Use linspace without scalar input?

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Andrew Poissant
Andrew Poissant el 29 de Ag. de 2017
Comentada: Ali Talha Atici el 19 de Abr. de 2022
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);
  3 comentarios
KSSV
KSSV el 29 de Ag. de 2017
n cannot be a decimal.....it should be an integer.
Andrew Poissant
Andrew Poissant el 29 de Ag. de 2017
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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Respuesta aceptada

Guillaume
Guillaume el 29 de Ag. de 2017
Editada: Guillaume el 29 de Ag. de 2017
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
  2 comentarios
Andrew Poissant
Andrew Poissant el 29 de Ag. de 2017
Thank you! Your answer was very helpful
Ali Talha Atici
Ali Talha Atici el 19 de Abr. de 2022
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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Más respuestas (2)

Jan
Jan el 29 de Ag. de 2017
Editada: Jan el 29 de Ag. de 2017
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);
  1 comentario
Andrew Poissant
Andrew Poissant el 29 de Ag. de 2017
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 el 29 de Ag. de 2017
Editada: Stephen23 el 29 de Ag. de 2017
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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