Firstly I'm new to matlab programming. Here is my doubt. I would like to break the loop when the cells in this matrix is either empty or a NaN. Matlab suggesting not to have isempty and isnan in same line. The error i get is "Attempted to access coordinates(10,1); index out of bounds because size(coordinates)=[9,8].". I'm messing up somewhere, could someone help on this? Thanks.
n_b=0;
n_nodes=0;
for j=1:2:10
for i=1:1:10
if (isempty(coordinates(i,j))) or (isnan(coordinates(i,j)))
break
else
disp(coordinates(i,j));
n_nodes=n_nodes+1;
end
end
n_b=n_b+1;
if (n_b==4) % break at four columns
break
end
end
The problem is with this lines:
for j=1:2:10
for i=1:1:10
You are trying to access coordinates that do not exist. Try this:
for j = 1:2:size(coordinates,2)
for i = 1:size(coordinates,1)
This will always stop before the matrix ends.
Also, delete the isempty() check from your condition, as it never fulfills, together with the or which is not a valid MATLAB syntax (should be ||).
Here is how you full code should look:
n_b = 0;
n_nodes = 0;
for j = 1:2:size(coordinates,2)
for i = 1:size(coordinates,1)
if isnan(coordinates(i,j))
break
else
disp(coordinates(i,j));
n_nodes = n_nodes+1;
end
end
n_b = n_b+1;
if (n_b==4) % break at four columns
break
end
end
Say I have a for loop in MATLAB:
scales = 5:5:95;
for scale = scales
do stuff
end
How can I get the iteration number inside a MATLAB for loop as concisely as possible?
In Python for example I would use:
for idx, item in enumerate(scales):
where idx is the iteration number.
I know that in MATLAB (like in any other language) I could create a count variable:
scales = 5:5:95;
scale_count = 0;
for scale = scales
scale_count = scale_count + 1;
do stuff
end
I could otherwise use find:
scales = 5:5:95;
for scale = scales
scale_count = find(scales == scale);
do stuff
end
But I'm curious to know whether there exists a more concise way to do it, e.g. like in the Python example.
Maybe you can use the following:
scales = 5:5:95;
for iter = 1:length(scales)
scale=scales(iter); % "iter" is the iteration number.
do stuff
end
Since for iterates over the columns of whatever you give it, another way of approximating multiple loop variables would be to use an appropriately constructed matrix:
for scale=[5:5:95; 1:19]
% do stuff with scale(1) or scale(2) as appropriate
end
(my personal preference is to loop over the indices as per Parag's answer and just refer to data(index) directly within the loop, without an intermediate. Matlab's syntax isn't very concise at the best of times - you just get used to it)
The MATLAB way is probably doing it with vectors.
For example suppose you want to find in a vector if there is a value that is equal to its position. You would generally do this:
a = [10 20 1 3 5];
found = 0;
for index = 1:length(a)
if a(index) == index
found = 1;
break;
end
end
Instead you can do:
found = any(a == 1:length(a));
In general
for i=1:length(a)
dostuff(a(i), i);
end
can be replaced with:
dostuff(a(i), 1:length(a))
it dostuff can be vectorized or
arrayfun(#dostuff, a, 1:length(a))
otherwise.
I am trying to make my program work. It is for prime numbers.
Below is my function then my main program
I believe this is the right formatting, (obviously I'm probably wrong cause its not working) and I have been trying to fix it to no avail. Any help will be most appreciated and an explanation of what I am doing wrong (I am pretty sure it has to do with subtle formatting) would be great too.
function [answer,primefactors ] = primey1 (N)
for i=2:(N-1)
A(i-1)=rem(N,i);
end
A;
if(all(A)==1)
answer=['Yes']
primefactors=[1,N]
elseif(all(A)==0)
answer=['No']
fac=[]
for i=2:N
q=0;
while N/i==floor(N/i)
N=N/i;
q=q+1;
end
if q>0
fac=[fac,i]
if N==1
break
primefactors=[fac]
end
end
end
end
endfunction
As noted by Magla, MATLAB comes with the primes and factor functions, which you can at least use to compare your implementation with and/or check your outcomes.
Anyway, as for your code, try this:
function [answer, primefactors] = primey1(N)
% Use vectorization for simple cases such as these:
A = rem(N,2:N-1);
if all(A)
answer = 'Yes';
primefactors = [1,N];
% Re-wrote this entire section. There were a bunch of things wrong with it:
% the order of commands was wrong, variables were not assigned for some N,
% etc. Just compare the two implementations
else
answer = 'No';
primefactors = [];
for ii = 2:N
q = 0;
while N/ii == floor(N/ii)
N = N/ii;
q = q+1;
end
if q > 0
primefactors = [primefactors,ii]; %#ok<AGROW>
if N==1
break;
end
end
end
end
end
Matlab has a factor function that does what your code is trying to do
p = factor(10)
returns 2, 5
and
p = factor(11)
returns 11.
All you have is to test for length
if length(p) == 1
is true for prime numbers.
The solution as offered by #Rody should do the job, in theory even more efficient than this, however to illustrate the concept of initialization, I would recommend you to initialize the output variables of your function right after the function start. In your case this would mean that I recommend starting the code like this:
function [answer, primefactors] = primey1(N)
% Function to determine whether a number is prime and which prime factors it has
% Assign default values
answer = 'No';
primefactors = [];
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