I have a regular grid in cylindrical co-ordinates (R, z, theta). At each grid point I have a value for the density at that point. I am looking for advice on how to interpolate the density values to get the value at a certain point within a 3D grid cell, using the values from the 8 grid points around it. Is there an R package that will make this simple?
I have tried to search for answers, but most seem to be related to the 'grid' plotting package.
e.g., some sample data:
R <- c(1,2,3)
z <- c(1,2,3)
th <- c(1,2,3)
dens <- array(rep(1, 3*3*3), dim=c(3, 3, 3))
dens[1,2,1] <- 2
How do I get the value of dens at (R,z,th) = (1.5,1.5,1.5) ?
Related
I need to create something like a spider chart in R without using any libraries. That’s my code for now. It creates a figure with points number equal to the length of vector ‘a’. However, I’d like each point to be at the distance from the coordinates center equal to a respective number in a vector, for example one point at a distance 1, another at 2, so on. Is it possible to do so?
a <- 1:6
angle <- seq(0, 2*pi, (2*pi)/length(a))
x <- cos(angle)
y <- sin(angle)
plot(x, y,
type = "l")
See ?stars:
a <- 1:6
stars(matrix(a, nrow=1), scale=FALSE)
For future reference, using R's built-in help search would have found this with ??spider
Suppose I have two datasets: (1) a data frame: coordinates of localities, each with ID; and (2) a linguistic distance matrix which reflects the linguistic distance between these localities.
# My data are similar to this structure
# dataframe
id <- c("A","B","C","D","E")
x_coor <- c(0.5,1,1,1.5,2)
y_coor <- c(5.5,3,7,6.5,5)
my.data <- data.frame(id = id, x_coor = x_coor, y_coor = y_coor)
# linguistic distance matrix
A B C D
B 308.298557
C 592.555483 284.256926
D 141.421356 449.719913 733.976839
E 591.141269 282.842712 1.414214 732.562625
Now, I want to visualize the linguistic distance between every two sites onto a map by the thickness or color of the line connect the adjacent localities in R.
Just like this:
enter image description here
My idea is to generate the delaunay triangulation by deldir or tripack package in R.
# generate delaunay triangulation
library(deldir)
de=deldir(my.data$x_coor,my.data$y_coor)
plot.deldir(de,wlines="triang",col='blue',wpoints = "real",cex = 0.1)
text(my.data$x_coor,my.data$y_coor,my.data$id)
this is the plot:
enter image description here
My question is how to reflect the linguistic distance by the thickness or color of the edges of triangles? Is there any other better method?
Thank you very much!
What you want to do in respect of the line widths can be done "fairly
easily" by the deldir package. You simply call plot.deldir() with the
appropriate value of "lw" (line width).
At the bottom of this answer is a demonstration script "demo.txt" which shows how to do this in the case of your example. In particular this script shows
how to obtain the appropriate value of lw from the "linguistic distance
matrix". I had to make some adjustments in the way this matrix was
presented. I.e. I had to convert it into a proper matrix.
I have rescaled the distances to lie between 0 and 10 to obtain the
corresponding values of the line widths. You might wish to rescale in a different manner.
In respect of colours, there are two issues:
(1) It is not at all clear how you would like to map the "linguistic
distances" to colours.
(2) Unfortunately the code for plot.deldir() is written in a very
kludgy way, whence the "col" argument to segments() cannot be
appropriately passed on in the same manner that the "lw" argument can.
(I wrote the plot.deldir() code a long while ago, when I knew far less about
R programming than I know now! :-))
I will adjust this code and submit a new version of deldir to CRAN
fairly soon.
#
# Demo script
#
# Present the linguistic distances in a useable way.
vldm <- c(308.298557,592.555483,284.256926,141.421356,449.719913,
733.976839,591.141269,282.842712,1.414214,732.562625)
ldm <- matrix(nrow=5,ncol=5)
ldm[row(ldm) > col(ldm)] <- vldm
ldm[row(ldm) <= col(ldm)] <- 0
ldm <- (ldm + t(ldm))/2
rownames(ldm) <- LETTERS[1:5]
colnames(ldm) <- LETTERS[1:5]
# Set up the example data. It makes life much simpler if
# you denote the "x" and "y" coordinates by "x" and "y"!!!
id <- c("A","B","C","D","E")
x_coor <- c(0.5,1,1,1.5,2)
y_coor <- c(5.5,3,7,6.5,5)
# Eschew nomenclature like "my.data". Such nomenclature
# is Micro$oft-ese and is an abomination!!!
demoDat <- data.frame(id = id, x = x_coor, y = y_coor)
# Form the triangulation/tessellation.
library(deldir)
dxy <- deldir(demoDat)
# Plot the triangulation with line widths proportional
# to "linguistic distances". Note that plot.deldir() is
# a *method* for plot, so you do not have to (and shouldn't)
# type the ".deldir" in the plotting command.
plot(dxy,col=0) # This, and plotting with "add=TRUE" below, is
# a kludge to dodge around spurious warnings.
ind <- as.matrix(dxy$delsgs[,c("ind1","ind2")])
lwv <- ldm[ind]
lwv <- 10*lwv/max(lwv)
plot(dxy,wlines="triang",col='grey',wpoints="none",
lw=10*lwv/max(lwv),add=TRUE)
with(demoDat,text(x,y,id,col="red",cex=1.5))
I'm working on some bioacoustical analysis and got stuck with an issue that I believe it can be worked out mathematically. I'll use an sound sample from seewavepackage:
library(seewave)
library(tuneR)
data(tico)
By storing a spectrogram (i.e. graphic representation of the sound wave tico) in an R object, we can now deal with the wave file computationally.
s <- spectro(tico, plot=F)
class(s)
>[1] "list"
length(s)
>[1] 3
The object created s consists in two numerical vectors x = s$time, y = s$freq representing the X and Y axis, respectively, and a matrix z = s$amp of amplitude values with the same dimensions of x and y. Z is a virtually a 3D matrix that can be plotted using persp3D (plot3D), plot_ly (plotly) or plot3d (rgl). Alternatively, the wave file can be plotted in 3D using seewave if one wishes to visualize it as an interative rgl plot.
spectro3D(tico)
That being said, the analysis I'm conducting aims to calculate contours of relative amplitude:
con <- contourLines(x=s$time, y=s$freq, z=t(s$amp), levels=seq(-25, -25, 1))
Select the longest contour:
n.con <- numeric(length(con))
for(i in 1:length(con)) n.con[i] <- length(con[[i]]$x)
n.max <- which.max(n.con)
con.max <- con[[n.max]]
And then plot the selected contour against the spectrogram of tico:
spectro(tico, grid=F, osc=F, scale=F)
polygon(x=con.max$x, y=con.max$y, lwd=2)
Now it comes the tricky part. I must find a way to "subset" the matrix of amplitude values s$amp using the coordinates of the longest contour con.max. What I aim to achieve is a new matrix containing only the amplitude values inside the polygon. The remaining parts of the spectrogram should then appear as blank spaces.
One approach I though it could work would be to create a loop that replaces every value outside the polygon for a given amplitude value (e.g. -25 dB). I once did an similar approach to remove the values below -30 dB and it worked out perfectly:
for(i in 1:length(s$amp)){if(s$amp[i] == -Inf |s$amp[i] <= -30)
{s$amp[i] <- -30}}
Another though would be to create a new matrix with the same dimensions of s$amp, subset s$amp using the coordinates of the contour, then replace the subset on the new matrix. Roughly:
mt <- matrix(-30, nrow=nrow(s$amp), ncol = ncol(s$amp))
sb <- s$amp[con.max$y, con.max$x]
new.mt <- c(mt, sb)
s$amp <- new.mt
I'll appreciate any help.
I want to create 50 concentric circles. I did it with python but now I want to do this in R. I have tried the symbols function but with no result. I want my circles to start from x,y coordinates and the radius of each circle to be 3times bigger than the previous.
step=1
for(i in seq(1,50,1)){
symbols (x, y, circles=50, col="grey")
step=step+3
}
From this I get one circle as a result.
I am new in programming so it is probably very simple. Should I use a specific package?
The beauty of R is that many things can be vectorized, including the imput to the 'symbols' function. Here's an example for you:
#vector of radii
#written in a way that's easily changable
n_circles <- 50
my_circles <- seq(1,by=1,length.out = n_circles)
#generate x and y
x <- rep(1,n_circles)
y <- rep(1, n_circles)
#plot
symbols(x,y,1:n_circles)
I am trying to make a surface plot for data that is in a very long list of x,y,z points. To do this, I am dividing the data into a grid of 10k squares and finding the max value of z within each square. From my understanding, each z value should be stored in a matrix where each element of the matrix corresponds to a square on the grid. Is there an easier way to do this than the code below? That last line is already pretty long and it is only one square.
x<-(sequence(101)-1)*max(eff$CFaR)/100
y<-(sequence(101)-1)*max(eff$EaR)/100
effmap<-matrix(ncol=length(x)-1, nrow=length(y)-1)
someMatrix <- max(eff$Cost[which(eff$EaR[which(eff$CFaR >= x[50] & eff$CFaR <x[51], arr.ind=TRUE)]>=y[20] & eff$EaR[which(eff$CFaR >= x[50] & eff$CFaR <x[51], arr.ind=TRUE)]< y[91])])
So this is my interpretation of what you are trying to accomplish...
df <- read.csv("effSample.csv") # downloaded from your link
df <- df[c("CFaR","EaR","Cost")] # remove unnecessary columns
df$x <- cut(df$CFaR,breaks=100,labels=FALSE) # establish bins: CFaR
df$y <- cut(df$EaR,breaks=100,labels=FALSE) # establish bins: EaR
df.max <- expand.grid(x=1:100,y=1:100) # template; 10,000 grid cells
# maximum cost in each grid cell - NOTE: most of the cells are *empty*
df.max <- merge(df.max,aggregate(Cost~x+y,df,max),all.x=TRUE)
z <- matrix(df.max$Cost,nr=100,nc=100) # Cost vector -> matrix
# colors based on z-value
palette <- rev(rainbow(20)) # palette of 20 colors
zlim <- range(z[!is.na(z)])
colors <- palette[19*(z-zlim[1])/diff(zlim) + 1]
# create the plot
library(rgl)
open3d(scale=c(1,1,10)) # CFaR and EaR range ~ 10 X Cost range
x.values <- min(df$CFaR)+(0:99)*diff(range(df$CFaR))/100
y.values <- min(df$EaR)+(0:99)*diff(range(df$EaR))/100
surface3d(x.values,y.values,z,col=colors)
axes3d()
title3d(xlab="CFaR",ylab="EaR",zlab="Cost")
The code above generates a rotatable 3D plot, so the image is just a screen shot. Notice how there are lots of "holes". This is (partially) because you provided only part of your data. However, it is important to realize that just because you imagine 10,000 grid cells (e.g., a 100 X 100 grid), does not mean that there will be data in every cell.