How I can create a vector with specified blanks?
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Hi,I have a time series of daily data for 4 years(4*365 data) and I would like to have a some of daily data to be blanks(gaps) to do some analysis on the rest. My problem is that how I can create a specified length gaps,for example 3 days,on my vector data.Is there any script to create artificial gaps with specific gap length?
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Image Analyst
2016 年 7 月 6 日
編集済み: Image Analyst
2016 年 7 月 7 日
Rita: Seems like you're having trouble so I put together this code for you:
% First we need to create sample data.
data = randi(9, 1, 300) % Create sample data.
% Get location of nans to put into the original sample data.
originalNanLocations = sort(randperm(length(data)-3, 10)) % Make 10 originall nans
data(originalNanLocations) = nan;
% Now we have our original data and we can begin...
% Find out where the original nans are, so we don't place other nans next to them
originalNanLocations = isnan(data);
% Find out how many nans we need to be in there.
% It should be about 10% of the number of elements.
numRequiredNans = floor(0.1 * length(data))
% Assign nans. Note, there may be overlap so that some stretches may be more than 3 long.
numNansCurrently = 0;
% Create a failsafe so we don't get into an infinite loop
maxIterations = 1000000;
iterationNumber = 1;
% Now loop, placing nans, until we get the required number of nans in the data.
while numNansCurrently < numRequiredNans && iterationNumber <= maxIterations
% Get location of starts of 15 "new" nan runs
nanStartLocation = randi(length(data)-4) + 1;
% Find the ending indexes of each run
nanEndLocation = nanStartLocation + 2;
% Get data from the start to the end plus one on either side of it.
% so we don't place other nans next to them,
% since that would create a stretch of 4 or more.
thisData = data((nanStartLocation-1):(nanEndLocation+1));
% Now find if out where the original nans are,
if any(isnan(thisData))
% Nans were found - skip this stretch of data.
continue;
end
% If we get here no nans were found in that location
% and we are free to assign new nans there.
data(nanStartLocation:nanEndLocation) = nan;
% Count the number of nans we have
numNansCurrently = sum(isnan(data));
iterationNumber = iterationNumber + 1;
end
data % Print to command window.
% Now the original nans will still be there and not be adjacent
% to any of the "new" nan stretches of three nans in a row.
% And the overall number of nans will be 10% of the elements
% or not more than 2 more than that.
その他の回答 (3 件)
Azzi Abdelmalek
2016 年 7 月 5 日
You can set your data to nan
Data=rand(4*365,1) % your data
idx_gap=10:12
Data(idx_gap)=nan
3 件のコメント
Azzi Abdelmalek
2016 年 7 月 5 日
Data=randi(10,20,1) % your data
freq=3
period=6
ii=1:period:numel(Data)
idx=bsxfun(@plus,ii',1:3)
Data(idx)=nan
Image Analyst
2016 年 7 月 5 日
Try this:
data = randi(9, 1, 300) % Create sample data.
% Get location of starts of 15 nan runs
nanStartLocations = sort(randperm(length(data)-3, 15))
% Find the ending indexes of each run
nanEndLocations = nanStartLocations + 3
% Assign nans. Note, there may be overlap so that some stretches may be more than 3 long.
for k = 1 : length(nanStartLocations)
data(nanStartLocations(k):nanEndLocations(k)) = nan;
end
data % Print to command window.
2 件のコメント
Image Analyst
2016 年 7 月 5 日
If you want to make sure that no runs of 3 nans overlay with any other run of 3 nans, then you're going to have to check for that while you're assigning them. Change the for to a while and then check the elements to see if any are already nan. If any are, then use "continue" to skip to the bottom of the while loop. If none are, then do the assignment and increment your count. Keep going until you've added the required number of nan runs. It's not hard. Give that a try. You might try isnan() to get a list of what elements are nan.
Javad Beyranvand
2022 年 5 月 2 日
Data=rand(4*365,1) % داده های شما idx_gap=10:12 داده(idx_gap)=nan
if true
% code
end
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