Get more exact x/y values from a curve fit

I have some some point coordinates, that i get from annotation in an image. Let's say these are my x and y values.

x = (1:50);
y = x + randi(10,1,50) + sin(x);

Now i fit these

fitted = fit(x',y','smoothingspline');

To get the y values, i do

y_fit(:,1) = feval(fitted, x);

Plot returns a fitted curve with 50 points, but i need a much smoother curve fit. I need around 1000 x/y values from 10 annotated points.

plot(x,y)
hold on
plot(x,y_fit)

Now in my actual problems, i have e.g. 10 x values distributed over 1000 pixels. If i eval my fitted function, only get 10 y values back, but i need much more precise coordinates, to get a very smooth line through these 10 points, and not just connect 10 points in a linear manner. Goal is to annotate around 20 points in an 5000x2000 pixel image, fit a circle/polygon through these 20 points and use poly2mask() to get a logical array. But the circle should not just be a connection of the 20 points, but a much smoother curve.

Answer 1

[Ameer Hamza]

What about fitting a low-order polynomial through your data-points

x = (1:50);
y = x + randi(10,1,50) + sin(x);

fitted = fit(x',y','poly4');
y_fit(:,1) = fitted(x); % this syntax is also valid

plot(x,y)
hold on
plot(x,y_fit)

[Stefan Lang]

This still gives me only 50 y values, which will be linearly connected when plotted. I need to get a smooth curve with many more than 50 points. Just like the EXACT function the curvefit gives me from point 1 to 50, so i can calculate e.g. y value for x point 48.24.

[Ameer Hamza]

You can interpolate at more points too, following define x_fit, which is different from x. You can use get the value at any point using function call syntax, for example, fitted(48.24).

x = (1:50);
y = x + randi(10,1,50) + sin(x);

fitted = fit(x',y','poly4');
x_fit = 1:0.1:50;
y_fit = fitted(x_fit); % this syntax is also valid

plot(x,y)
hold on
plot(x_fit, y_fit)

[Stefan Lang]

Now this seems to work, thanks a lot!

[Ameer Hamza]

I am glad to be of help!