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

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

[Arif Hoq]

its for the value of z. i guess, all the value of z are same

[Nicholas Kavouris]

no they are not, check the attached jpg, the values of z greatly vary

[Arif Hoq]

please attach your data

[Nicholas Kavouris]

z is thousands of cells long, attached is a plot of z

[Arif Hoq]

that's a figure, not the data. attach your variable Temp_F

[Nicholas Kavouris]

The variable in the loop is not Temp_F it is a convolution of the diff of Temp_F

[AndresVar]

@Nicholas Kavouris you should use z(i)

edit: or don't use a loop

temp_status=4*ones(size(z));
temp_status(z>=15) = 1;
temp_status(z>=-15 & z<15) = 2;
temp_status(z<-15) = 3;

[Nicholas Kavouris]

how does indexing the z comparison work?

[AndresVar]

@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)

Answer 1

[Image Analyst]

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;

[Nicholas Kavouris]

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]

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

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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.

Answer 2

[Simon Chan]

Just an addition:

x=diff(Temp_F);
z=conv(x,ones(300,1),"same");
temp_status = (z>15) + (z>=-15 & z<15)*2 + (z<-15)*3;

[Image Analyst]

Very nice! 😊

Answer 3

[Arif Hoq]

try this:

A=load('Temp_F.mat');
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