Replace percentage of data in a for loop
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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
댓글 수: 2
Geoff Hayes
2017년 6월 6일
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?
답변 (1개)
Walter Roberson
2017년 6월 6일
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
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