I need to reset the 'lower triangle' of a 3 dimentional matrix.
This means, that if the original matrix is:
C(:,:,1) = [1 2 3 ; 2 4 6 ; 3 6 9]
C(:,:,2) = [2 4 6 ; 4 8 12 ; 6 12 18]
C(:,:,3) = [3 6 9 ; 6 12 18 ; 9 18 27]
Then the resulting matrix should be:
C(:,:,1) = [1 2 3 ; 2 4 6 ; 3 6 9]
C(:,:,2) = [0 0 0 ; 4 8 12 ; 6 12 18]
C(:,:,3) = [0 0 0 ; 0 0 0 ; 9 18 27]
Any idea how such a thing csn be done?
(My original 3 dim matrix is large)
Thanks!
The built-in triu can't deal with this 3D array but you could do it in a simple loop.
for k = 2:size(C, 3)
C(1:k-1,:,k) = 0;
end
You can generate a 2D mask, permute its dimensions, and multiply with singleton expansion using bsxfun:
result = bsxfun(#times, C, permute((1:size(C,1)).'>=(1:size(C,2)), [1 3 2]));
Or, from version R2016b onwards, you can remove bsxfun thanks to automatic singleton expansion:
result = C .* permute((1:size(C,1)).'>=(1:size(C,2)), [1 3 2]);
Assuming you have a "square" 3-D matrix (i.e. NxNxN as you had it in your original post 3x3x3) you could also use reshaping and repmating:
Edit: since repmat is too slow, I swapped it for a bsxfun implementation.
[a,b,c] = size(C)
D = reshape(tril(ones(a)),[a,1,a]);
F = ones(1,size(E,1));
D = bsxfun(#times,D,F);
C(~D)=0;
I also did a quick comparison of the three proposed solutions. #LuisMendo's solution wouldn't work for me, there is a matrix dimension error in the >= comparison (sizes [1 N] and [N 1] are compared).
Between my and #Suever's solution, his is significantly faster:
Comparing the three methods with variable size Cs:
Suever's version (for loop):
Took 0.3529s to compute.
Took 0.0002s to compute size 3x3x3.
Took 0.0008s to compute size 10x10x10.
Took 0.0008s to compute size 50x50x50.
Took 0.0455s to compute size 250x250x250.
Took 0.3055s to compute size 500x500x500.
My version (reshape/repmat):
Took 0.9086s to compute.
Took 0.0522s to compute size 3x3x3.
Took 0.0042s to compute size 10x10x10.
Took 0.0017s to compute size 50x50x50.
Took 0.1060s to compute size 250x250x250.
Took 0.7445s to compute size 500x500x500.
Related
This question already has an answer here:
Linear indexing, logical indexing, and all that
(1 answer)
Closed 7 years ago.
The matlab help page for matrix indexing says:
Logical Indexing with a Smaller Array
In most cases, the logical indexing array should have the same number
of elements as the array being indexed into, but this is not a
requirement. The indexing array may have smaller (but not larger)
dimensions:
A = [1 2 3;4 5 6;7 8 9]
A =
1 2 3
4 5 6
7 8 9
B = logical([0 1 0; 1 0 1])
B =
0 1 0
1 0 1
isequal(numel(A), numel(B))
ans =
0
A(B)
ans =
4
7
8
What kind of crazy rule is matlab using here?
To understand this behavior, it's necessary to understand how matrices are stored in memory. Matlab stores matrices using column major layout. This means the 2d matrix:
A = 1 2 3
4 5 6
7 8 9
is stored in memory as one dimensional array going down the columns of A:
A = { array = [1 4 7 2 5 8 3 6 9]
n_rows = 3
n_cols = 3 }
The matrix B:
B = 0 1 0
1 0 1
is stored in memory as:
B = { array = [0 1 1 0 0 1]
n_rows = 2
n_cols = 3 }
Let's put the underlying representations next to eachother:
A.array = [1 4 7 2 5 8 3 6 9]
B.array = [0 1 1 0 0 1]
Using logical indexing, A(B) gives you [4, 7, 8] If you think a bit deeper, what's causing the unintuitive result is the combination of: (1) Matlab uses column major layout and (2) the number of columns in A and B are different.
Note: I'm using pseudo code here. A.array isn't valid code etc...
Bonus:
You can see what happens when the reshape command is called. The underlying data array doesn't change, just the n_rows and n_cols associated with the data array.
I want to create a matrix from a vector by concatenating the vector onto itself n times. So if my vector is mx1, then my matrix will be mxn and each column of the matrix will be equal to the vector.
Which of the following is the best/correct way, or maybe there is a better way I do not know?
matrix = repmat(vector, 1, n);
matrix = vector * ones(1, n);
Thanks
Here is some benchmarking using timeit with different vector sizes and repetition factors. The results to be shown are for Matlab R2015b on Windows.
First define a function for each of the considered approaches:
%// repmat approach
function matrix = f_repmat(vector, n)
matrix = repmat(vector, 1, n);
%// multiply approach
function matrix = f_multiply(vector, n)
matrix = vector * ones(1, n);
%// indexing approach
function matrix = f_indexing(vector,n)
matrix = vector(:,ones(1,n));
Then generate vectors of different size, and use different repetition factors:
M = round(logspace(2,4,15)); %// vector sizes
N = round(logspace(2,3,15)); %// repetition factors
time_repmat = NaN(numel(M), numel(N)); %// preallocate results
time_multiply = NaN(numel(M), numel(N));
time_indexing = NaN(numel(M), numel(N));
for ind_m = 1:numel(M);
for ind_n = 1:numel(N);
vector = (1:M(ind_m)).';
n = N(ind_n);
time_repmat(ind_m, ind_n) = timeit(#() f_repmat(vector, n)); %// measure time
time_multiply(ind_m, ind_n) = timeit(#() f_multiply(vector, n));
time_indexing(ind_m, ind_n) = timeit(#() f_indexing(vector, n));
end
end
The results are plotted in the following two figures, using repmat as reference:
figure
imagesc(time_multiply./time_repmat)
set(gca, 'xtick',1:2:numel(N), 'xticklabels',N(1:2:end))
set(gca, 'ytick',1:2:numel(M), 'yticklabels',M(1:2:end))
title('Time of multiply / time of repmat')
axis image
colorbar
figure
imagesc(time_indexing./time_repmat)
set(gca, 'xtick',1:2:numel(N), 'xticklabels',N(1:2:end))
set(gca, 'ytick',1:2:numel(M), 'yticklabels',M(1:2:end))
title('Time of indexing / time of repmat')
axis image
colorbar
Perhaps a better comparison is to indicate, for each tested vector size and repetition factor, which of the three approaches is the fastest:
figure
times = cat(3, time_repmat, time_multiply, time_indexing);
[~, fastest] = min(times, [], 3);
imagesc(fastest)
set(gca, 'xtick',1:2:numel(N), 'xticklabels',N(1:2:end))
set(gca, 'ytick',1:2:numel(M), 'yticklabels',M(1:2:end))
title('1: repmat is fastest; 2: multiply is; 3: indexing is')
axis image
colorbar
Some conclusions can be drawn from the figures:
The multiply-based approach is always slower than repmat
The indexing-based approach is similar to repmat. It tends to be faster for large values of vector size or repetition factor, and slower for small values.
Either method is correct if they provide you with the desired output.
However, depending on how you declare your vector you may get incorrect results with repmat that will be spotted if you use ones. For instance take this example
>> v = 1:10;
>> m = v * ones(1, n)
Error using *
Inner matrix dimensions must agree.
>> m = repmat(v, 1, n)
m =
Columns 1 through 22
1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2
Columns 23 through 44
3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4
Columns 45 through 50
5 6 7 8 9 10
ones provides an error to let you know you aren't doing the right thing but repmat doesn't. Whilst this example works correctly with both repmat and ones
>> v = (1:10).';
>> m = v * ones(1, n)
m =
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
6 6 6 6 6
7 7 7 7 7
8 8 8 8 8
9 9 9 9 9
10 10 10 10 10
>> m = repmat(v, 1, n)
m =
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
6 6 6 6 6
7 7 7 7 7
8 8 8 8 8
9 9 9 9 9
10 10 10 10 10
You can also do this -
vector(:,ones(1,n))
But, if I have to choose, repmat would be the go-to approach for me, as it is made exactly for this purpose. Also, depending on how you are going to use this replicated array, you can just avoid creating it altogether with bsxfun that does on-the-fly replication on its input arrays and some operation to be applied on the inputs. Here's a comparison on that - Comparing BSXFUN and REPMAT that shows bsxfun to be better than repmat in most cases.
Benchmarking
For the sake of performance, let's test out these. Here's a benchmarking code to do so -
%// Inputs
vector = rand(1000,1);
n = 1000;
%// Warm up tic/toc.
for iter = 1:50000
tic(); elapsed = toc();
end
disp(' ------- With REPMAT -------')
tic,
for iter = 1:200
A = repmat(vector, 1, n);
end
toc, clear A
disp(' ------- With vector(:,ones(1,n)) -------')
tic,
for iter = 1:200
A = vector(:,ones(1,n));
end
toc, clear A
disp(' ------- With vector * ones(1, n) -------')
tic,
for iter = 1:200
A = vector * ones(1, n);
end
toc
Runtime results -
------- With REPMAT -------
Elapsed time is 1.241546 seconds.
------- With vector(:,ones(1,n)) -------
Elapsed time is 1.212566 seconds.
------- With vector * ones(1, n) -------
Elapsed time is 3.023552 seconds.
Both are correct, but repmat is a more general solution for multi-dimensional matrix copying and is thus bound to be slower than an other solution. The specific 'homemade' solution of multiplying two vectors is possibly faster. It is probably even faster to do selecting instead of multiplying, i.e. vector(:,ones(n,1)) instead of vector*ones(1,n).
EDIT:
Type open repmat in your Command Window. As you can see, it is not a built-in function. You can see that it also makes use of ones (selecting) to copy matrices. However, since it is a more general solution (for scalars and multi-dimensional matrices and copies in multiple directions), you will find unnecessary if statements and other unnecessary code, effectively slowing things down.
EDIT:
Multiplying vectors with ones becomes slower for very large vectors. The unequivocal winner is using ones with selection, i.e. vector(:,ones(n,1)) (which should always be faster than repmat since it uses the same strategy).
Suppose A is a 3-D matrix as below (2 rows-2 columns-2 pages).
A(:,:,1)=[1,2;3,4];
A(:,:,2)=[5,6;7,8];
I want to have a vector, say "a", whose inputs are the average of diagonal elements of matrices on each page. So in this simple case, a=[(1+4)/2;(5+8)/2].
But I have difficulties in matlab to do so. I tried the codes below but failed.
mean(A(1,1,:),A(2,2,:))
You can use "partially linear indexing" in the two dimensions that define the diagonal, as follows:
Since partially linear indexing can only be applied on trailing dimensions, you first need to apply permute to rearrange dimensions, so that the first and second dimensions become second and third.
Now you leave the first dimension untouched, linearly-index the diagonals in the second and third dimensions (which effectly reduces those two dimensions to one), and apply mean along the (combined) second dimension.
Code:
B = permute(A, [3 1 2]); %// step 1: permute
result = mean(B(:,1:size(A,1)+1:size(A,1)*size(A,2)), 2); %// step 2: index and mean
In your example,
A(:,:,1)=[1,2;3,4];
A(:,:,2)=[5,6;7,8];
this gives
result =
2.5000
6.5000
You can use bsxfun for a generic solution -
[m,n,r] = size(A)
mean(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
Sample run -
>> A
A(:,:,1) =
8 4 1
7 6 3
1 5 8
A(:,:,2) =
1 7 6
8 5 2
1 2 7
A(:,:,3) =
6 2 8
1 1 6
1 4 5
A(:,:,4) =
8 1 6
1 5 1
9 2 7
>> [m,n,r] = size(A);
>> sum(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
ans =
22 13 12 20
>> mean(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
ans =
7.3333 4.3333 4 6.6667
This question pops up quite often in one form or another (see for example here or here). So I thought I'd present it in a general form, and provide an answer which might serve for future reference.
Given an arbitrary number n of vectors of possibly different sizes, generate an n-column matrix whose rows describe all combinations of elements taken from those vectors (Cartesian product) .
For example,
vectors = { [1 2], [3 6 9], [10 20] }
should give
combs = [ 1 3 10
1 3 20
1 6 10
1 6 20
1 9 10
1 9 20
2 3 10
2 3 20
2 6 10
2 6 20
2 9 10
2 9 20 ]
The ndgrid function almost gives the answer, but has one caveat: n output variables must be explicitly defined to call it. Since n is arbitrary, the best way is to use a comma-separated list (generated from a cell array with ncells) to serve as output. The resulting n matrices are then concatenated into the desired n-column matrix:
vectors = { [1 2], [3 6 9], [10 20] }; %// input data: cell array of vectors
n = numel(vectors); %// number of vectors
combs = cell(1,n); %// pre-define to generate comma-separated list
[combs{end:-1:1}] = ndgrid(vectors{end:-1:1}); %// the reverse order in these two
%// comma-separated lists is needed to produce the rows of the result matrix in
%// lexicographical order
combs = cat(n+1, combs{:}); %// concat the n n-dim arrays along dimension n+1
combs = reshape(combs,[],n); %// reshape to obtain desired matrix
A little bit simpler ... if you have the Neural Network toolbox you can simply use combvec:
vectors = {[1 2], [3 6 9], [10 20]};
combs = combvec(vectors{:}).' % Use cells as arguments
which returns a matrix in a slightly different order:
combs =
1 3 10
2 3 10
1 6 10
2 6 10
1 9 10
2 9 10
1 3 20
2 3 20
1 6 20
2 6 20
1 9 20
2 9 20
If you want the matrix that is in the question, you can use sortrows:
combs = sortrows(combvec(vectors{:}).')
% Or equivalently as per #LuisMendo in the comments:
% combs = fliplr(combvec(vectors{end:-1:1}).')
which gives
combs =
1 3 10
1 3 20
1 6 10
1 6 20
1 9 10
1 9 20
2 3 10
2 3 20
2 6 10
2 6 20
2 9 10
2 9 20
If you look at the internals of combvec (type edit combvec in the command window), you'll see that it uses different code than #LuisMendo's answer. I can't say which is more efficient overall.
If you happen to have a matrix whose rows are akin to the earlier cell array you can use:
vectors = [1 2;3 6;10 20];
vectors = num2cell(vectors,2);
combs = sortrows(combvec(vectors{:}).')
I've done some benchmarking on the two proposed solutions. The benchmarking code is based on the timeit function, and is included at the end of this post.
I consider two cases: three vectors of size n, and three vectors of sizes n/10, n and n*10 respectively (both cases give the same number of combinations). n is varied up to a maximum of 240 (I choose this value to avoid the use of virtual memory in my laptop computer).
The results are given in the following figure. The ndgrid-based solution is seen to consistently take less time than combvec. It's also interesting to note that the time taken by combvec varies a little less regularly in the different-size case.
Benchmarking code
Function for ndgrid-based solution:
function combs = f1(vectors)
n = numel(vectors); %// number of vectors
combs = cell(1,n); %// pre-define to generate comma-separated list
[combs{end:-1:1}] = ndgrid(vectors{end:-1:1}); %// the reverse order in these two
%// comma-separated lists is needed to produce the rows of the result matrix in
%// lexicographical order
combs = cat(n+1, combs{:}); %// concat the n n-dim arrays along dimension n+1
combs = reshape(combs,[],n);
Function for combvec solution:
function combs = f2(vectors)
combs = combvec(vectors{:}).';
Script to measure time by calling timeit on these functions:
nn = 20:20:240;
t1 = [];
t2 = [];
for n = nn;
%//vectors = {1:n, 1:n, 1:n};
vectors = {1:n/10, 1:n, 1:n*10};
t = timeit(#() f1(vectors));
t1 = [t1; t];
t = timeit(#() f2(vectors));
t2 = [t2; t];
end
Here's a do-it-yourself method that made me giggle with delight, using nchoosek, although it's not better than #Luis Mendo's accepted solution.
For the example given, after 1,000 runs this solution took my machine on average 0.00065935 s, versus the accepted solution 0.00012877 s. For larger vectors, following #Luis Mendo's benchmarking post, this solution is consistently slower than the accepted answer. Nevertheless, I decided to post it in hopes that maybe you'll find something useful about it:
Code:
tic;
v = {[1 2], [3 6 9], [10 20]};
L = [0 cumsum(cellfun(#length,v))];
V = cell2mat(v);
J = nchoosek(1:L(end),length(v));
J(any(J>repmat(L(2:end),[size(J,1) 1]),2) | ...
any(J<=repmat(L(1:end-1),[size(J,1) 1]),2),:) = [];
V(J)
toc
gives
ans =
1 3 10
1 3 20
1 6 10
1 6 20
1 9 10
1 9 20
2 3 10
2 3 20
2 6 10
2 6 20
2 9 10
2 9 20
Elapsed time is 0.018434 seconds.
Explanation:
L gets the lengths of each vector using cellfun. Although cellfun is basically a loop, it's efficient here considering your number of vectors will have to be relatively low for this problem to even be practical.
V concatenates all the vectors for easy access later (this assumes you entered all your vectors as rows. v' would work for column vectors.)
nchoosek gets all the ways to pick n=length(v) elements from the total number of elements L(end). There will be more combinations here than what we need.
J =
1 2 3
1 2 4
1 2 5
1 2 6
1 2 7
1 3 4
1 3 5
1 3 6
1 3 7
1 4 5
1 4 6
1 4 7
1 5 6
1 5 7
1 6 7
2 3 4
2 3 5
2 3 6
2 3 7
2 4 5
2 4 6
2 4 7
2 5 6
2 5 7
2 6 7
3 4 5
3 4 6
3 4 7
3 5 6
3 5 7
3 6 7
4 5 6
4 5 7
4 6 7
5 6 7
Since there are only two elements in v(1), we need to throw out any rows where J(:,1)>2. Similarly, where J(:,2)<3, J(:,2)>5, etc... Using L and repmat we can determine whether each element of J is in its appropriate range, and then use any to discard rows that have any bad element.
Finally, these aren't the actual values from v, just indices. V(J) will return the desired matrix.
I have around 100 1D arrays I'd like to merge to a matrix.
The arrays have 140 to 180 columns.
Is it possible to merge these 1 x (140-180) arrays to a matrix with a dimension of 100 (amount of arrays) x 180 ?
All the arrays contain numbers. I want to expand the 1x140 array to a 1x180 array by means of interpolation.
In a simplified form, it should be something like this:
A = [1 5 7 8 3]
B = [1 3 5]
result=
[1 5 7 8 3
1 2 3 4 5]
The array B (1x3) is expanded to an 1x5 matrix. And the values in between are interpolated.
Basically, I thought of using "vertcat" after all arrays are expanded by a same amount of columns.
Thanks in advance,
Koen
How about this?
array = {[1 5 7 8 3],[1 3 5]}; % example data
N = 5; % desired length (180 in your case)
aux = cellfun(#(v) interp1(linspace(0,1,length(v)),v,linspace(0,1,N)), array, 'uni', false);
result = cat(1,aux{:});
It uses linear interpolation. For your example, this gives
>> result
result =
1 5 7 8 3
1 2 3 4 5
Note that linear interpolation modifies all values of the vector except first and last, in general. For example, with N=5 the vector [1 3 4 5] would get interpolated to [1 2.5 3.5 4.25 5]. You could use other forms of interpolation by passing an extra argument to interp1, see help interp1.