1:2:8 = [1 3 5 7]
that is, the second number in the triplet is the stride between successive elements.
Your matrix conn has a 1 at the (row,col) where rollers are connected, and a 0 elsewhere. You can find the row and column indices of all the 1s like this:
[ri,ci] = find(conn==1)
You could then pick up the (row,col) locations of the 1s without the nest of loops and if statements that begins
for ii = 1:N
for jj = 1:N
if(ii~=jj)
if(conn(ii,jj) == 1)
I could go on, but won't, that's enough for one comment.
Please help me to improve the following Matlab code to improve execution time.
Actually I want to make a random matrix (size [8,12,10]), and on every row, only have integer values between 1 and 12. I want the random matrix to have the sum of elements which has value (1,2,3,4) per column to equal 2.
The following code will make things more clear, but it is very slow.
Can anyone give me a suggestion??
clc
clear all
jum_kel=8
jum_bag=12
uk_pop=10
for ii=1:uk_pop;
for a=1:jum_kel
krom(a,:,ii)=randperm(jum_bag); %batasan tidak boleh satu kelompok melakukan lebih dari satu aktivitas dalam satu waktu
end
end
for ii=1:uk_pop;
gab1(:,:,ii) = sum(krom(:,:,ii)==1)
gab2(:,:,ii) = sum(krom(:,:,ii)==2)
gab3(:,:,ii) = sum(krom(:,:,ii)==3)
gab4(:,:,ii) = sum(krom(:,:,ii)==4)
end
for jj=1:uk_pop;
gabh1(:,:,jj)=numel(find(gab1(:,:,jj)~=2& gab1(:,:,jj)~=0))
gabh2(:,:,jj)=numel(find(gab2(:,:,jj)~=2& gab2(:,:,jj)~=0))
gabh3(:,:,jj)=numel(find(gab3(:,:,jj)~=2& gab3(:,:,jj)~=0))
gabh4(:,:,jj)=numel(find(gab4(:,:,jj)~=2& gab4(:,:,jj)~=0))
end
for ii=1:uk_pop;
tot(:,:,ii)=gabh1(:,:,ii)+gabh2(:,:,ii)+gabh3(:,:,ii)+gabh4(:,:,ii)
end
for ii=1:uk_pop;
while tot(:,:,ii)~=0;
for a=1:jum_kel
krom(a,:,ii)=randperm(jum_bag); %batasan tidak boleh satu kelompok melakukan lebih dari satu aktivitas dalam satu waktu
end
gabb1 = sum(krom(:,:,ii)==1)
gabb2 = sum(krom(:,:,ii)==2)
gabb3 = sum(krom(:,:,ii)==3)
gabb4 = sum(krom(:,:,ii)==4)
gabbh1=numel(find(gabb1~=2& gabb1~=0));
gabbh2=numel(find(gabb2~=2& gabb2~=0));
gabbh3=numel(find(gabb3~=2& gabb3~=0));
gabbh4=numel(find(gabb4~=2& gabb4~=0));
tot(:,:,ii)=gabbh1+gabbh2+gabbh3+gabbh4;
end
end
Some general suggestions:
Name variables in English. Give a short explanation if it is not immediately clear,
what they are indented for. What is jum_bag for example? For me uk_pop is music style.
Write comments in English, even if you develop source code only for yourself.
If you ever have to share your code with a foreigner, you will spend a lot of time
explaining or re-translating. I would like to know for example, what
%batasan tidak boleh means. Probably, you describe here that this is only a quick
hack but that someone should really check this again, before going into production.
Specific to your code:
Its really easy to confuse gab1 with gabh1 or gabb1.
For me, krom is too similar to the built-in function kron. In fact, I first
thought that you are computing lots of tensor products.
gab1 .. gab4 are probably best combined into an array or into a cell, e.g. you
could use
gab = cell(1, 4);
for ii = ...
gab{1}(:,:,ii) = sum(krom(:,:,ii)==1);
gab{2}(:,:,ii) = sum(krom(:,:,ii)==2);
gab{3}(:,:,ii) = sum(krom(:,:,ii)==3);
gab{4}(:,:,ii) = sum(krom(:,:,ii)==4);
end
The advantage is that you can re-write the comparsisons with another loop.
It also helps when computing gabh1, gabb1 and tot later on.
If you further introduce a variable like highestNumberToCompare, you only have to
make one change, when you certainly find out that its important to check, if the
elements are equal to 5 and 6, too.
Add a semicolon at the end of every command. Having too much output is annoying and
also slow.
The numel(find(gabb1 ~= 2 & gabb1 ~= 0)) is better expressed as
sum(gabb1(:) ~= 2 & gabb1(:) ~= 0). A find is not needed because you do not care
about the indices but only about the number of indices, which is equal to the number
of true's.
And of course: This code
for ii=1:uk_pop
gab1(:,:,ii) = sum(krom(:,:,ii)==1)
end
is really, really slow. In every iteration, you increase the size of the gab1
array, which means that you have to i) allocate more memory, ii) copy the old matrix
and iii) write the new row. This is much faster, if you set the size of the
gab1 array in front of the loop:
gab1 = zeros(... final size ...);
for ii=1:uk_pop
gab1(:,:,ii) = sum(krom(:,:,ii)==1)
end
Probably, you should also re-think the size and shape of gab1. I don't think, you
need a 3D array here, because sum() already reduces one dimension (if krom is
3D the output of sum() is at most 2D).
Probably, you can skip the loop at all and use a simple sum(krom==1, 3) instead.
However, in every case you should be really aware of the size and shape of your
results.
Edit inspired by Rody Oldenhuis:
As Rody pointed out, the 'problem' with your code is that its highly unlikely (though
not impossible) that you create a matrix which fulfills your constraints by assigning
the numbers randomly. The code below creates a matrix temp with the following characteristics:
The numbers 1 .. maxNumber appear either twice per column or not at all.
All rows are a random permutation of the numbers 1 .. B, where B is equal to
the length of a row (i.e. the number of columns).
Finally, the temp matrix is used to fill a 3D array called result. I hope, you can adapt it to your needs.
clear all;
A = 8; B = 12; C = 10;
% The numbers [1 .. maxNumber] have to appear exactly twice in a
% column or not at all.
maxNumber = 4;
result = zeros(A, B, C);
for ii = 1 : C
temp = zeros(A, B);
for number = 1 : maxNumber
forbiddenRows = zeros(1, A);
forbiddenColumns = zeros(1, A/2);
for count = 1 : A/2
illegalIndices = true;
while illegalIndices
illegalIndices = false;
% Draw a column which has not been used for this number.
randomColumn = randi(B);
while any(ismember(forbiddenColumns, randomColumn))
randomColumn = randi(B);
end
% Draw two rows which have not been used for this number.
randomRows = randi(A, 1, 2);
while randomRows(1) == randomRows(2) ...
|| any(ismember(forbiddenRows, randomRows))
randomRows = randi(A, 1, 2);
end
% Make sure not to overwrite previous non-zeros.
if any(temp(randomRows, randomColumn))
illegalIndices = true;
continue;
end
end
% Mark the rows and column as forbidden for this number.
forbiddenColumns(count) = randomColumn;
forbiddenRows((count - 1) * 2 + (1:2)) = randomRows;
temp(randomRows, randomColumn) = number;
end
end
% Now every row contains the numbers [1 .. maxNumber] by
% construction. Fill the zeros with a permutation of the
% interval [maxNumber + 1 .. B].
for count = 1 : A
mask = temp(count, :) == 0;
temp(count, mask) = maxNumber + randperm(B - maxNumber);
end
% Store this page.
result(:,:,ii) = temp;
end
OK, the code below will improve the timing significantly. It's not perfect yet, it can all be optimized a lot further.
But, before I do so: I think what you want is fundamentally impossible.
So you want
all rows contain the numbers 1 through 12, in a random permutation
any value between 1 and 4 must be present either twice or not at all in any column
I have a hunch this is impossible (that's why your code never completes), but let me think about this a bit more.
Anyway, my 5-minute-and-obvious-improvements-only-version:
clc
clear all
jum_kel = 8;
jum_bag = 12;
uk_pop = 10;
A = jum_kel; % renamed to make language independent
B = jum_bag; % and a lot shorter for readability
C = uk_pop;
krom = zeros(A, B, C);
for ii = 1:C;
for a = 1:A
krom(a,:,ii) = randperm(B);
end
end
gab1 = sum(krom == 1);
gab2 = sum(krom == 2);
gab3 = sum(krom == 3);
gab4 = sum(krom == 4);
gabh1 = sum( gab1 ~= 2 & gab1 ~= 0 );
gabh2 = sum( gab2 ~= 2 & gab2 ~= 0 );
gabh3 = sum( gab3 ~= 2 & gab3 ~= 0 );
gabh4 = sum( gab4 ~= 2 & gab4 ~= 0 );
tot = gabh1+gabh2+gabh3+gabh4;
for ii = 1:C
ii
while tot(:,:,ii) ~= 0
for a = 1:A
krom(a,:,ii) = randperm(B);
end
gabb1 = sum(krom(:,:,ii) == 1);
gabb2 = sum(krom(:,:,ii) == 2);
gabb3 = sum(krom(:,:,ii) == 3);
gabb4 = sum(krom(:,:,ii) == 4);
gabbh1 = sum(gabb1 ~= 2 & gabb1 ~= 0)
gabbh2 = sum(gabb2 ~= 2 & gabb2 ~= 0);
gabbh3 = sum(gabb3 ~= 2 & gabb3 ~= 0);
gabbh4 = sum(gabb4 ~= 2 & gabb4 ~= 0);
tot(:,:,ii) = gabbh1+gabbh2+gabbh3+gabbh4;
end
end
I have the following code, pasted below. I would like to change it to only average the 10 most recently filtered images and not the entire group of filtered images. The line I think I need to change is: Yout(k,p,q) = (Yout(k,p,q) + (y.^2))/2;, but how do I do it?
j=1;
K = 1:3600;
window = zeros(1,10);
Yout = zeros(10,column,row);
figure;
y = 0; %# Preallocate memory for output
%Load one image
for i = 1:length(K)
disp(i)
str = int2str(i);
str1 = strcat(str,'.mat');
load(str1);
D{i}(:,:) = A(:,:);
%Go through the columns and rows
for p = 1:column
for q = 1:row
if(mean2(D{i}(p,q))==0)
x = 0;
else
if(i == 1)
meanvalue = mean2(D{i}(p,q));
end
%Calculate the temporal mean value based on previous ones.
meanvalue = (meanvalue+D{i}(p,q))/2;
x = double(D{i}(p,q)/meanvalue);
end
%Filtering for 10 bands, based on the previous state
for k = 1:10
[y, ZState{k}] = filter(bCoeff{k},aCoeff{k},x,ZState{k});
Yout(k,p,q) = (Yout(k,p,q) + (y.^2))/2;
end
end
end
% for k = 2:10
% subplot(5,2,k)
% subimage(Yout(k)*5000, [0 100]);
% colormap jet
% end
% pause(0.01);
end
disp('Done Loading...')
The best way to do this (in my opinion) would be to use a circular-buffer to store your images. In a circular-, or ring-buffer, the oldest data element in the array is overwritten by the newest element pushed in to the array. The basics of making such a structure are described in the short Mathworks video Implementing a simple circular buffer.
For each iteration of you main loop that deals with a single image, just load a new image into the circular-buffer and then use MATLAB's built in mean function to take the average efficiently.
If you need to apply a window function to the data, then make a temporary copy of the frames multiplied by the window function and take the average of the copy at each iteration of the loop.
The line
Yout(k,p,q) = (Yout(k,p,q) + (y.^2))/2;
calculates a kind of Moving Average for each of the 10 bands over all your images.
This line calculates a moving average of meanvalue over your images:
meanvalue=(meanvalue+D{i}(p,q))/2;
For both you will want to add a buffer structure that keeps only the last 10 images.
To simplify it, you can also just keep all in memory. Here is an example for Yout:
Change this line: (Add one dimension)
Yout = zeros(3600,10,column,row);
And change this:
for q = 1:row
[...]
%filtering for 10 bands, based on the previous state
for k = 1:10
[y, ZState{k}] = filter(bCoeff{k},aCoeff{k},x,ZState{k});
Yout(i,k,p,q) = y.^2;
end
YoutAvg = zeros(10,column,row);
start = max(0, i-10+1);
for avgImg = start:i
YoutAvg(k,p,q) = (YoutAvg(k,p,q) + Yout(avgImg,k,p,q))/2;
end
end
Then to display use
subimage(Yout(k)*5000, [0 100]);
You would do sth. similar for meanvalue