CODEWITHSUNDEEP
arraysmatlabindexing
sas4740
sas4740

Reputation: 4840

Selecting elements from an array in MATLAB

I know that in MATLAB, in the 1D case, you can select elements with indexing such as a([1 5 3]), to return the 1st, 5th, and 3rd elements of a. I have a 2D array, and would like to select out individual elements according to a set of tuples I have. So I may want to get a(1,3), a(1,4), a(2,5) and so on. Currently the best I have is diag(a(tuples(:,1), tuples(:,2)), but this requires a prohibitive amount of memory for larger a and/or tuples. Do I have to convert these tuples into linear indices, or is there a cleaner way of accomplishing what I want without taking so much memory?

Upvotes: 4

Views: 12192

Answers (3)

Brian L
Brian L

Reputation: 3251

Converting to linear indices seems like a legitimate way to go:

indices = tuples(:, 1) + size(a,1)*(tuples(:,2)-1);
selection = a(indices);

Note that this is also implement in the Matlab built-in solution sub2ind, as in nate'2 answer: 

a(sub2ind(size(a), tuples(:,1),tuples(:,2)))

however, 

a = rand(50);
tuples = [1,1; 1,4; 2,5];

start = tic;
for ii = 1:1e4
    indices = tuples(:,1) + size(a,1)*(tuples(:,2)-1); end
time1 = toc(start);


start = tic;
for ii = 1:1e4
    sub2ind(size(a),tuples(:,1),tuples(:,2)); end
time2 = toc(start);

round(time2/time1)

which gives

ans =   
    38

so although sub2ind is easier on the eyes, it's also ~40 times slower. If you have to do this operation often, choose the method above. Otherwise, use sub2ind to improve readability.

Upvotes: 6

Fantastic Mr Fox
Fantastic Mr Fox

Reputation: 33864

you can reference the 2D matlab position with a 1D number as in:

a = [3 4 5;
     6 7 8;
     9 10 11;];
a(1) = 3;
a(2) = 6;
a(6) = 10;

So if you can get the positions in a matrix like this:

a([(col1-1)*(rowMax)+row1, (col2-1)*(rowMax)+row2, (col3-1)*(rowMax)+row3])

note: rowmax is 3 in this case

will give you a list of the elements at col1/row1 col2/row2 and col3/row3.

so if 

row1 = col1 = 1
row2 = col2 = 2
row3 = col3 = 3

you will get:

[3, 7, 11]

back.

Upvotes: 0

bla
bla

Reputation: 26069

if x and y are vectors of the x y values of matrix a, then sub2und should solve your problem:

a(sub2ind(size(a),x,y))

For example

a=magic(3)

a =

 8     1     6
 3     5     7
 4     9     2

x = [3 1];
y = [1 2];


a(sub2ind(size(a),x,y))

ans =

 4     1

Upvotes: 3

Related Questions