# 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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Saeid am 27 Dez. 2022
Kommentiert: Voss am 28 Dez. 2022
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)
XY = 2×30
4 5 7 10 2 9 8 8 5 7 2 10 4 5 8 6 6 2 5 4 4 3 8 2 2 2 1 6 7 8 9 5 1 5 5 1 7 4 3 7 3 7 2 2 9 3 7 4 4 2 10 7 8 1 4 4 9 9 9 1
% NewArray
xy = [XY(1,1:m:end); mean(reshape(XY(2,:),m,[]),1)]
xy = 2×6
4.0000 9.0000 2.0000 6.0000 4.0000 2.0000 5.0000 4.4000 4.6000 4.0000 6.0000 6.4000
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Saeid am 27 Dez. 2022
Bearbeitet: Saeid am 27 Dez. 2022
Thanks Vos.. this will work perfectly!
Voss am 27 Dez. 2022
You're welcome!

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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)]
A1 = 2×30
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100 104 108 112 116 120
x = A1(1, 1:5:end)
x = 1×6
1 6 11 16 21 26
y = mean(reshape(A1(2,1:end),6,5),2)
y = 6×1
52 56 60 64 68 72
B1 = [x;y.']
B1 = 2×6
1 6 11 16 21 26 52 56 60 64 68 72
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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)]
A1 = 2×30
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100 104 108 112 116 120
your_method = mean(reshape(A1(2,:),6,5),2).'
your_method = 1×6
52 56 60 64 68 72
my_method = mean(reshape(A1(2,:),5,[]),1)
my_method = 1×6
12 32 52 72 92 112
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))
ans = 52
mean(A1(2,2:6:end)) % etc.
ans = 56

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