Given a vector of assorted positive integers, how to create a vector with the means of every 2 integers inserted between each other?
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I've managed to take the mean of the whole vector with the mean function, but this doesn't seem like the right path to take. I am thinking of: v= 2 6 8 3 1 9 4 5 7 m1= v(1,2:9) m2= [m1,v(end)] m3= [v;m2] vmeans=mean(m2)
there are quite a few intermediate steps just to get to the vector of the means by themselves.
any answers to get to these values better or how to insert those values in between each original integer are much appreciated
thanx
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David Young
am 9 Jun. 2011
result = interp1(v, linspace(1, length(v), length(v)*2-1), 'linear')
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David Young
am 12 Jun. 2011
Yes, to understand this it may help to note that, for example
linspace(1, 3, 5)
gives
[1 1.5 2 2.5 3]
- that is, the points in the index space of v for which you want to find values.
Weitere Antworten (3)
David Young
am 9 Jun. 2011
result(1:2:2*length(v)-1) = v;
result(2:2:2*(length(v)-1)) = conv(v, [1 1]/2, 'valid')
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Andrei Bobrov
am 9 Jun. 2011
m2 = v([2:end,end]);
m3 = [v;m2];
vmeans = mean(m2);
EDIT
vout = reshape([v;conv(v,[1 1],'valid')/2 0],1,[]);
vout = vout(1:end-1);
more only it case
vout = interp1(1:length(v),v,1:.5:length(v));
4 Kommentare
David Young
am 9 Jun. 2011
It's more efficient to divide the mask in the convolution by 2, rather than dividing the result of the convolution by 2.
linspace is preferable to using the colon operator if the output needs to be a definite length and the increment is not an integer.
David Young
am 9 Jun. 2011
v= [2 6 8 3 1 9 4 5 7]; % data
m = (v(1:end-1)+v(2:end))/2;
t = [v; [m 0]];
t = t(:);
result = t(1:end-1).'
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