Clear Filters
Clear Filters

looping till a condition is met.....

8 views (last 30 days)
Hi;
I have an array of 2400 values. I want to loop such that i need to break into chunks till the sum becomes (say for suppose 2). i want to add 1st value plus 2nd value so on till the condition is met and then after the condition is met start adding from next number again till the condition meets.
I need chunks of values whose sum is 2 and i don't know the the exact chunks which vary by the data and the limit i set for the condition to break.
limit = 2;
s = 0;
for i = 1:length(Work)
while 1
if s >= limit
break
end
s = s + Work(i);
end
W = Work(i)
end

Accepted Answer

Image Analyst
Image Analyst on 28 Dec 2016
I believe this will do what you want. Give it a try. It scans along work getting the cumulative sum from that index onwards. Then it finds the first index where that cumulative sum exceeds 2. When that happens, it records the sum for that chunk (typically in the range 2-3) and the original index where that chunk ends.
% Create sample data.
Work = rand(2400,1);
% Define the limit of sums that the chunks need to surpass.
sumLimit = 2;
% Preallocate more memory than needed for W
sumValues = zeros(size(Work));
indexLocations = zeros(size(Work));
chunk = 1; % Index for what chunk we're working on.
index = 1;
while index < length(Work) && chunk < length(Work)
% Find the sums from this index onwards to the end of the Work vector.
cdf = cumsum(Work(index:end));
% Find the first time the sum exceeds the limit.
limitLocation = find(cdf > sumLimit, 1, 'first');
% Remember, limitLocation is relative to Index, not to element 1.
% If no remaining sum is more than sumLimit, just break out of the loop.
if isempty(limitLocation)
break;
end
% Save this sum value.
sumValues(chunk) = cdf(limitLocation);
% Save this index value. The index value relative to index 1, not relative to limitLocation.
indexLocations(chunk) = limitLocation + index - 1;
% Move pointer to the next element.
index = indexLocations(chunk) + 1;
% Increment what chunk number we're working on.
chunk = chunk + 1;
end
% Trim trailing zeros from sumValues and indexLocations.
sumValues(sumValues == 0) = []
indexLocations(indexLocations == 0) = []
  6 Comments
sri satya ravi
sri satya ravi on 8 Jan 2017
Edited: sri satya ravi on 8 Jan 2017
for k = 1 : length(Values) thisArray = Values{k}; % Variable number of elements metCondition(k) = Window_EXH_temp{chunk}<200(thisArray); % Returns true or false end % Extract only those chunks meeting condition: Values = Values(metCondition);
Hi; For my dataset i got Window_EXH_temp as 1*3518. each column has different size dataset again. so i wanted to check for condition when the value in Window_EXH_temp in a chunk is less than 200 exclude that whole chunk and move on to the next one. how should i write the condition.
If you want me to attach my whole code I can do that.
Thanks a lot for the help
Image Analyst
Image Analyst on 8 Jan 2017
Try this:
function metCondition = CheckChunkCondition(vector)
metCondition = true; % Initialize
if any(vector < 200)
metCondition = false;
end

Sign in to comment.

More Answers (1)

the cyclist
the cyclist on 27 Dec 2016
Edited: the cyclist on 27 Dec 2016
Here's one way. The core idea is that one tests whether the current value of W is over the limit, and if so then move the "pointer" index to the next element.
% Some pretend data
Work = rand(2400,1);
limit = 2;
% Preallocate
W = zeros(size(Work));
C = zeros(size(Work));
idx = 1;
for i = 1:length(Work)
C(idx) = C(idx) + 1;
if W(idx) >= limit
idx = idx + 1;
end
W(idx) = W(idx) + Work(i);
end
% Trim the excess from W and C
W(idx+1:end) = [];
C(idx+1:end) = [];
  4 Comments
sri satya ravi
sri satya ravi on 27 Dec 2016
Thanks..... I did try it and it worked. W is showing an array of 36 numbers. So my data is split into 36 segments of values close to 2. Can i know if there is a way to find the point at which the condition is met...i.e first 20 values till the condition is met or 25 values like that.
the cyclist
the cyclist on 27 Dec 2016
I've edited the code to track that in the variable C.

Sign in to comment.

Categories

Find more on Startup and Shutdown in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!