How can I interpolate just one point on a given function while saving time?
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Hi all,
I have a function 
and I have a collection of N couple of points 
, namely 
and 
respecting the function f. I want to evaluate my function in a single point 
out of set X, so 
. Until now I'm using the Matlab function 
= interp1(
,'spline'). But maybe this methods requires a certain computational effort. Since I always evaluate my function on a single point (not a vector of points), is there an alternative function (or method) to interp1 to do this task that allows to save time?
and I have a collection of N couple of points , namely and respecting the function f. I want to evaluate my function in a single point out of set X, so . Until now I'm using the Matlab function = interp1(,'spline'). But maybe this methods requires a certain computational effort. Since I always evaluate my function on a single point (not a vector of points), is there an alternative function (or method) to interp1 to do this task that allows to save time?Thank you in advance!
4 个评论
Jan
2021-3-19
Do you really need a spline?
Walter Roberson
2021-3-19
Will you be doing this repeatedly with the same X and Y?
Is your X sorted? Is it monotonic (no duplicates, no reversals) ?
Renato Quartullo
2021-3-19
编辑：Renato Quartullo
2021-3-19
Image Analyst
2021-3-20
Everything takes computation effort. Please state how long it takes, and how fast you need it to be. You can use tic and toc to time it.
采纳的回答
Jan
2021-3-22
3 times faster without overhead, if you fokus on the neighborhood:
X = -360:0.5:359.5;
Y = rand(size(X));
xi = 7.59;
tic
for k = 1:1e4
yi1 = interp1(X, Y, xi,'spline');
end
toc
tic
index = find(xi >= X, 1, 'last');
XX = X(index-8:index+8);
for k = 1:1e4
F = griddedInterpolant(XX, Y(index-8:index+8), 'spline');
yi2 = F(xi);
end
toc
yi1 - yi2 % Of course a difference: RAND are very noisy input data
% Elapsed time is 1.554598 seconds.
% Elapsed time is 0.513496 seconds.
1 个评论
Michal
2023-8-25
You should check out conditions: 1 <= index-8 & index+8 <= length(X)
更多回答（1 个）
Walter Roberson
2021-3-19
With the X being sorted and known increments, for any xstar you can directly compute the corresponding X bin that does not exceed xstar as xstar*2+721. Call that idx
jdxstart = min(max(1,idx-2), length(X) - 4);
jdx = jdxstart:jdxstart+4;
pp = spline(X(jdx), y(idx))
ystar = ppval(pp, xstar);
That is, we extract X y "near" xstar and spline only there to reduce the work.
6 个评论
Renato Quartullo
2021-3-20
Jan
2021-3-20
interp1 and spline have a lot of overhead. If you post some code, which shows unequivocally, how you call these functions, it is possible to suggest a lean version, which avoids the overhead. Because your system is tiny, it is e.g. a waste of time to set it up as sparse matrix as in spline().
Renato Quartullo
2021-3-22
This is the function, or, better, the points that I called Y versus X. X is a collection of points described as before, so they are -360:0.5:359.5. For example I want to evaluate the value relative to x* = 7.59. Until now I'm using:
y* = interp1(X,Y,x*,'spline');
This command takes on average 80 microseconds (tested on 100'000 instances). Is this time improvable in your opinion?
Renato Quartullo
2021-3-22
Renato Quartullo
2021-3-22
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