# 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]?

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xiyou fu 2016 年 9 月 17 日
コメント済み: David Goodmanson 2016 年 9 月 19 日
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]?

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### 採用された回答

David Goodmanson 2016 年 9 月 18 日
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.
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David Goodmanson 2016 年 9 月 19 日
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.

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### その他の回答 (3 件)

Teja Muppirala 2016 年 9 月 17 日
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
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Walter Roberson 2016 年 9 月 18 日
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.

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Image Analyst 2016 年 9 月 17 日
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?
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Image Analyst 2016 年 9 月 18 日
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?

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Andrei Bobrov 2016 年 9 月 19 日
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);
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David Goodmanson 2016 年 9 月 19 日
cool!

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