Iterating through a dataset and creating a vector with according values

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So I have a 10 000 X 1 vector (let's call it vector A) of different values ranging from -1 to 1. I want to create another 10 000 X 1 vector (let's call it vector B) that has values corresponding to the ones in the first vector. I want to create a loop that does this:
Everytime a value in vector A is bigger than 0.5, add the value 3 to vector B with the same corresponding position.
Everytime a value in vector A is smaller than -0.5, add the value 1 to vector B with the same corresponding position.
Everytime a value in vector A is between -0.5 and 0.5, add the value 2 to vector B with the same corresponding.
In that case, the resulting matrix should look like this:
[A] [B]
-0.9 1
-0.3 2
0.6 3
-0.2 2
-0.7 1
I know I need to use a loop but I struggle with this. Thank you so much for your help.
  2 Comments
Dyuman Joshi
Dyuman Joshi on 20 Sep 2022
What about border values i.e. 0.5 and -0.5, where value will you assign to corresponding B elements?

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Answers (1)

dpb
dpb on 20 Sep 2022
Rarely need loops for such things as this with MATLAB; either logical addressing or table lookup almost always comes to the rescue -- the latter (using the builtin interpolation routine) solution here would be
A=[-0.9;-0.3;0.6;-0.2;-0.7];
B=interp1([-1,-0.5,0.5,1],[1,2,3,3],A,'previous')
B = 5×1
1 2 3 2 1
Alternatively, another lookup solution, with less overhead...
B=discretize(A,[-1,-0.5,0.5,1])
B = 5×1
1 2 3 2 1
  4 Comments
Dyuman Joshi
Dyuman Joshi on 20 Sep 2022
After you get the values, you can get the values as such
str={'GT';'IN';'ST'};
A=[-0.9;-0.3;0.6;-0.2;-0.5;0.5];
B=interp1([-1,-0.5,0.5,1],[1,2,3,3],A,'previous');
%cell array
C=str(B)
C = 6×1 cell array
{'GT'} {'IN'} {'ST'} {'IN'} {'IN'} {'ST'}
%char array
D=cell2mat(C)
D = 6×2 char array
'GT' 'IN' 'ST' 'IN' 'IN' 'ST'
dpb
dpb on 20 Sep 2022
Edited: dpb on 20 Sep 2022
If the border edge cases are that important, use bkpt-eps(bkpt) to add the necessary granularity in either solution.
There are multiple other answers I've posted (at least one pretty recently) that illustrate...

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