Find all instances of a condition in x , replace corresponding y values with ymax (of subset or a give value) via for loop..data attached

 %% Trying to find id of repeted values ​​in x, corresponding to h_w id, repalicng them with max (h_w) or given value, 
% Following is my attept but no luck,
% because I want to skip x values ​​already sorted by for loop in
% subsequent rounds
h_new linspace = (8.5,5.9, length (H_W));
% H_W (:) = 0;
for ii = 1: length (h_new)
[id, v] = find (x == x (ii)); 
H_W (id) = h_new (ii);
repeatedVal_h = H_W (id);
end
 
% The problem is loop, say once it locates all instances of x = 10, in x and has replaced in h_w, 
% it must ignore, next x = 10 in for loop.
% I hope I made clear

[Bob Thompson]

So let me rephrase this and see if I understand.

You want to identify all instances of a value in x and replace the corresponding values of H_W with the index loop of h_new.

This part of the code is working, but the problem is that for repeating values of x the code does not skip the later instances of the same value, causing the values of H_W to change multiple times.

Does that about right?

Answer 1

[BlueBird]

Hi Bob! I have attached the data (see attachment), I want only to replace value corresponding to non unique ones, e.g. look at a the short example in the table .

[Bob Thompson]

What I'm getting from this is that you want to replace all of the indices of x which correspond to a certain value (say 2) with a desired value (say 100). My code will do this, you just need to change what h_new(hcount) values are. Something like this.

uniques = unique(x);
hcount = 1;
for ii = 1:length(uniques);
    [id, ~] = find(x == uniques(ii));
    H_W(id) = 100*hcount;
    repeatedVal_h = 100*hcount;
    hcount = hcount + 1;
end

I don't expect that to do exactly what you're looking for, I'm just giving an example of how you can change your desired value within my sample I provided.

[BlueBird]

.Thank you it solve my probelm for now.

Yes, it does exactly what I want, very slow though.

Answer 2

[BlueBird]

Thanx Bob for responding to a SOS signal:

Yes, Exactly, Let me make it more clear:

What I wnat to do is :

locate all the instance of say x =10; find indices , then replace these indices in h_w , replace them with a desired value, which may be single value or some linearly decreasing etc.

but the loop must go on and find indices of x=11, replace, corresponding value with next next desired value:

etc.

The length of vectors must remain intact, like you cant delete repating values etc.

Waiting

x h_w desired

____ ______ ____

2 10 100

2 10.077 100

4 10.154 400

4 10.231 400

3 10.308 300

2 10.385 100

3 10.462 300

4 10.538 400

5 10.615 500

3 10.692 300

2 10.769 100

4 10.846 400

6 10.923 600

2 11 100

Answer 3

[Bob Thompson]

Instead of looping through all values of h_new, you might try looping through all unique values of x.

uniques = unique(x);
hcount = 1;
for ii = 1:length(uniques);
    [id, ~] = find(x == uniques(ii));
    H_W(id) = h_new(hcount);
    repeatedVal_h = h_new(hcount);
    hcount = hcount + 1;
end

Without seeing your data I cannot confirm, but I suspect this will drastically reduce the number of h_new values you progress through, but should eliminate looping over duplicates.