Linearly spaced vectors

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Ragaar Ashnod
Ragaar Ashnod on 2 Mar 2011
For academic interest, I wish to further improve the speed and memory performance of the included code. The function's purpose, which I've provided below, is to generate a series of linearly spaced vectors using only an input vector for defining the upper and lower limits.
EXAMPLE:
x = [0
12.872
21.453
30.132
40.371
57.631
77.816
100];
increment = 2
y = f(x,increment)
An important aspect to note is that, with the exception of edge conditions, the generated outputs are even and monotonically increasing:
y = [0, ..., 12, 12.872,
12.872, 14, ..., 20, 21.453,
21.453, 22, ..., 30, 30.132,
...
57.631, 58, ..., 76, 77.816,
77.816, 78, ..., 100];
The method currently being used is as follows:
function y = multi_linspace(x,increment)
% Multi_linspace generates a vector of linearily spaced vectors.
input_length = length(x);
fractional = @(x)([fix(x)*Inf^(x-fix(x)),x]);
for i=1:input_length
if (i ~= input_length)
buffer{i}=ceil(x(i))+mod(ceil(x(i)),2):increment:x(i+1);
edge(1,:)=fractional(x(i));
edge(2,:)=fractional(x(i+1));
id = isinf(edge(:,1));
prefix = id(1); suffix = id(2);
if prefix
% x(i) == edge(1,2)
buffer{i} = [x(i),buffer{i}];
end
if suffix
% x(i+1) == edge(2,2)
buffer{i} = [buffer{i},x(i+1)];
end
end
end
y=[buffer{:}]';
end

Accepted Answer

Ragaar Ashnod
Ragaar Ashnod on 2 Mar 2011
The speed of a single call execution is on average better with the original code. However, it is [clearly] difficult to read and difficult for matlab's internal process(es) to optimize for subsequent calls. Inspired by, and barely improving upon, the cyclist's code the following code provides significant improvement for subsequent calls. And sort is likely faster only because it is a built-in function.
function y = multi_linspace2(x,increment)
y = (x(1):increment:x(end))';
y = sort([y;x(2:end-1);x(2:end-1)]);
end

More Answers (3)

the cyclist
the cyclist on 2 Mar 2011
I made no attempt to understand the complexity of your code, but I did check that the following reproduces your result exactly, for the test case you gave.
I don't know if it's faster, but it is shorter and simpler. But maybe you are testing special cases, which I am not.
function y = multi_linspace2(x,increment)
y = (x(1):increment:x(end))';
for ii=2:length(x)-1
y = [y(y<x(ii)); x(ii); x(ii); y(y>=x(ii))];
end
end
  1 Comment
Ragaar Ashnod
Ragaar Ashnod on 2 Mar 2011
Was actually showing pretty good results with mine, but like you said it is really complex.
Your post helped inspire a potential follow-up, though.

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Sean de Wolski
Sean de Wolski on 2 Mar 2011
x2 = [x(1:end-1), x(2:end)]
x2 = num2cell(x2,2)
y = cellfun(@(x)x(1):increment:x(2),x2,'uni',false)
%Scd
  2 Comments
Ragaar Ashnod
Ragaar Ashnod on 2 Mar 2011
This was the first path I attempted. The values generated progress the decimal places included and that is something I needed to avoid.
Sean de Wolski
Sean de Wolski on 2 Mar 2011
Just change the function handle:
y = cellfun(@(x)[x(1), ceil(x(1)):increment:floor(x(2)), x(2)],x2,'uni',false)

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Oleg Komarov
Oleg Komarov on 2 Mar 2011
Have you seen mcolon, could be a good benchmark.
  1 Comment
Ragaar Ashnod
Ragaar Ashnod on 2 Mar 2011
Sadly, I had never heard of it before. Thanks for the link!

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