# Forming loop to perform same action

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SUBHAJIT KAR on 4 Apr 2020
Commented: dpb on 5 Apr 2020
I am having problem creating the loop for the task.
Thres = 0.17;
istrue1= ((x>=928 & x<=1139) & y>=Thres)|((x>=2527 & x<=2856) & y>=Thres)|((x>=4113 & x<=4376) & y>=Thres)|((x>=5464 & x<=5643) & y>=Thres)|((x>=7254 & x<=7604) & y>=Thres);
TP = sum(istrue1);
Using the command I can find out the value of TP for a particular value of thres. But I want to vary the value of thres from 0:0.1:1 and want to get 10 values of TP simultaneously rather than changing Thres each time. I know it is easy but not for me. Please help.

dpb on 4 Apr 2020
Thres = 0.17;
istrue1=((x>= 928 & x<=1139) & y>=Thres) | ...
((x>=2527 & x<=2856) & y>=Thres) | ...
((x>=4113 & x<=4376) & y>=Thres) | ...
((x>=5464 & x<=5643) & y>=Thres) | ...
((x>=7254 & x<=7604) & y>=Thres);
is same as
istrue1=(x>= 928 & x<=1139 | ...
x>=2527 & x<=2856 | ...
x>=4113 & x<=4376 | ...
x>=5464 & x<=5643 | ...
x>=7254 & x<=7604) & y>=Thres);
Is place for my "syntactic sugar" helper utility iswithin...
istrue1=(iswithin(x, 928,1139) | ...
iswithin(x,2527,2856) | ...
iswithin(x,4113,4376) | ...
iswithin(x,5464,5643) | ...
iswithin(x,7254,7604)) & y>=Thres);
to reduce the clutter of all the tests conditions by moving into the function.
function flg=iswithin(x,lo,hi)
% returns T for values within range of input
% SYNTAX:
% [log] = iswithin(x,lo,hi)
% returns T for x between lo and hi values, inclusive
flg= (x>=lo) & (x<=hi);
end

SUBHAJIT KAR on 5 Apr 2020
In this case How to input various ranges for x? The function will take only one lo and hi.
dpb on 5 Apr 2020
As shown, it doesn't replace the individual tests, just hides the clutter.
I don't see any brilliant vectorizing available here; could build a function that looped over an array of lo, hi intervals or write the loop construct at this level would probably be my choice if there are variable numbers of regions and do this much. It's a situation haven't run into in my own code so hadn't attacked it, specifically.

Image Analyst on 4 Apr 2020
Edited: Image Analyst on 4 Apr 2020
Rather than be so complicated, break it into parts, like
term1 = (x>=928 & x<=1139)
term2 = y>=Thres
term3 = (x>=2527 & x<=2856)
and so on. Then look at the size and values of each term. This is just basic debugging... Then
put Thresh into a loop
Thres = [0 : 0.1 : 1];
for k = 1 : length(Thres)
thisThres = Thres(k);
istrue1 = ...... function thisThresh instead of Thres
end

SUBHAJIT KAR on 4 Apr 2020
Thanks Everyone. Your help make MATLAB so much interesting. Please help me on this.Since x is a vector of length 8000 and I need 10 threshold values Thres= 0:0.1:1 Can I preallocate a vector z = zeros(10,8000) to store the result of istrue1 and consequently find the values of each TP?
Image Analyst on 4 Apr 2020
You can preallocate using the lengths of the x and Thres vectors like this:
z = zeros(length(Thres), length(x));
By the way, the length of Thres is not 10:
>> length( 0:0.1:1 )
ans =
11
SUBHAJIT KAR on 5 Apr 2020
Many Thanks. It is working Perfectly. I have modified the code as :
Thres = [0 : 0.1 : 1];
z = zeros(1,length(Thres));
for k = 1 : length(Thres)
thisThres = Thres(k);
istrue1 = ((x>=928 & x<=1139) & y>=thisThres)|((x>=2527 & x<=2856) & y>=thisThres)|((x>=4113 & x<=4376) & y>=thisThres)|((x>=5464 & x<=5643) & y>=thisThres)|((x>=7254 & x<=7604) & y>=thisThres);
TP = sum(istrue1);
z(k) = TP;
end