Dynamically adding element to an array

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Darren Lim
Darren Lim 2021년 1월 6일
댓글: Darren Lim 2021년 1월 6일
Dear Community,
Still new to matlab and exploring, It may be a simple problem but I am stuck for sometime on trying to come out with a clean logic .
Say i have an array (N X 2 )
signal price
1 29.21
-1 29.41
-1 29.43
-1 29.63
1 29.01
0 29.21
1 28.62
1 28.82
I wanted to add a row if the consecutive row in column 1 is -1 to 1 or -1 to 1 , the row to be added is a 0 and column 2 will be the price in the next row.
Basically a 1 is a long signal and -1 is a short signal and 0 is exit a position (in finance). The problem with the original data is when the scenario is from long to short ( 1 to -1) , i need to exit it before shorting , so i need to add a row with 0 and the price to exit.
The result I am expecting will be :
signal price
1 29.21
0 29 41 (added row)
-1 29.41
-1 29.43
-1 29.63
0 29.01 (added row)
1 29.01
0 29.21
1 28.62
1 28.82
The code I attempted :
spread = y; %price
temp=NaN(length(spread), 2); % initialise N X 2 array
temp(:,2)=spread; % put price into 2nd column
%numUnits is a single row of signals (eg : -1,0,1...)
for i = 2:numel(numUnits)
temp(i,1)=numUnits(i);
compare = abs(numUnits(i)-numUnits(i-1));
if compare == 2
temp = [temp(1:i-1,:); temp(i:i,:); temp(i:end,:)];
temp(i,1)=0;
%temp(i,2)=i;
end
end
The result is incorrect because the for loop is always operating on the original NumUnits array and not the Temp where I am adding the extra rows. I dont have an elegant solution to add dynamically an array and operate it on the fly.
Anyone able to help or give some clue? Thanks in advance!
Darren

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Timo Dietz
Timo Dietz 2021년 1월 6일
편집: Timo Dietz 2021년 1월 6일
priceArray= ...
[1 29.21
-1 29.41
-1 29.43
-1 29.63
1 29.01
0 29.21
1 28.62
1 28.82]
y=priceArray(:, 2);
numUnits = priceArray(:, 1);
% YOUR CODE WITH MODIFICATIONS:
spread = y; %price
temp=NaN(2*length(spread), 2); % initialise N X 2 array TWICE the size of the origin
%temp(:,2)=spread; % put price into 2nd column
%numUnits is a single row of signals (eg : -1,0,1...)
idx = 1; % introduce second index variable
for i = 2:numel(numUnits)
%temp(i,1)=numUnits(i);
compare = abs(numUnits(i)-numUnits(i-1))
if compare == 2
temp(idx, :) = priceArray(i-1, :);
idx = idx + 1;
temp(idx, :) = [0 priceArray(i, 2)];
else
temp(idx, :) = priceArray(i-1, :);
end
idx = idx + 1;
end
temp(idx, :) = priceArray(i, :); % add last row which always equals the origin
temp = temp(1:idx, :)
Try these modifications. The point is that you have to introduce a second index variable since your temp array is bigger than
the origin. So you can't use one common index. Further, the initialization of temp must reserve more rows - I chose twice
the size for simplicity (+50% should suffice already).
  댓글 수: 1
Darren Lim
Darren Lim 2021년 1월 6일
thank you Timo, this works! on my learning journey and this helps a lot!

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