Use linspace without scalar input?

조회 수: 18 (최근 30일)
Andrew Poissant
Andrew Poissant 2017년 8월 29일
댓글: Ali Talha Atici 2022년 4월 19일
I want to use linspace that goes from two non-scalar terms. Is there a way to use linspace with specified number of points, n, for decimal values for inputs? Example code is below for what I am looking for.
x = 0.01;
y = [1.01, 3.01];
n = 10;
dxy = linspace(x, y, n);
  댓글 수: 3
KSSV
KSSV 2017년 8월 29일
n cannot be a decimal.....it should be an integer.
Andrew Poissant
Andrew Poissant 2017년 8월 29일
My apologies. n will be an integer since it is the number of points. I want to use decimal values for x and y. So part of the output I am looking for would be dxy = [0.01 0.11 0.21 ... 1.01; 0.01 0.11 0.21 ... 3.01]. Hope this makes things clearer.

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채택된 답변

Guillaume
Guillaume 2017년 8월 29일
편집: Guillaume 2017년 8월 29일
x = 0.01;
y = [1.01, 3.01]; %would be better as a column vector as you wouldn't to tranpose it in the arrayfun
n = 10;
dxy = cell2mat(arrayfun(@(e) linspace(x, e, n), y', 'UniformOutput', false))
Or using a loop:
x = 0.01;
y = [1.01, 3.01]; %would be better as a column vector as you wouldn't to tranpose it in the arrayfun
n = 10;
dxy = zeros(numel(y), n);
for row = 1:numel(y)
dxy(row, :) = linspace(x, y(row), n);
end
or using Jan's idea but with R2016b or later syntax:
x = 0.01;
y = [1.01, 3.01]; %would be better as a column vector as you wouldn't the transpose in the calculation of dxy
n = 10;
dxy = (0:n-1) .* (y'-x)/n + x
  댓글 수: 2
Andrew Poissant
Andrew Poissant 2017년 8월 29일
Thank you! Your answer was very helpful
Ali Talha Atici
Ali Talha Atici 2022년 4월 19일
Thank you, Guillaume,
I have a little update to the last script to match the results.
x = 0.01;
y = [1.01, 3.01]; %would be better as a column vector as you wouldn't the transpose in the calculation of dxy
n = 10;
dxy = (0:n-1) .* (y'-x)/(n-1) + x

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추가 답변 (2개)

Jan
Jan 2017년 8월 29일
편집: Jan 2017년 8월 29일
x = 0.01;
y = [1.01, 3.01];
n = 10;
dxy = [linspace(x, y(1), n); ...
linspace(x, y(2), n)];
Or:
Step = repmat((y(:) - x(:)) / (n - 1), 1, n);
Step(:, 1) = x(:);
dxy = cumsum(Step);
  댓글 수: 1
Andrew Poissant
Andrew Poissant 2017년 8월 29일
How would I generalize the first option if I have many values in the y vector? I tried the following:
for i = 1:length(y)
dxy(i) = linspace(x, y(i), n);
end
but got the error "Subscripted assignment dimension mismatch."

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Stephen23
Stephen23 2017년 8월 29일
편집: Stephen23 2017년 8월 29일
An efficient general solution for any size y using linspace and bsxfun:
>> x = 0.01;
>> y = [1.01, 3.01];
>> n = 10;
>> bsxfun(@plus,x*linspace(1,0,n),bsxfun(@times,y(:),linspace(0,1,n)))
ans =
0.010000 0.121111 0.232222 0.343333 0.454444 0.565556 0.676667 0.787778 0.898889 1.010000
0.010000 0.343333 0.676667 1.010000 1.343333 1.676667 2.010000 2.343333 2.676667 3.010000
MATLAB versions with implicit expansion could probably do this (untested):
x.*linspace(1,0,n) + y(:).*linspace(0,1,n)

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