how to reduce the use of repeated for loops as i explained below???

5 次查看（过去 30 天）
M.Prasanna kumar
M.Prasanna kumar 2019-8-22
评论： M.Prasanna kumar 2019-8-23
when SS =1 , KK loop should not take the value from 1 to 3
when SS = 2:23 , KK loop should not run for SS-1:SS+1
when SS =24 , KK loop should not take the value from 22 to 24
to calculate the variable "dist" i used three sets of for loops. is it possible to reduce
I am able to get the answer but i dont want to repeat the same code for three times. how to resolve the problem???
for ss = 1
for kk = setdiff(1:no_receivers+1,1:3)
for ii = 1:no_gridsx
dist(ii,1:250,kk,ss) = sqrt((apert_x(1,ss)-X(ii,1:250)).^2+(apert_y(1,ss)-Y(ii,1:250)).^2)+sqrt((X(ii,1:250)-apert_x(1,kk)).^2+(Y(ii,1:250)-apert_y(1,kk)).^2);
end
end
end
for ss = 2:23
for kk = setdiff(1:no_receivers+1,ss-1:ss+1)
for ii = 1:no_gridsx
dist(ii,1:250,kk,ss) = sqrt((apert_x(1,ss)-X(ii,1:250)).^2+(apert_y(1,ss)-Y(ii,1:250)).^2)+sqrt((X(ii,1:250)-apert_x(1,kk)).^2+(Y(ii,1:250)-apert_y(1,kk)).^2);
end
end
end
for ss = 24
for kk = setdiff(1:no_receivers+1,22:24)
for ii = 1:no_gridsx
dist(ii,1:250,kk,ss) = sqrt((apert_x(1,ss)-X(ii,1:250)).^2+(apert_y(1,ss)-Y(ii,1:250)).^2)+sqrt((X(ii,1:250)-apert_x(1,kk)).^2+(Y(ii,1:250)-apert_y(1,kk)).^2);
end
end
end

请先登录，再回答此问题。

采纳的回答

Rik
Rik 2019-8-22
Use a temporary variable to adapt your range. The code below show one such approach.
%intitialize for testing
no_gridsx=10;
no_receivers=26;
ss_max=24;
dist=NaN(no_gridsx,250,no_receivers+1,ss_max);
for ss = 1:ss_max
skips=ss-1:ss+1;
if ss==1%correct for edge cases
skips=skips+1;
elseif ss==ss_max
skips=skips-1;
end
for kk = setdiff(1:no_receivers+1,skips)
for ii = 1:no_gridsx
dist(ii,1:250,kk,ss) = ...
sqrt((apert_x(1,ss)-X(ii,1:250)).^2+(apert_y(1,ss)-Y(ii,1:250)).^2) ...
+sqrt((X(ii,1:250)-apert_x(1,kk)).^2+(Y(ii,1:250)-apert_y(1,kk)).^2);
end
end
end
  4 个评论
显示 2更早的评论
Stephen23
Stephen23 2019-8-23
编辑：Stephen23 2019-8-23
"why did you subtract ss==24 & add ss==1 , what is the logic behind this"
(ss-1:ss+1)-(ss==24)+(ss==1)
%^^^^^^^^^^ creates a vector, exactly like you did
% ^^^^^^^^^ when ss==24 subtracts 1 from the vector
% ^^^^^^^^ when ss==1 adds 1 to the vector
With setdiff this provides exactly the vectors described on the first three lines of your question.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Deep Learning Toolbox 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by