Given a vector of assorted positive integers, how to create a vector with the means of every 2 integers inserted between each other?
1 回表示 (過去 30 日間)
古いコメントを表示
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
0 件のコメント
採用された回答
David Young
2011 年 6 月 9 日
result = interp1(v, linspace(1, length(v), length(v)*2-1), 'linear')
2 件のコメント
David Young
2011 年 6 月 12 日
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.
その他の回答 (3 件)
David Young
2011 年 6 月 9 日
result(1:2:2*length(v)-1) = v;
result(2:2:2*(length(v)-1)) = conv(v, [1 1]/2, 'valid')
0 件のコメント
Andrei Bobrov
2011 年 6 月 9 日
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 件のコメント
David Young
2011 年 6 月 9 日
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
2011 年 6 月 9 日
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).'
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Creating and Concatenating Matrices についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!