I plotted an expression curve, i.e.curve(-log((1-x)/0.9999)/x,ylim=c(0,4)).
However, I want to see the reverse relationship, i.e. y changes over x instead of x changes over y. Are there any R function can plot it automatically? Or a function that can solve the equation?
There are two obvious choices:
(i) derive the inverse function algebraically (trivial in this case),
That is, take y=-log((1-x)/0.9999) and make x the subject of the equation (which would require straightforward algebraic manipulation suitable for a question on math.SE if it's not obvious how to proceed)...
... and then use curve on the result of that, or
(ii) use plot rather than curve to plot a set of (x,y) pairs (set type="l" to get a curve), and simply interchange which is x and which is y in the call to plot.
Related
I want to plot in Maple the solutions to the equation (x-y)^2+(1-z)^2=0.
However, implicitplot3d is not able to plot them, at least using the default arguments. Any recommendations?
I know a priori that the set of solutions is going to be a curve contained in a plane, because I want to plot solutions of equations of the form 'f(x,y)^2+(z-1)^2=0'. Where 'f(x,y)' is a polynomial.
If x, y, and z are all real then those two squares must both equal zero, and thus z=1.
In that case you can simply utilize the implicitplot command for a 2-D plot of f(x,y)=0, and if you wish you can transform that to a 3-D plot with z=1.
restart;
with(plots,display): with(plots,implicitplot):
with(plottools,transform):
eqn := (x-y)^2+(1-z)^2 = 0:
P2D := implicitplot(eval(eqn,z=1)):
display(transform((x,y)->[x,y,1])(P2D),
labels=[x,y,z]);
eqn := (x^2-y)^2+(1-z)^2 = 0:
P2D := plots:-implicitplot(eval(eqn,z=1)):
display(transform((x,y)->[x,y,1])(P2D),
labels=[x,y,z]);
I have a similar line graph plotted using R plot function (plot(df))
I want to get distance of the whole line between two points in the graph (e.g., between x(1) and x(3)). How can I do this?
If your function is defined over a fine grid of points, you can compute the length of the line segment between each pair of points and add them. Pythagoras is your friend here:
To the extent that the points are not close enough together that the function is essentially linear between the points, it will tend to (generally only slightly) underestimate the arc length.
Note that if your x-values are stored in increasing order, these ẟx and ẟy values can be obtained directly by differencing (in R that's diff)
If you have a functional form for y as a function of x you can apply the integral for the arc length -- i.e. integrate
∫ √[1+(dy/dx)²] dx
between a and b. This is essentially just the calculation in 1 taken to the limit.
If both x and y are parametric functions of another variable (t, say) you can simplify the parametric form of the above integral (if we don't forget the Jacobian) to integrating
∫ √[(dx/dt)²+(dy/dt)²] dt
between a and b
(Note the direct parallel to 1.)
if you don't have a convenient-to-integrate functional form in 2. or 3. you can use numerical quadrature; this can be quite efficient (which can be handy when the derivative function is expensive to evaluate).
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>
t = 0:%pi/50:10*%pi;
plot3d(sin(t),cos(t),t)
When I execute this code the plot is done but the line is not visible, only the box. Any ideas which property I have to change?
Thanks
The third argument should, in this case, be a matrix of the size (length arg1) x (length arg2).
You'd expect plot3d to behave like an extension of plot and plot2d but it isn't quite the case.
The 2d plot takes a vector of x and a vector of y and plots points at (x1,y1), (x2,y2) etc., joined with lines or not as per style settings. That fits the conceptual model we usually use for 2d plots - charting the relationship of one thing as a function of another, in most cases (y = f(x)). THere are other ways to use a 2d plot: scatter graphs are common but it's easy enough to produce one using the two-rows-of-data concept.
This doesn't extend smoothly to 3d though as there are many other ways you could use a 3d plot to represent data. If you gave it three vectors of coordinates and asked it to draw a line between them all what might we want to use that for? Is that the most useful way of using a 3d plot?
Most packages give you different visualisation types for the different kinds of data. Mathematica has a lot of 3d visualisation types and Python/Scipy/Mayavi2 has even more. Matlab has a number too but Scilab, while normally mirroring Matlab, in this case prefers to handle it all with the plot3d function.
I think of it like a contour plot: you give it a vector of x and a vector of y and it uses those to create a grid of (x,y) points. The third argument is then a matrix whose dimensions match those of the (x,y) grid holding the z-coordinates of each point. The first example in the docs does what I think you're after:
t=[0:0.3:2*%pi]';
z=sin(t)*cos(t');
plot3d(t,t,z);
The first line creates a column vector of length 21
-->size(t)
ans =
21. 1.
The second line computes a 21 x 21 matrix of products of the permutations of sin(t) with cos(t) - note the transpose in the cos(t') element.
-->size(z)
ans =
21. 21.
Then when it plots them it draws (x1,y1,z11), (x1,y2,x12), (x2,y2,z22) and so on. It draws lines between adjacent points in a mesh, or no lines, or just the surface.
I am doing some caclulations with extreme velocities and the only way to solve my system equations is to do it graphically. Once I have plotted my curve, I would like to develop a function that enter an x-value and the function itself plots a line from this x-value up to the corresponding point of the curve and from this point, another line over y-value. Like this I would get my y-value that would be the solution of my system equations.
Here is my code. The function Vr_Vmed is the expression of my final equation. In fact, n=4 and Tr=50 and x is the variable.
par(font=10,font.axis=10,font.lab=10,font.main=11,font.sub=10)
curve(Vr_Vmed(x,n,Tr),xlim=c(1,2.5),ylim=c(1,17),
xaxs="i",yaxs="i",xaxt="n",yaxt="n",lwd=2,
xlab="K Weibull",ylab="Vref / Vmed",usr=c(1,2.5,1,17),
main="Vref Estimation")
axis(1,at=c(seq(1,2.5,0.1)),xaxp=c(1,2.5,1))
axis(2,at=c(seq(1,17,1)))
If you just want to add lines to your plot,
you can use lines or segments.
f <- function(x) {
y <- Vr_Vmed(x,n,Tr)
lines(c(x,x,0),c(0,y,y))
}
f(2)
(But that does not "solve" anything: your Vr_med function
aparently does all the work.)