2d grid interpolation to receive x-value

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Ahmed Chahbaz
Ahmed Chahbaz 2018-5-17
评论： John D'Errico 2018-5-17
Hello, I have a 2D-Grid. Assuming:
x=[0:0.05:0.5]
y=[0:5:20]
The grid is filled with a 5x11 DataSet, which contains random data (call it data).
Now I could interpolate using
interp2(x,y,data,x_i,y_i)
To receive the corresponding interpolated value to [x_i, y_i] from my dataSet.
What I need to do, is to interpolate an x value by passing y_i and a dataPoint from my dataset.
Something like:
[x_i]=interp2(y,data,y_i,data(j,k))
i,j,k const.
Thanks in advance!
  2 个评论
Ahmed Chahbaz
Ahmed Chahbaz 2018-5-17
My data set consists of a height profile (data), which depends of temperature(y) and time (x).
Given one data value data(j,k) and a temperature point (y_i), I want to retrieve the interpolated value of my time (x_i).

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回答（2 个）

KSSV
KSSV 2018-5-17
YOu should use scatteredInterpolant for this case. How about this?
x=[0:0.05:0.5]
y=[0:5:20] ;
data = rand(length(y),length(x)) ;
[X,Y] = meshgrid(x,y) ;
F = scatteredInterpolant(Y(:),data(:),X(:)) ;
iwant = F(Y(1,7),rand(1,7))
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Ahmed Chahbaz
Ahmed Chahbaz 2018-5-17
编辑：Ahmed Chahbaz 2018-5-17
@KSSV Thanks for your solution approach, but it is not working for me. I think it is because my data is a 5x11 matrice and scatteredInterpolant requires a column-vector format.
@John Thanks for your help. Waiting for your answer.
John D'Errico
John D'Errico 2018-5-17
There are serious potential problems here, ones that may not be solved using scatteredInterpolant as KSSV suggests.
Lets look at what you are trying to do, using some random data.
x = [0:0.05:0.5];
y = [0:5:20];
data = rand(numel(y),numel(x));
data
data =
0.49836 0.75127 0.95929 0.84072 0.34998 0.35166 0.28584 0.075854 0.12991 0.16218 0.60198
0.95974 0.2551 0.54722 0.25428 0.1966 0.83083 0.7572 0.05395 0.56882 0.79428 0.26297
0.34039 0.50596 0.13862 0.81428 0.25108 0.58526 0.75373 0.5308 0.46939 0.31122 0.65408
0.58527 0.69908 0.14929 0.24352 0.61604 0.54972 0.38045 0.77917 0.011902 0.52853 0.68921
0.22381 0.8909 0.25751 0.92926 0.47329 0.91719 0.56782 0.93401 0.33712 0.16565 0.74815
Yeah, I know, not very creative. But it is early in the morning for me, and you did not give us much to go on. And your data surely has SOME structure to it.
Now, let me pick a value for y to use here. We can think of this problem as a reverse interpolation at a specific y. I'll pick the value for y as 17.
yi = 17;
What does this surface look like along the line when y is fixed at 17? INTERP2 will give us that.
datapred = interp2(X,Y,data,x,repmat(yi,size(x)))
datapred =
0.44069 0.77581 0.19258 0.51782 0.55894 0.69671 0.4554 0.8411 0.14199 0.38338 0.71279
plot(x,datapred,'bx-')
grid on
xlabel x
ylabel 'data @ yi == 17'
Now, if you pick any given value of data, can you infer the value of x that produced it? Can you use reverse interpolation to infer x now? BE VERY CAREFUL HERE!
One thing you CANNOT do is simply swap the axes, and try to use an interpolant!!!!!!!
Thus, at datapred of 0.61604 (which lived at data(4,5) in my random set), for example, what value of x would we predict?
H = refline(0,data(4,5));set(H,'color','r')
So there are actually 7 locations where the red and blue lines intersect. That happens because our data was purely random. If the surface was nice and smooth and monotonic, things would be a bit better. There might be a unique solution then. But even so, the use of scatteredInterpolant as suggested by KSSV will be a problem.
When you use this form for scatteredInterpolant, what happens?
F = scatteredInterpolant(Y(:),data(:),X(:));
ezsurf(@(u,v) F(u,v),[min(y),max(y),0,1])
F(yi,data(4,5))
ans =
0.31909
So this gives ONE solution. But remember, there were 7 possible solutions. The function handle returned by scatteredInterpolant just gives one number. To be honest, I'm not even positive I can trust that one number that was returned.
plot(x,datapred,'bx-')
H = refline(0,data(4,5));set(H,'color','g')
hold on
plot(F(yi,data(4,5)),data(4,5),'ro')
In fact, the red dot there is not exactly on the blue curve. There is a slight error because the scattered interpolant is doing a 2-d interpolation, whereas the blue curve was created in a subtly different way.
Again, you cannot just swap the variables in an interpolation problem. The result will be random and sometimes significantly so.

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John D'Errico
John D'Errico 2018-5-17
编辑：John D'Errico 2018-5-17
There are serious potential problems here, ones that may not be solved using scatteredInterpolant as KSSV suggests.
Lets look at what you are trying to do, using some random data.
x = [0:0.05:0.5];
y = [0:5:20];
data = rand(numel(y),numel(x));
data
data =
0.49836 0.75127 0.95929 0.84072 0.34998 0.35166 0.28584 0.075854 0.12991 0.16218 0.60198
0.95974 0.2551 0.54722 0.25428 0.1966 0.83083 0.7572 0.05395 0.56882 0.79428 0.26297
0.34039 0.50596 0.13862 0.81428 0.25108 0.58526 0.75373 0.5308 0.46939 0.31122 0.65408
0.58527 0.69908 0.14929 0.24352 0.61604 0.54972 0.38045 0.77917 0.011902 0.52853 0.68921
0.22381 0.8909 0.25751 0.92926 0.47329 0.91719 0.56782 0.93401 0.33712 0.16565 0.74815
Yeah, I know, not very creative. But it is early in the morning for me, and you did not give us much to go on. And your data surely has SOME structure to it.
Now, let me pick a value for y to use here. We can think of this problem as a reverse interpolation at a specific y. I'll pick the value for y as 17.
yi = 17;
What does this surface look like along the line when y is fixed at 17? INTERP2 will give us that.
datapred = interp2(X,Y,data,x,repmat(yi,size(x)))
datapred =
0.44069 0.77581 0.19258 0.51782 0.55894 0.69671 0.4554 0.8411 0.14199 0.38338 0.71279
plot(x,datapred,'bx-')
grid on
xlabel x
ylabel 'data @ yi == 17'
Now, if you pick any given value of data, can you infer the value of x that produced it? Can you use reverse interpolation to infer x now? BE VERY CAREFUL HERE!
One thing you CANNOT do is simply swap the axes, and try to use an interpolant!!!!!!!
Thus, at datapred of 0.61604 (which lived at data(4,5) in my random set), for example, what value of x would we predict?
H = refline(0,data(4,5));set(H,'color','r')
So there are actually 7 locations where the red and blue lines intersect. That happens because our data was purely random. If the surface was nice and smooth and monotonic, things would be a bit better. There might be a unique solution then. But even so, the use of scatteredInterpolant as suggested by KSSV will be a problem.
How can we solve this, in a way that will always work, for any data? You need to recognize there may be multiple solutions.
A nice way, that will be robust and return all possible solutions in the event that multiple solutions do exist, is Doug Schwarz's intersections utility. You can download it from the file exchange from this link:
fast-and-robust-curve-intersections
xi = intersections(x,datapred,[min(x),max(x)],repmat(data(4,5),1,2))
xi =
0.026164
0.063696
0.22072
0.26671
0.32083
0.3661
0.48532
As you can see, it found all 7 possible solutions.
plot(x,datapred,'bx-')
H = refline(0,data(4,5));set(H,'color','g')
hold on
plot(xi,data(4,5),'ro')
If data is a much more well-behaved, essentially monotonic surface, then intersections will still work, and usually produce one value. (Monotonicity is a concept that must be defined very carefully in multiple dimensions.)
  2 个评论
Ahmed Chahbaz
Ahmed Chahbaz 2018-5-17
Hello Mr.D'Errico.
Thanks for your detailed explanation.
My data set consists of a height profile (data), which depends of temperature(y) and time (x) and is monotone increasing.
Given one data value for my height [data(j,k)] and a temperature value [y_i], I want to retrieve the interpolated value of my timestamp [x_i].
In your example this would mean you could freely choose at the beginning [y_i] (which is what you did) and also one data point from your random created data set.
Given Those two values I want to retrieve the corresponding x value. Basically, I think this is some kind of reversed Look-Up Table ? Any suggestions how to solve this?
John D'Errico
John D'Errico 2018-5-17
But is that not exactly what I showed you how to solve?
If your surface is a smooth, well-behaved one, then you will see one solution, if any solution exists. So what is the question here? I gave you exactly the solution you wanted.
1. Use interp2 to generate a set of interpolations at yi.
2. Then use intersections to solve for xi.
Do you really desperately need a way in ONE line of code instead of two simple lines? Sigh. I'll need to create a more well-behaved surface.
x = [0:0.05:0.5];
y = [0:5:20];
data = exp(x + y'/20);
That example surface presumes your version of MATLAB is reasonably current.
yi = 17;
xi = fzero(@(u) interp2(x,y,data,u,yi) - data(4,5),[min(x),max(x)]);
xi =
0.0922306852282017
exp(xi + yi/20)
ans =
2.56569830498033
interp2(x,y,data,xi,yi)
ans =
2.58570965931585
data(4,5)
ans =
2.58570965931585
Did it work? Yes. But you need to recognize that if there happen to be multiple solutions, then fzero will NOT return all of them. You will arbitrarily get at most one of them, and it may indeed fail.

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