Reduce a large XY array to a much smaller xy array where the x data is diluted to a much smaller vector and y values are the mean of the ones inbetween
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I need to dilute a very large XY array where e.g. length(X)=10000 and I want to reduce the length of this array to a much smaller number. It is of course possible to reshape the array but this requires eleiminating all the elements inbetween. I need the y elements that make up the new y array to be an average of the elements that have been eliminated.
This means:
OldArray=XY=[X0 X1 X2 X3 X4 X5 X6 X7 X8 ... Xn; Y1 Y2 Y3 Y4 Y5 Y6 Y7 Y8 ... Yn];
NewArray=xy=[x1 x2 x3 ... xm; y1 y2 y3 ... ym]; m is a fraction of n
where
x1=X0; x2=X5; x3=X10; x4=15 ...
but
y1=mean(y0:y4); y2=mean(y5:y9); y3=mean(y10:y14); ...
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Akzeptierte Antwort
Voss
am 27 Dez. 2022
I'll assume that "Y1 Y2 ... Yn" should be "Y0 Y1 ... Yn" and that "y1=mean(y0:y4); y2=mean(y5:y9); y3=mean(y10:y14); ..." should be "y1=mean(Y0:Y4); y2=mean(Y5:Y9); y3=mean(Y10:Y14); ...".
n = 30;
m = 5;
% OldArray
XY = randi(10,2,n)
% NewArray
xy = [XY(1,1:m:end); mean(reshape(XY(2,:),m,[]),1)]
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Weitere Antworten (1)
Sulaymon Eshkabilov
am 27 Dez. 2022
Eg. taking a matrix of 2-by-30:
A1 = [1:30; 4*(1:30)]
x = A1(1, 1:5:end)
y = mean(reshape(A1(2,1:end),6,5),2)
B1 = [x;y.']
3 Kommentare
Sulaymon Eshkabilov
am 28 Dez. 2022
Saeid,
This given answer is completely correct, but you are not reading it correctly. All the best.
Voss
am 28 Dez. 2022
@Sulaymon Eshkabilov: I interpret "y1=mean(y0:y4);", etc., to mean that the 1st element of the 2nd row of the desired result should be the mean of the first 5 elements of the 2nd row of the input matrix, and so on similarly for the subsequent elements of the 2nd row of the result (mean of elements 6 through 10 of 2nd row of input matrix, then elements 11 through 15, and so on).
Using your example input matrix, the mean of the first 5 elements of the 2nd row (4, 8, 12, 16, 20) is 12, not 52, which is the mean of (4, 28, 52, 76, 100):
A1 = [1:30; 4*(1:30)]
your_method = mean(reshape(A1(2,:),6,5),2).'
my_method = mean(reshape(A1(2,:),5,[]),1)
Do you interpret "y1=mean(y0:y4);" to mean that the 1st element of the 2nd row should be the mean of [y0 y6 y12 y18 y24], i.e., elements 1, 7, 13, 19, and 25 of the 2nd row of the input matrix?
mean(A1(2,1:6:end))
mean(A1(2,2:6:end)) % etc.
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