how to store the conditional loop data

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Chaudhary P Patel
Chaudhary P Patel 2022-4-29
评论： Jan 2022-4-29
sir/ madam please correct me.
ut=rand(15,1);
uy=([0.84;0.384;0.784]);
eval (['U_y','=[u_y;u_y;u_y;u_y;u_y]']);
for n=1:15
if ut(n,1)<=U_y(n,1)
ut(n,1)=ut(n,1);
else
ut(n,1)=0;
end
utt(:,1)=ut(n,1); %% i want to store the ut as a utt in a column
end
  4 个评论
显示 2更早的评论
Star Strider
Star Strider 2022-4-29
My code tests each element of ‘ut’ against ‘U_y’ and sets it to zero if the condition is satisfied.
What result do you want?
Jan
Jan 2022-4-29
@Chaudhary P Patel: The line ut(n,1)=ut(n,1) replaced the n.th element of the vector ut by the n.th element of the vector ut. This is a waste of time only.
An option to avoid this (see others in the already provided answers):
for n = 1:15
if ut(n) <= U_y(n)
utt(n) = ut(n);
else
utt(n) = 0;
end
end
Or with a proper pre-allocation, to avoid the time-consuming growing of arrays:
utt = zero(15, 1); % The default value is 0 now
for n = 1:15
if ut(n) <= U_y(n)
utt(n) = ut(n);
end
end

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采纳的回答

Star Strider
Star Strider 2022-4-29
Ran in:
If I understand the code correctly, it can be simply stated as:
ut=rand(15,1);
u_y=([0.84;0.384;0.784]);
U_y = [u_y;u_y;u_y;u_y;u_y]; % Create Column Vector
utt = ut; % Copy 'ut' to 'utt'
utt(ut>U_y) = 0 % Set Appropriate Elements To Zero
utt = 15×1
0 0 0.6317 0.7311 0.1069 0.5889 0.5515 0 0.2595 0.3677
No loops!
.

更多回答（1 个）

Jan
Jan 2022-4-29
Your cleaned code:
ut = rand(15, 1);
uy = [0.84; 0.384; 0.784];
U_y = repmat(uy, 5, 1);
for n = 1:15
if ut(n) > U_y(n)
ut(n) = 0;
end
utt(n) = ut(n);
% ^ insert an index here
end
Remember, that X(k,1) can be written as X(k), which looks a little bit cleaner.
Instead of the loop, you can write:
utt = ut .* (ut <= U_y);
The expression in the parentheses is a vector of 1s and 0s and the elementwise multiplication set the wanted elements to 0.

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