I am attempting to plot a two-dimensional function of x and y below in R. I am not good at coding, so it's possible I'm missing something easy, but I got it to work just fine yesterday. It was displaying the 3D plot perfectly and I was even messing with different colors and point sizes for a while with no issues. I save, exit, and then I open the file today and for some reason I am getting an error. I didn't change anything and I have no clue why it won't work now.
library(rgl)
set.seed(27)
x <- runif(100000,-1,1)
y <- runif(100000,-1,1)
## Two-dimensional function of interest
example <- function(x, y){
f_xy <- x^2 + 2*(y^2) - 0.3*cos(3*pi*x) - 0.4*cos(4*pi*y) + 0.7
return(f_xy)
}
example(x,y)
plot3d(
x = x, y = y, z = example(x,y),
type = "p",
size = 1.3,
xlab = "X",
ylab = "Y",
zlab = "f(x,y)",
)
I get the following error message:
Error in example(x, y) :
dims [product 4] do not match the length of object [100000]
I have narrowed it down to the cosine function being the problem I think. If I mess with
f_xy <- x^2 + 2*(y^2) - 0.3*cos(3*pi*x) - 0.4*cos(4*pi*y) + 0.7
and get rid of the 3pi for x and the 4pi for y then it gives me a plot, but it's obviously not the one I need since I have now changed the equation.
Any ideas on how I can fix this?
I just figured out how to use cospi(3*x) instead of cos(3*pi*x) and now it works. I honestly have no idea why it worked yesterday without me doing this.
Related
I am trying to plot the function described below. However, when I do plot it, the x-axis shows 0 up to 1000, probably due to my definition of x0, instead of 0 up to 10. However, if I just use 1:10 as my x-values, the plot is not smooth at all. My question is: how do I set the x-axis to give me the real values that correspond to the function at that point?
PS: I have seen solutions where people replace the x-axis fully with their own values, but I am looking for a method that just uses the actual values of x.
eq <- function(x){
4*x^3 - 72*x^2 + 288*x
}
x0 <- seq(1, 10, by = 0.01)
plot(eq(x0),type='l',main="Plot of 4*x^3 - 72*x^2 + 288*x", xlab = "x", ylab = "4*x^3 - 72*x^2 + 288*x")
You have missed your x0 values. R deduced the x values based on the number of elements in eq(x0) (A sequence 1:1000 was used by default). By the way, you can improve your labels with expression
plot(x = x0,
y = eq(x0),
type='l',
main=expression(paste("Plot of ", 4*x^3 - 72*x^2 + 288*x)),
xlab = expression(x),
ylab = expression(4*x^3 - 72*x^2 + 288*x))
It works fine
I tried to run the following script to change the x axis. I followed other answers given in Stack Overflow; however I don’t get what I’m looking for. The original x axis goes from 0 to 1 but I want to change it from 0 to 20.
x<-c(0,0.1,0.2,0.25,0.30,0.35,0.40,0.60,0.90,1)
y<-c(0,0.014,0.41165,0.258,0.57,0.57,0.1803,0.5844,0.10185,0.085)
da<-cbind(x,y)
dat=data.frame(da)
plot(y~x,data=dat,xaxt="n")
pas=c(0,1,2,3,4,5,6,7,8,10,11,12,13,14,15,16,17,18,19,20)
axis(1, at=1:20,labels=pas)
This script produces a figure with just a 0 at the last mark. I need a x axis from 0 to 20. Thanks
I am not sure why you are trying to do this but the following would work.
x <- c(0,0.1,0.2,0.25,0.30,0.35,0.40,0.60,0.90,1)
y <- c(0,0.014,0.41165,0.258,0.57,0.57,0.1803,0.5844,0.10185,0.085)
dat <- data.frame(x = x, y = y)
x.range <- c(0,20)
x.labels <- x.range[1]:x.range[2]
pas <- seq(0, 1, length.out = length(x.labels))
plot(y ~ x, data = dat, xaxt = "n", xlim = c(0,1))
axis(1, at = pas, labels = x.labels, cex.axis=0.65)
You may have missed the xlim/ylim.
Did you mean y axis? Then just use :
plot(y~x,data=dat,ylim=c(0,20))
plot(y~x,data=dat,xlim=c(0,20))# you need to remove the 'xaxt="n"' to see the x-axis
Although either way I am not sure this is the best idea given the range of the data.
Using ggplot2
library(ggplot2)
qplot(x,y,data=dat)+ scale_x_continuous(labels=c("0"="fake0","0.25"="fake5","0.5"="fake10",".75"="fake15","1"="fake20"))# replace 'fake#' with anything
Although I am not sure why are you trying to do this.
I apologize first for bringing what I imagine to be a ridiculously simple problem here, but I have been unable to glean from the help file for package 'polynom' how to solve this problem. For one out of several years, I have two vectors of x (d for day of year) and y (e for an index of egg production) data:
d=c(169,176,183,190,197,204,211,218,225,232,239,246)
e=c(0,0,0.006839425,0.027323127,0.024666883,0.005603878,0.016599262,0.002810977,0.00560387 8,0,0.002810977,0.002810977)
I want to, for each year, use the poly.calc function to create a polynomial function that I can use to interpolate the timing of maximum egg production. I want then to superimpose the function on a plot of the data. To begin, I have no problem with the poly.calc function:
egg1996<-poly.calc(d,e)
egg1996
3216904000 - 173356400*x + 4239900*x^2 - 62124.17*x^3 + 605.9178*x^4 - 4.13053*x^5 +
0.02008226*x^6 - 6.963636e-05*x^7 + 1.687736e-07*x^8
I can then simply
plot(d,e)
But when I try to use the lines function to superimpose the function on the plot, I get confused. The help file states that the output of poly.calc is an object of class polynomial, and so I assume that "egg1996" will be the "x" in:
lines(x, len = 100, xlim = NULL, ylim = NULL, ...)
But I cannot seem to, based on the example listed:
lines (poly.calc( 2:4), lty = 2)
Or based on the arguments:
x an object of class "polynomial".
len size of vector at which evaluations are to be made.
xlim, ylim the range of x and y values with sensible defaults
Come up with a command that successfully graphs the polynomial "egg1996" onto the raw data.
I understand that this question is beneath you folks, but I would be very grateful for a little help. Many thanks.
I don't work with the polynom package, but the resultant data set is on a completely different scale (both X & Y axes) than the first plot() call. If you don't mind having it in two separate panels, this provides both plots for comparison:
library(polynom)
d <- c(169,176,183,190,197,204,211,218,225,232,239,246)
e <- c(0,0,0.006839425,0.027323127,0.024666883,0.005603878,
0.016599262,0.002810977,0.005603878,0,0.002810977,0.002810977)
egg1996 <- poly.calc(d,e)
par(mfrow=c(1,2))
plot(d, e)
plot(egg1996)
I am in the midst of solving a problem in Reconstructing (or recovering) a probability distribution function when only the moments of the distribution are known. I have written codes in R for it and although the logic seems right to me, I am not getting the output that I want.
The equation I am trying to using as the approximated (or reconstructed or recovered) CDF is what you see in the image below. I am writing codes for the right hand side of the equation and equating that to a vector that I call F in my codes.
The link to paper that contains the original equation can be found here.
http://www.sciencedirect.com/science/article/pii/S0167715208000187
It is marked as equation (2) in the paper.
Here is the code I wrote.:
#R Codes:
alpha <- 50
T <- 1
x <- seq(0, T, by = 0.1)
# Original CDF equation
Ft <- (1-log(x^3))*(x^3)
plot(x, Ft, type = "l", ylab = "", xlab = "")
# Approximated CDF equation using Moment type reconstruction
k<- floor(alpha*y/T)
for(i in 1:length(k))
{
for(j in k[i]:alpha)
{
F[x+1] <- (factorial(alpha)/(factorial(alpha-j)*factorial(j-k)*factorial(k)))*(((-1)^(j-k))/(T^j))*((9/(j+3))^2)
}
}
plot(x[1:7], F, type = "l", ylab = "", xlab = "")
Any help will be appreciated here because the approximation and the graph obtained using my codes is grossly different from the original curve.
It seems clear that your problem is in here.
F[x+1] <- (factorial(alpha)/(factorial(alpha-j)*factorial(j-k)*factorial(k)))*(((-1)^(j-k))/(T^j))*((9/(j+3))^2)
You are trying to get something varying in x, yes? So how can you get that, if the right hand side of this equation has nothing varying in x, while the left hand side has an assignment using non-integer indices?
alpha <- 30 #In the exemple you try to reproduce, they use an alpha of 30 if i understood correctly (i'm a paleontologist not a mathematician so this paper's way beyond my area of expertise :) )
tau <- 1 #tau is your T (i changed it to avoid confusion with TRUE)
x <- seq(0, tau, by = 0.001)
f<-rep(0,length(x)) #This is your F (same reason as above for the change).
#It has to be created as a vector of 0 before your loop since the whole idea of the loop is that you want to proceed by incrementation.
#You want a value of f for each of your element of x so here is your first loop:
for(i in 1:length(x)){
#Then you want the sum for all k going from 1 to alpha*x[i]/tau:
for(k in 1:floor(alpha*x[i]/tau)){
#And inside that sum, the sum for all j going from k to alpha:
for(j in k:alpha){
#This sum needs to be incremented (hence f[i] on both side)
f[i]<-f[i]+(factorial(alpha)/(factorial(alpha-j)*factorial(j-k)*factorial(k)))*(((-1)^(j-k))/(tau^j))*(9/(j+3)^2)
}
}
}
plot(x, f, type = "l", ylab = "", xlab = "")
I woud like to understand how levelplot works. I have almost no experience with plots and R.
What confuses me, is how should I interpret for example x~y*z ?
Lets assume I have a function, and I would like to show how often certain value occurs by using 3d plot. I would have hence x = x, y = f(x) and z = count. How to obtain such simple plot by using levelplot (or something else if it is not appriopriate).
In addition, should I group "count" myself - 3 columns in my data from, or just have 2 columns - x and f(x) and have duplications?
Hope my question is clear, I tried to read levelplot documentation, however I could not find any tutorial that teaches basics.
The following example is from the ?levelplot documentation.
The formula z~x*y means that z is a function of x, y and the interaction between x and y. Had the function been z~x+y it would have meant that z is a function of x and y, ignoring any interaction.
You can read more about the formula interface in the help for ?formula.
x <- seq(pi/4, 5 * pi, length.out = 100)
y <- seq(pi/4, 5 * pi, length.out = 100)
r <- as.vector(sqrt(outer(x^2, y^2, "+")))
grid <- expand.grid(x=x, y=y)
grid$z <- cos(r^2) * exp(-r/(pi^3))
levelplot(z~x*y, grid, cuts = 50, scales=list(log="e"), xlab="",
ylab="", main="Weird Function", sub="with log scales",
colorkey = FALSE, region = TRUE)