# Why does this loop return the same value for the variable each time?

4 ビュー (過去 30 日間)
Nicholas Kavouris 2022 年 3 月 18 日
コメント済み: Image Analyst 2022 年 3 月 18 日
Why does this loop return '4' every iteration for temp status? attached is a plot of Z, which should not return temp_status(i) as 4 each iteration as many values in z are above 15 and others below-15?
temp_status=[];
x=diff(Temp_F);
z=conv(x,ones(300,1),"same");
figure()
y=plot(z)
for i=1:length(z)
if z>=15
temp_status(i)=1;
elseif z>=-15 & z<15
temp_status(i)=2;
elseif z<-15
temp_status(i)=3;
else
temp_status(i)=4;
end
end
##### 9 件のコメント7 件の古いコメントを表示7 件の古いコメントを非表示
Nicholas Kavouris 2022 年 3 月 18 日
how does indexing the z comparison work?
AndresVar 2022 年 3 月 18 日

@Nicholas Kavouris it seems weird at first but it's a really cool feature in matlab. Once you use it a lot you will hate other languages that don't have this feature.
Here is a toy example,
temp_status = [4 4 4 4]; % for example all start as 4
z=[10 12 15 16]; % the comparable
idxs = z>=15 % the conditional indexes
idxs = 1×4 logical array
0 0 1 1
temp_status(idxs) = 1 % matlab recognizes logical indexing and only applies the assignment to the true "1" indices.
temp_status = 1×4
4 4 1 1
Only the 3rd and 4th values were changed from 4 to 1 because only indeces 3 and 4 were true (1)

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### 採用された回答

Image Analyst 2022 年 3 月 18 日
Try
temp_status=[];
x=diff(Temp_F);
z=conv(x,ones(300,1),"same");
figure()
y=plot(z)
indexes = z > 15;
temp_status(indexes) = 1;
indexes = (z >= -15) & (z <= 15);
temp_status(indexes) = 2;
indexes = z < -15;
temp_status(indexes) = 3;
##### 2 件のコメントなしを表示なしを非表示
Nicholas Kavouris 2022 年 3 月 18 日

this works, but I do not understand how indexing the z comparision works in terms of creating the desired cell value, could you explain? and how is this simpler than looping over all the data
Image Analyst 2022 年 3 月 18 日
It's simpler than looping because it's faster and less lines, and vectorized. Basically the indexes is a mask that is true where the condition is true and false where the condition is false. Then the next line uses that mask as a logical vector so the assignment takes place only where the mask is true. You should learn how to do it this way. This is the real power of MATLAB. All experienced MATLAB programmers would do it that way. No serious MATLABer would use a for loop for something as simple as this. If you learn "logical indexing" and "linear indexing" and how they can be used for vectorized operations, you'll be glad you did. Invest 2 hours
You'll be glad you did because it will make you a better programmer. And you won't be using a loop all the time like those poor C programmers.

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### その他の回答 (2 件)

Simon Chan 2022 年 3 月 18 日
x=diff(Temp_F);
z=conv(x,ones(300,1),"same");
temp_status = (z>15) + (z>=-15 & z<15)*2 + (z<-15)*3;
##### 1 件のコメント-1 件の古いコメントを表示-1 件の古いコメントを非表示
Image Analyst 2022 年 3 月 18 日
Very nice! 😊

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Arif Hoq 2022 年 3 月 18 日
try this:
temp_status=[];
Temp_F=A.Temp_F;
x=diff(Temp_F);
z=conv(x,ones(300,1));
figure()
y=plot(z);
for i=1:length(z)
if z(i)>=15
temp_status(i)=1;
elseif z(i)<=-15
temp_status(i)=3;
elseif z(i) > -15 & z(i) <15
temp_status(i)=2;
else
temp_status(i)=4
end
end
T=temp_status';
Z=z';
matrix=[Z(1:6) T(1:6)]
matrix = 6×2
0 2.0000 1.8000 2.0000 0 2.0000 1.8000 2.0000 0 2.0000 1.8000 2.0000

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