I'm trying to get the middle point of X Y and Weight values, but I didn't found the correct formula to get that.
For example in the picture I've added a few points with a few weights and I need to found the middle point base on the weights.
How about
sum (Wi Xi)/ sum (Wi)
Related
Given a set of coordinates corresponding to a closed shape, I want to calculate the total absolute curvature, which requires calculating the curvature for each point, taking the absolute value, and summing them. Simple enough.
I used the answer to this question to calculate the curvature from a matrix of x y coordinates (xymat) and get what I thought would be the total absolute curvature:
sum(abs(predict(smooth.spline(xymat), deriv = 2)$y))
The problem is that total absolute curvature has a minimum value of 2*pi and is exactly that for circles, but this code is evaluating to values less than 2*pi:
library(purrr)
xymat <- map_df(data.frame(degrees=seq(0:360)),
function(theta) data.frame(x = sin(theta), y = cos(theta)))
sum(abs(predict(smooth.spline(xymat), deriv = 2)$y))
This returns 1.311098 instead of the expected value of 6.283185.
If I change the df parameter of smooth.spline to 3 as in the previous answer, the returned value is 3.944053, still shy of 2*pi (the df value smooth.spline calculated for itself was 2.472213).
Is there a better way to calculate curvature? Is smooth.spline parameterized by arc length or will incorporating it (somehow) rescue this calculation?
Okay, a few things before we begin. You're using degrees in your seq, which will give you incorrect results (0 to 360 degrees). You can check that this is wrong by taking cos(360) in R, which isn't 1. This is explained in the documentation for the trig functions under Details.
So let's change your function to this
xymat <- map_df(data.frame(degrees=seq(0,2*pi,length=360)),
function(theta) data.frame(x = sin(theta), y = cos(theta)))
If you plot this, this indeed looks like a circle.
Let's actually restrict this to the lower half of the circle. If you put a spline through this without understanding the symmetry and looking at the plot, chances are that you'll get a horizontal line through the circle.
Why? because the spline doesn't know that it's symmetric above and below y = 0. The spline is trying to fit a function that explains the "data", not trace an arc. It splits the difference between two symmetric sets of points around y = 0.
If we restrict the spline to the lower half of the circle, we can use y values between 1 and -1, like this:
lower.semicircle <- data.frame(predict(smooth.spline(xymat[91:270,], all.knots = T)))
And let's fit a spline through it.
lower.semicircle.pred<-data.frame(predict(smooth.spline(lower.semicircle, all.knots = T)))
Note that I'm not using the deriv function here. That is for a different problem in the cars example to which you linked. You want total absolute curvature and they are looking at rate of change of curvature.
What we have now is an approximation to a lower semicircle using splines. Now you want the distance between all of the little sequential points like in the integral from the wikipedia page.
Let's calculate all of the little arc distances using a distance matrix. This literally calculates the Euclidean distances between each point to every other point.
all.pairwise.distances.in.the.spline.approx<-dist(lower.semicircle.pred, diag=F)
dist.matrix<-as.matrix(all.pairwise.distances.in.the.spline.approx)
seq.of.distances.you.want<-dist.matrix[row(dist.matrix) == col(dist.matrix) + 1]
This last object is what you need to sum across.
sum(seq.of.distances.you.want)
..which evaluates to [1] 3.079 for the lower semicircle, around half of your 2*pi expected value.
It's not perfect but splines have problems with edge effects.
I am currently trying to understand the calculation of the mean curvature for a 3D surface, where one coordinate is a function of the other two coordinates.
Looking at wikipedia https://en.wikipedia.org/wiki/Mean_curvature#Surfaces_in_3D_space under "[For the special case of a surface defined as a function of two coordinates, e.g. z = S(x,y)]" they give this formula:
mean curvature
What i don't understand here is the div(z - S) . If z = S(x,y) then i would think that z is the same as S and thus z - S equals 0.
I tried to follow the cited literature but i didn't find what i was looking for.
Apparently i misunderstand something here and z is not the same as S?
Any help would be appreciated.
z-S(x,y) is a function of 3 variables, the gradient of which is (-S_x,-S_y,1), see the second line. Then you normalize this gradient vector and compute the divergence of the normalized vector field.
I am working on a project of interpolating sample data {(x_i,y_i)} where the input domain for x_i locates in 4D space and output y_i locates in 3D space. I need generate two look up tables for both directions. I managed to generate the 4D -> 3D table. But the 3D -> 4D one is tricky. The sample data are not on regular grid points, and it is not one to one mapping. Is there any known method to treat this situation? I did some search online, but what I found is only for 3D -> 3D mapping, which are not suitable for this case. Thank you!
To answer the questions of Spektre:
X(3D) -> Y(4D) is the case 1X -> nY
I want to generate a table that for any given X, we can find the value for Y. The sample data is not occupy all the domain of X. But it's fine, we only need accuracy for point inside the domain of sample data. For example, we have sample data like {(x1,x2,x3) ->(y1,y2,y3,y4)}. It is possible we also have a sample data {(x1,x2,x3) -> (y1_1,y2_1,y3_1,y4_1)}. But it is OK. We need a table for any (a,b,c) in space X, it corresponds to ONE (e,f,g,h) in space Y. There might be more than one choice, but we only need one. (Sorry for the symbol confusing if any)
One possible way to deal with this: Since I have already established a smooth mapping from Y->X, I can use Newton's method or any other method to reverse search the point y for any given x. But it is not accurate enough, and time consuming. Because I need do search for each point in the table, and the error is the sum of the model error with the search error.
So I want to know it is possible to find a mapping directly to interpolate the sample data instead of doing such kind of search in 3.
You are looking for projections/mappings
as you mentioned you have projection X(3D) -> Y(4D) which is not one to one in your case so what case it is (1 X -> n Y) or (n X -> 1 Y) or (n X -> m Y) ?
you want to use look-up table
I assume you just want to generate all X for given Y the problem with non (1 to 1) mappings is that you can use lookup table only if it has
all valid points
or mapping has some geometric or mathematic symmetry (for example distance between points in X and Yspace is similar,and mapping is continuous)
You can not interpolate between generic mapped points so the question is what kind of mapping/projection you have in mind?
First the 1->1 projections/mappings interpolation
if your X->Y projection mapping is suitable for interpolation
then for 3D->4D use tri-linear interpolation. Find closest 8 points (each in its axis to form grid hypercube) and interpolate between them in all 4 dimensions
if your X<-Y projection mapping is suitable for interpolation
then for 4D->3D use quatro-linear interpolation. Find closest 16 points (each in its axis to form grid hypercube) and interpolate between them in all 3 dimensions.
Now what about 1->n or n->m projections/mappings
That solely depends on the projection/mapping properties which I know nothing of. Try to provide an example of your datasets and adding some image would be best.
[edit1] 1 X <- n Y
I still would use quatro-linear interpolation. You still will need to search your Y table but if you group it like 4D grid then it should be easy enough.
find 16 closest points in Y-table to your input Y point
These points should be the closest points to your Y in each +/- direction of all axises. In 3D it looks like this:
red point is your input Y point
blue points are the found closest points (grid) they do not need to be so symmetric as on image .
Please do not want me to draw 4D example that make sense :) (at least for sober mind)
interpolation
find corresponding X points. If there is more then one per point chose the closer one to the others ... Now you should have 16 X points and 16+1 Y points. Then from Y points you need just to calculate the distance along lines from your input Y point. These distances are used as parameter for linear interpolations. Normalize them to <0,1> where
0 means 'left' and 1 means 'right' point
0.5 means exact middle
You will need this scalar distance in each of Y-domain dimension. Now just compute all the X points along the linear interpolations until you get the corresponding red point in X-domain.
With tri-linear interpolation (3D) there are 4+2+1=7 linear interpolations (as on image). For quatro-linear interpolation (4D) there are 8+4+2+1=15 linear interpolations.
linear interpolation
X = X0 + (X1-X0)*t
X is interpolated point
X0,X1 are the 'left','right' points
t is the distance parameter <0,1>
If I draw a line from let's say: (2,3) to (42,28), how can I get all points on the line in a Point list? I tried using the slope, but I can't seem to get the hang of it.
To be clear: I would like all the pixels that the line covers. So I can make the line 'clickable'.
This is a math question. The equation of a line is:
y = mx + c
So you need to figure out the gradient (m) and the intercept (c) and then plug in values for x to get values for y.
But what do you mean by "all the points on a line"? There is an infinite number of points if x and y are real numbers.
You can use the formula (x-x1)/(x1-x2) = (y-y1)/(y1-y2). And you know the points with x values ranging from 2 to 42 are on the line and their associated y values have to be found. If any of the resulting y value is not an integer then it should be approximated rightly. And if two consecutive y values differ by more than 1 then the missing y values should be mapped to the last x value.
Here is the pseudo code (tried to capture the crux of the algorithm)
prevY = y1
for(i=x1+1;i<=x2;++i)
{
y = computeY(i);
if(diff(y,prevY)>1) dump_points(prevY,y,i);
prevY = y;
dump_point(i,y);
}
dump_points(prevY,y2,x2);
I am probably not covering all the cases here (esp. not the corner ones). But the idea is that for one value of x there would could be many values of y and vice versa depending on the slope of the line. The algorithm should consider this and generate all the points.
Hi well I got a problem about scaling shapes.Well I m trying to scale two similar shapes.It is in 2d and each shape has n points .I found a statement like this from a paper I read
"The size of a shape is the root mean square distance between the shape points and
it's centroid."
So from this point if I calculate the size of both shapes S1 and S2 and lets say S1=xS2 so if I create scaling matrix like this
[x 0]
[0 x]
(i just wrote 2x2 matrix i know it should be different) and if I mulitply it with S2 are their shapes aligned? Thx
Well I think I found a solution .It is done using a scale metric instead of real scale value.
if the shape 3 points (x1,y1) (x2,y2) (x3,y3) a scale metric S is square root of sum of each points squared values like
mean x=(x1+x2+x3)/3
mean y=(y1+y2+y3)/3
S=((x1-x)^2 +(y1-y)^2+(x2-x)^2 +(y2-y)^2+(x3-x)^2 +(y3-y)^2)^1/2
and if this scale metric is calculated for both shapes there will be an equation like this S1=AS2
and if all points of shape 2 is multiplied with value of A they will have similar shapes.
It reminds me of Fant's Resampling Algorithm, smooth...
Not sure it fits your question.
If I'm understanding what you wrote, then multiplying your matrix by a shape, (say, S2) will scale Each of S2's points by a factor of x.
This says nothing about their alignment. this paper might help your understanding if you want to do it efficiently