How to replace every value with the index of the value to the left of it

This way the left most column is 0. I initially tried using a while loop but with a really big matrix, it takes way too long. so if I had a matrix

x= [ 2 3  4  5  6 ; 
     8 9 10 11 12 ] 

the matrix i would want is

[ 0 1 3 5 7  9; 
  0 2 4 6 8 10 ]

I'm kind of at a crossroads here.

Answer 1

[DGM]

There's this:

x = [ 2 3 4 5 6 ; 8 9 10 11 12 ]

x = 2×5

2 3 4 5 6 8 9 10 11 12

idx = reshape(1:numel(x),size(x));
out = [zeros(size(x,1),1) idx(:,1:end-1)]

out = 2×5

0 1 3 5 7 0 2 4 6 8

[sunny blue]

I see! Why would we do 1: numel(x) in the reshape function?

[DGM]

Doing that gets us an index array that can be manipulated as needed

x = [ 2 3 4 5 6 ; 8 9 10 11 12 ]; % same thing
idx = reshape(1:numel(x),size(x))

idx = 2×5

1 3 5 7 9 2 4 6 8 10

You could also do the concatenation first and reduce the line count, but I don't think it's as easy to read:

x = [ 2 3 4 5 6 ; 8 9 10 11 12 ]; % same thing
out2 = reshape([zeros(1,size(x,1)) 1:numel(x)-size(x,1)],size(x))

out2 = 2×5

0 1 3 5 7 0 2 4 6 8

[Jan]

The output needs to have one column more than the input. Maybe this matchs the needs:

out = reshape([0, 0, 1:numel(x)], 2, []);

[sunny blue]

ok thank you!