# [Solved] Upsampling multidimensional array with interpn

18 ビュー (過去 30 日間)
Johnny Scalera 2020 年 7 月 9 日 15:36

Hi all,
I'm trying to upsample a multidimensional array. Here is the code that I used for a monodimensional array:
sf1 = 100; %original sampling frequency
time = (0:(1/sf1):10-(1/sf1))'; %time array
x = randn(length(time), 1);
sf2 = 200; %final sampling frequency
upTime = (0:(1/sf2):10-(1/sf2))';
y = interp1(time, x, upTime);
Now I would like to upsample a 5D array, such as:
z = [randn(length(time), 1) randn(length(time), 1) randn(length(time), 1) randn(length(time), 1) randn(length(time), 1)];
How Can I use the interpn function properly?
Thank You all!

### 採用された回答

John D'Errico 2020 年 7 月 9 日 21:33

That is NOT a 5 dimensional array.
While you may THINK of it as a set of points that live in 5 dimensions, it is NOT a 5-d array. You CANNOT use interpn to interpolate that data. NOT. Period. NOT.
THIS is a 5-dimensional array:
A = rand(2,3,4,5,6);
size(A)
ans =
2 3 4 5 6
You can then use tools like griddatan or scatteredInterpolant to interpolate such data.
If you want to view the data as 5 independent streams of ONE dimensional data, then you can still use interp1 to interpolate. For example:
A = rand(100,5);
t = (1:100)';
interp1(t,A,7.25)
ans =
0.35284 0.051081 0.32818 0.34913 0.75639
Only you know what the array represents. By themselves, numbers are just numbers. Only you know what they represent.

#### 2 件のコメント

Johnny Scalera 2020 年 7 月 10 日 7:49
You are right! This is the correct solution
y1 = interp1(time, z, upTime);
Thank you!
John D'Errico 2020 年 7 月 10 日 9:34
Excellent. It almost had to be one of those two cases.