create y using x, without loop

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Jebidiah Light
Jebidiah Light 2019-5-8
编辑： Jebidiah Light 2019-5-9
size x = 2,3
size y = 1,157
y(3:52,1) = 1; y(53:102) = 2;... I need this but in one line of code that doesn't involve a loop.
X is dynamically growing because it is pulling data from another array. In X, columns 1&2 represent the index of rows needed for Y; however, column 3 is data inputted to Y for each of those rows respectively. Columns 1&2 will always be ordered consecutively column 3 will not.
x =
[ 3 52 1
53 102 2
103 157 98]
y =
[0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 98 98 98...]
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Jebidiah Light
Jebidiah Light 2019-5-8
input 1's from column 3 to 52 then 2's from column 53 to 102 using x as index and data for y, which is 1,102.
Adam Danz
Adam Danz 2019-5-8
编辑：Adam Danz 2019-5-8
Your update helped with problem #2 (now we understand y is a row vector). Problem #1 still needs addressed though I think I have a guess...

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采纳的回答

Adam Danz
Adam Danz 2019-5-8
编辑：Adam Danz 2019-5-9
No loop method (assuming consecutiveness)
This solution assumes the indices defined in columns 1&2 are consecutive and without gaps.
y = [zeros(1,x(1,1)-1), repelem(x(:,3),x(:,2)-x(:,1)+1,1)'];
No loop method (the silly method)
This method works even when indices are not consecutive and have gaps.
x = [ 3 52 1
53 102 2];
xc = mat2cell(x,ones(1,size(x,1)),size(x,2));
yc = cellfun(@(x)ones(1,x(2)-x(1)+1).*x(3), xc, 'UniformOutput',false);
xIdx = cellfun(@(x)x(1):x(2), xc,'UniformOutput', false);
y = zeros(1,max(max(x(:,[1,2]))));
y(cell2mat(xIdx')) = cell2mat(yc');
Loop method (the better & fastest method)
This method works even when indices are not consecutive and have gaps.
x = [ 3 52 1
53 102 2];
y = zeros(1,max(max(x(:,[1,2]))));
for i = 1:size(x,1)
y(x(i,1):x(i,2)) = x(i,3);
end
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Adam Danz
Adam Danz 2019-5-9
编辑：Adam Danz 2019-5-9
The median difference is less than 1 microsecond (0.00001 seconds).
Jebidiah Light
Jebidiah Light 2019-5-9
BUT THE MICROSECONDS ADAM!!! Lol. I thought I reaccepted your answer earlier. Unless someone comes up with a better solution, I'm going to consider this matter closed. I really do appreciate all the time you have exhausted for this inquisition.

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更多回答（1 个）

Steven Lord
Steven Lord 2019-5-8
This is complicated code, but it does satisfy the "one line, no loop" requirement. You will need to substitute 102 for the desired length of the vector and [3 53] with the first column of x.
y = cumsum(subsasgn(zeros(1, 102), substruct('()', {[3 53]}), 1))
You'd probably want to break it apart into pieces to understand what this line is doing, then add a paragraph of comments before putting this line in your code so the next person reading the code can understand it.
The phrase "for loop" is not (necessarily) a four letter word in MATLAB. Use for if it's the right tool for the job, use something else if that other tool is the right one.
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Adam Danz
Adam Danz 2019-5-9
My 3rd solution is listed first in the answer (just FYI).

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