I Have my function below, the idea being that X is a 3x3 extract from T to be used in the loop, it correctly extracts the 3 rows but for some reason produces far too many columns, see example below.
function T = tempsim(rows, cols, topNsideTemp, bottomTemp, tol)
T = zeros(rows,cols);
T(1,:) = topNsideTemp;
T(:,1) = topNsideTemp;
T(:,rows) = topNsideTemp;
T(rows,:) = bottomTemp;
S = [0 1 0; 1 1 1; 0 1 0];
X = zeros(3,3);
A = zeros(3,3);
for ii = 2:(cols-1);
jj = 2:(rows-1);
X = T([(ii-1) ii (ii+1)], [(jj-1) jj (jj+1)])
A = X.*S;
T = (sum(sum(A)))/5
end
test sample
EDU>> T = tempsim(5,4,100,50,0)
X =
100 100 100 100 100 100 100 100 100
100 0 0 0 0 0 0 0 100
100 0 0 0 0 0 0 0 100
ans =
100 100 100 100 100 100 100 100 100
100 0 0 0 0 0 0 0 100
100 0 0 0 0 0 0 0 100
??? Error using ==> times
Matrix dimensions must agree.
Error in ==> tempsim at 14
A = X.*S;
any thoughts on how to fix this?
There's no need to preallocate X and A if you do a complete assignment anyway. Then, you replace T with a scalar inside the loop, which makes you run into problems in the next iteration. What I'm guessing you want could look something like this:
function T = tempsim(rows, cols, topNsideTemp, bottomTemp, tol)
T = zeros(rows,cols);
T(1,:) = topNsideTemp;
T(:,1) = topNsideTemp;
T(:,rows) = topNsideTemp;
T(rows,:) = bottomTemp;
S = [0 1 0; 1 1 1; 0 1 0];
for ii = 1:(cols-2);
for jj = 1:(rows-2);
X = T(ii:ii+2, jj:jj+2);
A = X.*S;
T(ii,jj) = (sum(sum(A)))/5;
end
end
Although I'm not sure if you really mean to do that – you're working on T while modifying it. As a wild guess, I suspect you might be looking for something like
conv2(T, S/5, 'same')
instead, perhaps after making your fixed-temp borders twice as thick and re-setting them after the call (since conv2 does zero-padding at the outer borders).
Here:
jj = 2:(rows-1);
X = T([(ii-1) ii (ii+1)], [(jj-1) jj (jj+1)])
jj becomes [2 3 4]
so X is
T([1 2 3], [ [2 3 4]-1 [2 3 4] [2 3 4]+1 ])
You probably missed a for loop.
Related
I trying to make a program that uses the for loop to calculate the set of index D. But I have a problem because the length of index are not the same.
Example:
z = [0 0 0 0 0 0 1]
v(1,:) = [1 0 0 0 1 0 1]
v(2,:) = [0 1 0 0 1 1 1]
v(3,:) = [0 0 1 0 1 1 0]
v(4,:) = [0 0 0 1 0 1 1]
v(1,:) = find(v(1,:)~=z);
v(2,:) = find(v(2,:)~=z);
v(3,:) = find(v(3,:)~=z);
v(4,:) = find(v(4,:)~=z);
we obtain :
D(1,:) = [1 5];
D(2,:) = [2 5 6];
D(3,:) = [3 5 6 7];
D(4,:) = [4 6];
Code :
for aa = 1:4
D(aa,:) = [find(v(aa,:)~=z)];
end
not work because length(D(1,:))~=length(D(2,:))~=length(D(3,:))
How I can use a loop to determine set of index D?
Thank you for any help!
One solution can be using cell like the following:
for aa = 1:4
D{aa} = [find(v(aa,:)~=z)];
end
You can use the matrix D, but initialize it beforehand like:
D = ones(size(v)) + length(z)
Then fill it like:
for ii = 1:size(z,1)
D(ii,v(ii,:)~=z) = find(v(ii,:)~=z);
end
Notice, I added the length of v to the matrix of ones, such that you are sure that the predefined numbers in the matrix are larger than any index, hence the min() will not freak out.
Hello everybody I have a very simple problem, I have too many data y, p and r. So I want to calculate it in a single code.
This an example of my code if I breakdown into separate code
y1=45
y2=56
y3=67
p1=34
p2=45
p3=56
r1=23
r2=34
r3=45
Ryaw1=[cosd(y1) -sind(y1) 0;
sind(y1) cosd(y1) 0;
0 0 1]
Rpitch1=[cosd(p1) 0 sind(p1);
0 1 0;
-sind(p1) 0 cos(p1)]
Rroll1=[1 0 0;
0 cosd(r1) -sind(r1);
0 sind(r1) cosd(r1)]
R1=Ryaw1*Rpitch1*Rroll1
Coordinate1=R1*X0
Ryaw2=[cosd(y2) -sind(y2) 0;
sind(y2) cosd(y2) 0;
0 0 1]
Rpitch2=[cosd(p2) 0 sind(p2);
0 1 0;
-sind(p2) 0 cos(p2)]
Rroll2=[1 0 0;
0 cosd(r2) -sind(r2);
0 sind(r2) cosd(r2)]
R2=Ryaw2*Rpitch2*Rroll2
Coordinate2=R2*X0
Ryaw3=[cosd(y3) -sind(y3) 0;
sind(y3) cosd(y3) 0;
0 0 1]
Rpitch3=[cosd(p3) 0 sind(p3);
0 1 0;
-sind(p3) 0 cos(p3)]
Rroll3=[1 0 0;
0 cosd(r3) -sind(r3);
0 sind(r3) cosd(r3)]
R3=Ryaw3*Rpitch3*Rroll3
Coordinate3=R3*X0
Coordinate=[Cooedinate1 Coordinate2 Coordinate3]
The goals is to find "Coordinate" (in matrix - combined from Coordinate1, Coordinate2, Coordinate3, .... ,Coordinate..) from every y, p and r data with the same "X0" as a single primary data for calculation.
Sorry for my bad english,
Thanks :)
Use vectors and matrices instead of individual scalars. These are indexed in almost the same way as you had before, i.e. y1 becomes y(1).
Then you can easily loop over the code 3 times and save the repetition.
See my commented code below.
% Define some X0. This should be a column vector.
X0 = [1; 2; 3];
% Make y,p,r into 3 element vectors
y = [45 56 67];
p = [34 45 56];
r = [23 34 45];
% Make R, Ryaw, Rpitch and Rroll 3x3x3 matrices
R = zeros(3,3,3);
Ryaw = zeros(3,3,3);
Rpitch = zeros(3,3,3);
Rroll = zeros(3,3,3);
% Make Coordinate a 3x3 matrix
Coordinate = zeros(3,3);
% Loop k from 1 to 3
% For each 3x3x3 matrix, the kth 3x3 matrix is equivalent to your Ryawk, Rpitchk, Rrollk, Rk
for k = 1:3
Ryaw(:,:,k) = [cosd(y(k)) -sind(y(k)) 0
sind(y(k)) cosd(y(k)) 0
0 0 1];
Rpitch(:,:,k)= [cosd(p(k)) 0 sind(p(k))
0 1 0
-sind(p(k)) 0 cos(p(k))];
Rroll(:,:,k) = [1 0 0
0 cosd(r(k)) -sind(r(k))
0 sind(r(k)) cosd(r(k))];
R(:,:,k) = Ryaw(:,:,k)*Rpitch(:,:,k)*Rroll(:,:,k);
Coordinate(:,k) = R(:,:,k)*X0;
end
disp(Coordinate)
I'd like to create a 4x12 matrix which is very similar to a upper triangle matrix, it looks like this:
1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 1 1 1
So my question is. What is the most efficient way to create it? no loops, no cellfun. Thanks.
One vectorized approach -
nrows = 4;
ncols = 12;
row_idx = repmat(1:nrows,ncols/nrows,1)
out = bsxfun(#le,[1:nrows]',row_idx(:).')
The Matlab R2015a and later approach using the newly introduced repelem:
n = 4;
m = 3;
out = repelem(triu(ones(n)),1,m);
out =
1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 1 1 1
It even seems faster than the bsxfun approach, though I can't believe this ;)
Benchmark
Unfortunately I couldn't consider andrew's solution as it is not complete and I didn't got it totally.
function [t] = bench()
n = 4;
m = 12;
t = zeros(3,15);
for ii = 1:15
fcns = {
#() thewaywewalk(ii*n,ii*m);
#() Divakar(ii*n,ii*m);
#() LuisMendo(ii*n,ii*m);
};
% timeit
for jj = 1:100;
t(:,ii) = t(:,ii) + cellfun(#timeit, fcns);
end
end
plot(1:15,t(1,:)); hold on;
plot(1:15,t(2,:)); hold on;
plot(1:15,t(3,:)); hold on;
xlabel('Matrix size: n = x*4, m = x*12')
ylabel('timing')
legend({'thewaywewalk','Divakar','Luis Mendo'},'location','northwest')
end
function Z = thewaywewalk(n,m)
Z = repelem(triu(ones(n)),1,m/n);
end
function Z = Divakar(n,m)
row_idx = repmat(1:n,m/n,1);
Z = bsxfun(#le,[1:n]',row_idx(:).');
end
function Z = LuisMendo(n,m)
Z = reshape(repmat(permute(triu(ones(n,n)), [1 3 2]), [1 m/n 1]), [n m]);
end
First bottomline - small matrices:
The new repelem does a very good job, but also the reshape(repmat(permute... does not disappoint. The bsxfun approach stays a little behind for some medium-sized matrices, before it becomes the leader for large matrices:
Second bottomline - big matrices:
As predicted by Divakar, bsxfun is the fastest for larger matrices, actually as expected as bsxfun is always the fastest! Interesting that the other two align perfectly, on could guess they almost work the same internally.
Create an upper triangular matrix of ones, permute second and third dimensions, repeat along second dimension, and reshape into desired shape:
m = 4;
n = 12;
result = reshape(repmat(permute(triu(ones(m,m)), [1 3 2]), [1 n/m 1]), [m n]);
depending on your matlab version
m = 4;
n = 12;
dec2bin(bitshift(num,-1*[0:n/m:n-1])) %this prints out a string
these should be logical arrays (i don't have either of these so I cant test it)
decimalToBinaryVector(bitshift(num,-1*[0:n/m:n-1]))
de2bi(bitshift(num,-1*[0:n/m:n-1]))
I have the following code that includes 3 iterated for loops in order to create an upper diagonal matrix, I plan on performing on large data set many times and want to make as computationally efficient as possible.
data = magic(3);
n = size(data,1);
W = zeros(n,n);
for i = 1:n
for j = i:n
if i==j
W(i,j)=0;
else
for k = 1:n
temp(1,k) = (data(i,k)-data(j,k))^2;
sumTemp = sumTemp + temp(1,k);
end
W(i,j)=sqrt(sumTemp);
end
temp = 0;
sumTemp = 0;
end
end
Answer should look like:
[0 6.4807 9.7980
0 0 6.4807
0 0 0]
I am working it hard right now, but figure I would throw it out there in case anyone has any suggestions that would save me hours of fiddling around.
This is hat I have at the moment:
data = magic(3);
n = size(data,1);
W = zeros(n,n);
for i = 1:n
for j = i+1:n
W(i,j)= norm(data(i,:)-data(j,:))
%W(i,j)= sqrt(sum((data(i,:)-data(j,:)).^2));
end
end
What I did:
vecorized the inner loop
removed www, which is unused
changed 2nd loop, start at i+1 because nothing is done for i=j
Replaced sqrt((a-b).^2) with norm(a-b)
And now the "full" vectorization:
data = magic(3);
n = size(data,1);
W = zeros(n,n);
tri=triu(ones(n,n),1)>0;
[i,j]=find(tri);
W(tri)=arrayfun(#(i,j)norm(data(i,:)-data(j,:)),i,j)
Here is a straightforward solution with bsxfun:
Wfull = sqrt(squeeze(sum(bsxfun(#minus,data,permute(data,[3 2 1])).^2,2)))
W = triu(Wfull)
Use this where data is N-by-D, where N is the number of points and D is dimensions. For example,
>> data = magic(3);
>> triu(sqrt(squeeze(sum(bsxfun(#minus,data,permute(data,[3 2 1])).^2,2))))
ans =
0 6.4807 9.7980
0 0 6.4807
0 0 0
>> data = magic(5); data(:,end-1:end)=[]
data =
17 24 1
23 5 7
4 6 13
10 12 19
11 18 25
>> triu(sqrt(squeeze(sum(bsxfun(#minus,data,permute(data,[3 2 1])).^2,2))))
ans =
0 20.8087 25.2389 22.7376 25.4558
0 0 19.9499 19.0263 25.2389
0 0 0 10.3923 18.3576
0 0 0 0 8.5440
0 0 0 0 0
>>
I want to generate a matrix that is "stairsteppy" from a vector.
Example input vector: [8 12 17]
Example output matrix:
[1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1]
Is there an easier (or built-in) way to do this than the following?:
function M = stairstep(v)
M = zeros(length(v),max(v));
v2 = [0 v];
for i = 1:length(v)
M(i,(v2(i)+1):v2(i+1)) = 1;
end
You can do this via indexing.
A = eye(3);
B = A(:,[zeros(1,8)+1, zeros(1,4)+2, zeros(1,5)+3])
Here's a solution without explicit loops:
function M = stairstep(v)
L = length(v); % M will be
V = max(v); % an L x V matrix
M = zeros(L, V);
% create indices to set to one
idx = zeros(1, V);
idx(v + 1) = 1;
idx = cumsum(idx) + 1;
idx = sub2ind(size(M), idx(1:V), 1:V);
% update the output matrix
M(idx) = 1;
EDIT: fixed bug :p
There's no built-in function I know of to do this, but here's one vectorized solution:
v = [8 12 17];
N = numel(v);
M = zeros(N,max(v));
M([0 v(1:N-1)]*N+(1:N)) = 1;
M(v(1:N-1)*N+(1:N-1)) = -1;
M = cumsum(M,2);
EDIT: I like the idea that Jonas had to use BLKDIAG. I couldn't help playing with the idea a bit until I shortened it further (using MAT2CELL instead of ARRAYFUN):
C = mat2cell(ones(1,max(v)),1,diff([0 v]));
M = blkdiag(C{:});
A very short version of a vectorized solution
function out = stairstep(v)
% create lists of ones
oneCell = arrayfun(#(x)ones(1,x),diff([0,v]),'UniformOutput',false);
% create output
out = blkdiag(oneCell{:});
You can use ones to define the places where you have 1's:
http://www.mathworks.com/help/techdoc/ref/ones.html