This code works perfectly in Octave, but not in Matlab. But why? Is there any workaround? Thanks.
a = [0; 5; 10];
b = [3 5 7];
a >= b
Octave behavior:
0 0 0
1 1 0
1 1 1
Matlab behavior:
Error using >
Matrix dimensions must agree.
Use bsxfun:
>> bsxfun( #ge, a, b )
ans =
0 0 0
1 1 0
1 1 1
bsxfun is so much FUN!
Related
I need to check if a matrix is a diagonal block matrix or not. Is there any easy way to check it? Especially, that would be perfect if a simple function such as isdiag()exist.
A function isdiag actually exists for R2014a and later. It works for square and non-square matrices. For earlier versions of Matlab see an alternative below.
Here is a quick demo:
A = [1 0 0;
0 2 0;
0 0 3];
isdiag(A)
B = [1 0 0;
0 2 3;
0 0 4];
isdiag(B)
C = [1 0 0 0;
0 2 0 0;
0 0 3 0];
isdiag(C)
D = [1 0 0 0;
0 2 0 3;
0 0 4 0];
isdiag(D)
... and we get:
ans =
1
ans =
0
ans =
1
ans =
0
In R2014a and later, you can as well use isbanded with a lower and upper bandwidth of 0. This will give you exactly the same result:
isbanded(M,0,0)
For versions before R2014a, you can use a combination of find and all:
[t1,t2] = find(M);
res = all(t1==t2)
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]))
can I know how can I replace values in specific matrix position without using for loop in MATLAB? I initialize matrix a that I would like to replace its value on specified row and column for each no. This has to be done a few time within num for loop. The num for loop is important here because I would want the update the value in the original code.
The real code is more complicated, I am simplifying the code for this question.
I have the code as follow:
a = zeros(2,10,15);
for num = 1:10
b = [2 2 1 1 2 2 2 1 2 2 2 2 1 2 2];
c = [8.0268 5.5218 2.9893 5.7105 7.5969 7.5825 7.0740 4.6471 ...
6.3481 14.7424 13.5594 10.6562 7.3160 -4.4648 30.6280];
d = [1 1 1 2 1 1 1 1 1 1 3 1 6 1 1];
for no = 1:15
a(b(no),d(no),no) = c(1,no,:)
end
end
A sample output for, say no 13 is as follows:
a(:,:,13) =
Columns 1 through 8
0 0 0 0 0 7.3160 0 0
0 0 0 0 0 0 0 0
Columns 9 through 10
0 0
0 0
Thank you so much for any help I could get.
It can be done using sub2ind, which casts the subs to a linear index.
Following your vague variable names, it would look like this (omitting the useless loop over num):
a = zeros(2,10,15);
b = [2 2 1 1 2 2 2 1 2 2 2 2 1 2 2];
d = [1 1 1 2 1 1 1 1 1 1 3 1 6 1 1];
c = [8.0268 5.5218 2.9893 5.7105 7.5969 7.5825 7.0740 4.6471 ...
6.3481 14.7424 13.5594 10.6562 7.3160 -4.4648 30.6280];
% // we vectorize the loop over no:
no = 1:15;
a(sub2ind(size(a), b, d, no)) = c;
Apart from the sub2ind based approach as suggested in Nras's solution, you can use a "raw version" of sub2ind to reduce a function call if performance is very critical. The related benchmarks comparing sub2ind and it's raw version is listed in another solution. Here's the implementation to solve your case -
no = 1:15
a = zeros(2,10,15);
[m,n,r] = size(a)
a((no-1)*m*n + (d-1)*m + b) = c
Also for pre-allocation, you can use a much faster approach as listed in Undocumented MATLAB blog post on Preallocation performance with -
a(2,10,15) = 0;
The function sub2ind is your friend here:
a = zeros(2,10,15);
x = [2 2 1 1 2 2 2 1 2 2 2 2 1 2 2];
y = [1 1 1 2 1 1 1 1 1 1 3 1 6 1 1];
z = 1:15;
dat = [8.0268 5.5218 2.9893 5.7105 7.5969 7.5825 7.0740 4.6471 ...
6.3481 14.7424 13.5594 10.6562 7.3160 -4.4648 30.6280];
inds = sub2ind(size(a), x, y, z);
a(inds) = dat;
Matlab provides a function 'sub2ind' may do what you expected.
with variable as the same you posted:
idx = sub2ind(size(a),b,d,[1:15]); % return the index of row a column b and page [1:15]
a(idx) = c;
how can I achieve the matrix C with following conditions:
A(i,j) is any B(k) , then A(i,j) = B(k)
else A(i,j) = 0
example:
A = [1 0 3 6 3 4; 2 0 3 1 8 2];
B = [1;2;3];
C = [1 0 3 0 3 0; 2 0 3 1 0 2]
Thank you!
A.*ismember(A,B)
Well not much to explain, ismember fits exactly your condition. Thus ismember(A,B) is 1 for all values you want to copy.
If your concern is to avoid products, you may try:
A(~ismember(A,B))=0;
depending on what you're looking for, from the matlab documentation
[LIA,LOCB] = ismember(A,B) also returns an array LOCB containing the
highest absolute index in B for each element in A which is a member of
B and 0 if there is no such index.
a = [9 9 8 8 7 7 7 6 6 6 5 5 4 4 2 1 1 1]
b = [1 1 1 3 3 3 3 3 4 4 4 4 4 9 9 9]
[lia1,locb1] = ismember(a,b)
% returns
lia1 = [1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 1]
locb1 = [16 16 0 0 0 0 0 0 0 0 0 0 13 13 0 3 3 3]
Another possibility, without ismember:
A .* any(bsxfun(#eq, A, permute(B,[3 2 1])),3)
Thank you guys!
I followed your advise and extended the code for my task. I guess my question was unclear formulated:
Following your advice, the Matrix C will give only 1 or 0.
I need the actual value of A in it.
In my following solution the code just rewrites A in the end. I gues there are many smarter solutions, but it works.
A = [1 0 3 6 3 4; 2 0 3 1 8 2];
B = [1;2;3];
[N M] = size (A);
C = ismember(A,B);
for i = 1:N
for j = 1:M
if C(i,j) == 0;
A(i,j) = 0;
end
end
end
Given the matrix with coordinates in 3D space and values for two variables (say a and b) in two matrices I would like to merge rows for same points into a common matrix.
To clearly explain the matter, let's say we have matrices
A=[posX, posY, posZ, a]
and
B=[posX, posY, posZ, b]
and would like to combine them into
AB = [posX, posY, posZ, a, b]
for example
A = [0 0 1 1; 0 1 0 4; 5 0 12 8];
B = [0 0 0 5; 0 1 0 3; 5 11 7 7];
would give
AB = [0 0 0 0 5; 0 0 1 1 0; 0 1 0 4 3; 5 0 12 8 0; 5 11 7 0 7];
In order to do that I first created
ATemp = [A, zeros(length(A,0)]
and
BTemp = [B(:, [1 2 3]), zeros(length(B),1), B(:,4)]
and then tried to use functions accumarray and grpstats but haven't managed to form the AB matrix.
I would be very thankful if anyone suggested the way to get the desired matrix.
AB=union(A(:,1:3),B(:,1:3),'rows');
AB(ismember(AB,A(:,1:3),'rows'),4)=A(:,4);
AB(ismember(AB(:,1:3),B(:,1:3),'rows'),5)=B(:,4)
[edit] This solution is only valid if each (x,y,z)-point occurs only once in each matrix. If there are several, there is a dimension mismatch in the second line (and/or the third).