Replace percentage of data in a for loop

Hello,

I'm hoping to find out how to replace a defined percentage of data? I'm currently using a for loop with an if statement and would like to add another component quantifying the percentage replaced.

This is my current code:

for i=1:53280
LULC1=data(i,21);
LCC=data(i,20);
if LULC1==12||LULC1==15
    data(i,3)=16;
end
end

[Geoff Hayes]

Tracy - please clarify what you mean by how to replace a defined percentage of data. In your above code, is it a certain percentage of the elements of data that will be replaced? So once you reach that percentage then you can exit the for loop?

[Tracy Campbell]

I am hoping to replace 15%, 25%, and 50% of the data that fits the conditions listed in the if statements. Any help is appreciated!

Answer 1

[Walter Roberson]

percent_to_replace = 17.3;    %for example

row_matches = find( ismember(data(:,21), [12 15]) );
num_matches = length(row_matches);
num_to_replace = round(num_matches * percent_to_replace / 100);
which_to_replace = row_matches( randperm(num_matches, num_to_replace) );

data(which_to_replace, 3) = 16;

This uses the percentage as a fixed percentage to replace, that to within round-off, exactly that portion will be replaced. There is an alternative to that, which is to treat the percentage as a probability that any given one will be replaced.

percent_to_replace = 17.3;    %for example

which_to_replace = ismember(data(:,21), [12 15]) & (rand(size(data,1),1) < percent_to_replace/100);

data(which_to_replace, 3) = 16;

This would replace the given percentage on average

[Tracy Campbell]

Thank you so much for the answer. I have a follow-up question. When I run the script, it provides me with the information in which rows fit the description, but does not actually replace the data. Am I missing something? Thank you.

[Walter Roberson]

The line

data(which_to_replace, 3) = 16;

does the replacement.