I have a vector like x=[2 3 1 5],how could I generate a vector like [1 2 1 2 3 1 1 2 3 4 5]?

I have a vector like x=[2 3 1 5],how could I generate a vector like [1 2 1 2 3 1 1 2 3 4 5]?

Answer 1

[David Goodmanson]

Assuming the rule you want is what Teja thought, then if you have room for a couple of arrays that could get pretty large depending on your data, you could try

m = max(x);
n = length(x);
A = repmat((1:m)',1,n);  % columns of consecutive integers
M = repmat(x,m,1);       % rows of copies of x
A(A>M) = nan;
A = A(:);                % concatenate columns
A(isnan(A)) = [];        % keep good ones 
A = A';

For x of length 1e7 with random entries from 1 to 10, this takes about 5 sec on my PC.

[xiyou fu]

It seems to work well in my case, thanks!

[David Goodmanson]

You're welcome. I should point out that since the numerical calculation here is simple, for a very large problem of this type you could gain speed and save memory by declaring variables to be unsigned integers such as uint8 or uint16, whatever is compatible with the entries in vector x. Uints do not support NaN, but you can use zero as an indicator instead. In the example I used, with uint8 the speed goes up by a factor of 2.5 and the memory is reduced by a factor of 8.

Answer 2

[Teja Muppirala]

x = [2 3 1 5]
y = arrayfun(@(a)1:a,x,'uniform',0)
y = cat(2,y{:})

y = 1 2 1 2 3 1 1 2 3 4 5

[xiyou fu]

Good! but when the length of x is large, it seems to be a little slow, is there any more efficient ways ?

[Walter Roberson]

Yes, you could write a mex file to do the work.

You can calculate the size of the output as sum() of the inputs, and you can pre-allocate that, and you can have a loop to fill it in, but it is not at all clear that it would end up being more efficient than what Teja shows.

Answer 3

[Image Analyst]

What rule are you using to generate that output? Here are several ways:

% Define input data.
x=[2 3 1 5]
% Generate a vector [1 2 1 2 3 1 1 2 3 4 5]:
output1 = [x(3), x(1), x(3), x(1), x(2), x(3), 1:length(x), x(end)]
output2 = [x([3, 1, 3, 1, 2, 3]), 1:length(x), x(end)]
output3 = [1 2 1 2 3 1 1 2 3 4 5];
output4 = [x([3, 1, 3, 1, 2, 3]), 1:5]

I'm sure there is a near infinite number of other rules that can get you that output vector. So what is your rule?

[xiyou fu]

I am sorry I didn't make the question clearly. In fact ,I have a for loop in my code, in the loop I can get a vector x (it's length will change with every loop), let's first make x=[2 3 1 5] for example, I want to generate another vector, which has the form like [1 2 1 2 3 1 1 2 3 4 5]. How can I do it in a efficient way? Thank you for your help.

[Image Analyst]

Tell me the rule. I don't have the Crystal Ball Toolbox yet and since you didn't share your for loop code, I don't know what you tried to do. You still haven't told me the rule even though I explicitly asked.

I gave you 4 ways to build that vector. What's wrong with them?

Answer 4

[Andrei Bobrov]

x = x(:);
ons = ones(sum(x),1);
ii = cumsum(x)+1;
ons(ii(1:end-1)) = ons(ii(1:end-1)) - x(1:end-1);
out = cumsum(ons);

[xiyou fu]

The solution is good! Thank you!

[David Goodmanson]

cool!