creating a matrix where the element of the second column is smaller than the element of the first column

I would like to create a matrix where the element of the second column is smaller than the element of the first column. For example,

A = [1 1; 2 1; 2 2; 3 1; 3 2; 3 3]

I would like to know a simpler way rather than going through loops.

Answer 1

[Guillaume]

In terms of clarity and speed, a loop is probably the best, I'd implement it as:

n = 4; %largest number
result = zeros(n*(n+1)/2, 2);
row = 1;
for i = 1:n
  result(row:row+i-1, :) = [repmat(i, i, 1), (1:i)'];
  row = row+i;
end

You can implement the above with arrayfun as James has done, but it's likely to be slower.

Another fancy non-loopy way of getting the result:

n = 4;
result = [repelem((1:n)', 1:n), flipud(nonzeros(hankel(n:-1:1)))]

The use of hankel to obtain the 2nd column is fairly obscure so I wouldn't recommend using that.

[Guillaume]

For comparison, a quick performance test on my machine (R2017a) of the 3 solutions proposed:

For some reason they all slow down suddenly at around n=350. For smaller n, the hankel version is actually faster, but for larger n the performance degrades quadratically (probably). The arrayfun version is always significantly slower than the loop version.

[Guillaume]

New comparison that includes all the valid proposed solutions, on a different machine and different version (R2017b):

For low n Roger's answer is the fastest. nchoose2 (without the sort) and my hankel version are on par. Jame's arrayfun solution is consistently slower than the explicit loop and nchoose2 with sorting is a disaster. The explicit loop is never much slower than the fastest solution and for n>500 (on that machine) is the best solution.

I'd say go with a loop as it's by far the clearest as to what it does.

Answer 2

[James Tursa]

E.g.,

n = largest number (e.g., 3)
result = cell2mat(arrayfun(@(x)[ones(x,1)*x,(1:x)'],1:n,'uni',false)');

[alpedhuez]

Would you explain how this line works?

[James Tursa]

Start with the function:

@(x)[ones(x,1)*x,(1:x)']

For an integer input x, this creates a 2-column matrix. The first column is all the number x, and the second column is the numbers 1 through x. E.g.,

>> f = @(x)[ones(x,1)*x,(1:x)']
f = 
    @(x)[ones(x,1)*x,(1:x)']
>> f(1)
ans =
     1     1
>> f(2)
ans =
     2     1
     2     2
>> f(3)
ans =
     3     1
     3     2
     3     3
>> f(4)
ans =
     4     1
     4     2
     4     3
     4     4

The @(x) anonymous function is fed into the arrayfun( ) function as the first argument.

The second argument to arrayfun( ) is the array 1:n. So, for each number in the second argument, arrayfun( ) will execute the @(x) function with this number as the input. The output of each "iteration" of arrayfun( ) is a matrix and not a scalar. So to gather up all of these outputs into a single cell array result, we also give the last two arguments 'uni' and false.

The cell2mat call simply concatenates all of the cell array results that came from the arrayfun( ) call into a single matrix. The transpose operator ' is used to ensure the concatenation happens vertically (1:n is a row vector so the cells come out as a row vector as well, but we want them stacked vertically).

Answer 3

[Image Analyst]

Here are a couple of ways, one by sorting and one by replacing the second column:

A = [1 1; 2 1; 2 2; 3 1; 3 2; 3 3]
% Sort descending.
ASorted = sort(A, 2, 'descend')
% Replace second row
A2 = A;
A2(:, 2) = A(:, 1) - 1; % Column 2 is one less than col 1

Answer 4

[Roger Stafford]

Here's another way:

N = 10; % Choose any integer N (largest number)
n = (1:N*(N+1)/2)';
c1 = round(sqrt(2*n-3/4));
c2 = n-(c1-1).*c1/2;
A = [c1,c2];

[Image Analyst]

And another way.

A(:, 2) = -inf;

or if "smaller" means closer to zero than closer to minus infinity:

A(:, 2) = 0;

[Roger Stafford]

@Image Analyst: I was just expanding on the example alpedhuez gave.

[Image Analyst]

Oh, I see now. You and Guillaume were generalizing the matrix alpedhuez gave to larger versions, assuming some algorithm, while I didn't do that and just took the "A" as the given matrix, A = [1 1; 2 1; 2 2; 3 1; 3 2; 3 3], which he needed to create/modify a matrix from that given matrix where the second column was less than the first. I can see there is a possibly ambiguity in interpretation when he gives A and says he needs to create a matrix, that hopefully the poster can clear up.

Answer 5

[Jos (10584)]

Another approach:

n = 7 ;
M = [n+1-nchoose2(1:n) ; repmat(1:n,2,1).'] % voila!
M_sorted = sortrows(M)    % in sorted order, but why bother

NCHOOSE2 is my very fast simple utility function to get combinations of 2 elements from a vector (like nchoosek(V,2), but way more efficient). It can be downloaded from the File Exchange:

https://uk.mathworks.com/matlabcentral/fileexchange/20144-nchoose2

Answer 6

[Jos (10584)]

And here is a one-liner (using only MatLab functions).

n = 5 ;
M = flipud(nchoosek(n:-1:0,2) + [0 1]) % + expansion works in later ML releases

Replace NCHOOSEK by NCHOOSE2 (see my other answer) for an efficient improvement :D