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I have models consisting of thousands of cylinders which are stored in a data frame with start-coordinates, end-coordinates, lengths and radii. I want to simulate the shading they would create in the real world, given a certain position of a light source. As a result, I would like to have a raster which contains the information whether the ground is shaded or not (on the xy-plane). Is there a way to do this in R? Even when handling several thousand cylinder objects at the same time?
Here a mockup for a single cylinder:
I usually draw my cylinders with the rgl package, but it would be also okay if I have to use another package. I figured I might be able to use the packages rayrender or raytracing, but I don't know to export the shaded ground from the view to an array or raster.
Edit: Code for creating & plotting some cylinders:
library(rgl)
# some cylinders
cylinder <- structure(list(radius = c(0.01, 0.01, 0.01, 0.02, 0.01, 0.01,
0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01,
0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01,
0.01, 0.01), length = c(0.07, 0.13, 0.08, 0.08, 0.1, 0.08, 0.09,
0.08, 0.07, 0.15, 0.02, 0.09, 0.12, 0.12, 0.08, 0.26, 0.1, 0.09,
0.08, 0.02, 0.12, 0.11, 0.08, 0.06, 0.06, 0.19, 0.05, 0.1, 0.09,
0.09), start_X = c(0.62, 0.61, 0.62, 0.63, 0.64, 0.65, 0.65,
0.63, 0.63, 0.63, 0.63, 0.64, 0.63, 0.69, 0.79, 0.81, 0.92, 0.97,
1.03, 1.07, 1.08, 1.15, 1.24, 1.3, 1.34, 0.61, 0.5, 0.47, 0.4,
0.37), start_Y = c(0.13, 0.11, 0.09, 0.09, 0.09, 0.08, 0.09,
0.08, 0.07, 0.08, 0.07, 0.07, 0.07, 0.05, 0.02, 0.04, 0.02, 0.01,
0.04, 0.05, 0.05, 0.1, 0.15, 0.19, 0.22, 0.07, 0.13, 0.16, 0.17,
0.26), start_Z = c(361, 361.07, 361.2, 361.29, 361.36, 361.46,
361.54, 361.62, 361.7, 361.77, 361.9, 361.92, 361.78, 361.88,
361.98, 362.04, 362.26, 362.35, 362.39, 362.46, 362.48, 362.56,
362.62, 362.65, 362.66, 361.76, 361.91, 361.95, 362.01, 362.07
), axis_X = c(-0.09, 0.05, 0.12, 0.14, 0.1, -0.03, -0.15, -0.07,
-0.07, -0.2, 0.52, -0.62, 0.43, 0.54, 0.16, 0.35, 0.53, 0.43,
0.76, 0.58, 0.63, 0.74, 0.66, 0.56, 0.79, -0.61, -0.65, -0.64,
-0.33, -0.7), axis_Y = c(-0.12, -0.09, -0.01, -0.08, -0.08, 0.01,
-0.11, -0.14, -0.04, -0.06, 0.06, 0.59, -0.14, -0.14, 0.38, -0.22,
0.1, 0, 0.14, 0.15, 0.47, 0.45, 0.46, 0.67, 0.48, 0.28, 0.2,
0, 0.55, 0.16), axis_Z = c(0.99, 0.99, 0.99, 0.99, 0.99, 1, 0.98,
0.99, 1, 0.98, 0.85, 0.53, 0.89, 0.83, 0.91, 0.91, 0.84, 0.9,
0.64, 0.8, 0.61, 0.5, 0.59, 0.48, -0.39, 0.74, 0.74, 0.77, 0.77,
0.69)), row.names = c(NA, 30L), class = "data.frame")
# calculate end points of cylinders
# (cylinders have starting coordinates, a length and a direction as unit vector)
cylinder$end_X = cylinder$start_X + cylinder$axis_X * cylinder$length
cylinder$end_Y = cylinder$start_Y + cylinder$axis_Y * cylinder$length
cylinder$end_Z = cylinder$start_Z + cylinder$axis_Z * cylinder$length
# prepare cylinders
cylinder_list <- lapply(1:nrow(cylinder), function(i) {
cyl <- cylinder3d(
center = cbind(
c(cylinder$start_X[i], cylinder$end_X[i]),
c(cylinder$start_Y[i], cylinder$end_Y[i]),
c(cylinder$start_Z[i], cylinder$end_Z[i])),
radius = cylinder$radius[i],
closed = -2)
cyl
})
# plot cylinders
open3d()
par3d(windowRect = c(50,50,650, 650))
shade3d(shapelist3d(cylinder_list, plot = FALSE), color = "blue")
I would like the light source to be a point source from infinite distance, as I would like to simulate sunlight. As the sun is so far away, I would just assume the light beams to be parallel to each other.
I believe this does what you want:
library(rgl)
# some cylinders
# ... deleted ...
# cylinder$start_Z <- cylinder$start_Z - 361.5
# calculate end points of cylinders
# (cylinders have starting coordinates, a length and a direction as unit vector)
cylinder$end_X = cylinder$start_X + cylinder$axis_X * cylinder$length
cylinder$end_Y = cylinder$start_Y + cylinder$axis_Y * cylinder$length
cylinder$end_Z = cylinder$start_Z + cylinder$axis_Z * cylinder$length
# prepare cylinders
cylinder_list <- lapply(1:nrow(cylinder), function(i) {
cyl <- cylinder3d(
center = cbind(
c(cylinder$start_X[i], cylinder$end_X[i]),
c(cylinder$start_Y[i], cylinder$end_Y[i]),
c(cylinder$start_Z[i], cylinder$end_Z[i])),
radius = cylinder$radius[i],
closed = -2)
cyl
})
cylinder_list <- shapelist3d(cylinder_list, plot = FALSE)
# plot cylinders
open3d()
#> glX
#> 1
par3d(windowRect = c(50,50,650, 650))
# shade3d(cylinder_list, color = "blue")
# Suppose Sun is in direction xyz
sun <- c(0,0,1) # straight up
# Get a matrix that projects sun to c(0,0,0),
# and leaves anything with z = 0 alone
M <- rbind( cbind( diag(2), c(-sun[1]/sun[3], -sun[2]/sun[3])), 0)
# Replot the shadows of the cylinders
shadows <- transform3d(cylinder_list, t(M))
shade3d(shadows, col = "gray", lit = FALSE)
# decorate3d() # Add axes
par3d(userMatrix = diag(4)) # Display flat.
lowlevel() # And show it in reprex
# Use snapshot3d() to get PNG file containing the final image
Created on 2022-07-12 by the reprex package (v2.0.1)
It should include your definition of the cylinders dataframe, but I left that out to make it easier to read.
If you want to show the cylinders as well as the shadows, uncomment the shade3d() call on cylinder_list. You'll probably also want to make the Z values smaller; I subtracted 361.5 to put them near zero. If you don't do this, the graph will be extremely long and thin, and you won't be able to see anything.
The idea of the code is to project the sun vector to c(0,0,0), while leaving anything with Z==0 alone. This flattens the cylinders into objects that fit in the Z == 0 plane. Setting the userMatrix to the identity matrix then displays the shadow. Use the snapshot3d() function to make this into a raster image (in a PNG file).
I have some data of x values called "norm". I want to plot a histogram and then plot the density function of a beta with parameters 3.5 and 3 onto the histogram. The main goal of this is to show that the beta fits the norm data. I also need the scale of the y axis to match for both the density and the histogram. I got a plot at one point but the density function was very low because the histogram counted so the y axis went to 30 in my case and obviously the density line was <1.
Here is my code:
x <- seq(0,1, len = 115)
db_trial = dbeta(x, 3.5, 3.0)
ggplot(data = norm)+
geom_line(aes(x,db_trial), col = "red", stat = "density")+
geom_histogram(aes(y = ...density...), bins = 10, alpha = .2, fill =
"green", col = "red")
Here is my data set norm which is just the presidents data set in R but divided by 100.
# dput(norm)
structure(list(approval_rate = c(0.87, 0.82, 0.75, 0.63, 0.5,
0.43, 0.32, 0.35, 0.6, 0.54, 0.55, 0.36, 0.39, 0.69, 0.57, 0.57,
0.51, 0.45, 0.37, 0.46, 0.39, 0.36, 0.24, 0.32, 0.23, 0.25, 0.32,
0.32, 0.59, 0.74, 0.75, 0.6, 0.71, 0.61, 0.71, 0.57, 0.71, 0.68,
0.79, 0.73, 0.76, 0.71, 0.67, 0.75, 0.79, 0.62, 0.63, 0.57, 0.6,
0.49, 0.48, 0.52, 0.57, 0.62, 0.61, 0.66, 0.71, 0.62, 0.61, 0.57,
0.72, 0.83, 0.71, 0.78, 0.79, 0.71, 0.62, 0.74, 0.76, 0.64, 0.62,
0.57, 0.8, 0.73, 0.69, 0.69, 0.71, 0.64, 0.69, 0.62, 0.63, 0.46,
0.56, 0.44, 0.44, 0.52, 0.38, 0.46, 0.36, 0.49, 0.35, 0.44, 0.59,
0.65, 0.65, 0.56, 0.66, 0.53, 0.61, 0.52, 0.51, 0.48, 0.54, 0.49,
0.49, 0.61, 0.68, 0.44, 0.4, 0.27, 0.28, 0.25, 0.24, 0.24, 0.01
)), .Names = "approval_rate", row.names = c(NA, -115L), class = "data.frame")
This returns an error "Stat_bin requires the following missing aesthetics: x". What am I doing wrong. I am a novice with ggplot2.
It's usually better to use stat_function for this type of thing. Note that I'm technically using an anonymous function that wraps dbeta, so you can adjust the height of the curve via multiplication.
g <- ggplot(data = norm, aes(x = approval_rate))+
geom_histogram() +
stat_function(fun = function(x) dbeta(x, shape1 = 3.5, shape2 = 3.0) * 5, color = 'red')
print(g)
Could anyone tell why running hist on matrix CHh gives error "Error in plot.window(xlim, ylim, "", ...) : need finite 'ylim' values"? If I eliminate the min function the error disappears. Yet I don't understand why that represents a problem. Thank you.
CFh <-structure(c(-0.64, 0.34, 0.65, 0.26, -0.64, 0.92, -0.64, -0.1, -0.41, -0.36, 0.16, 0.92, 1.43, -0.41, 0.65, 0.28, 0.47, 0.35, -0.54, 0.65, 0.28, -0.1, 0.92, -0.36, 0.25, 0.34, -0.34, 0.07, 0.65, 0, -0.04, 0.47, 0.78, 0.47, 1.43, -0.23, -0.41, 0.28, 0.62, 0.35, -0.34, -0.23, -0.36, 0.28, 0.26, 0.03, 0.28, 0.07, 0.47, 0.63, 0.35, 0.47, 0, -0.28, 0.34, 0.16, 0.62, -0.04, 0.03, -0.41, -0.34, -0.64, -0.32, -0.28, -0.04, -0.36, 0.34, 0.47, 0.63, 0.62, 0, -0.04, -0.23, 0.65, -0.04, 0.47, -0.64, 0, -0.34, 0.28, -0.1, -0.28, 0.35, -0.34, -0.04, 0.63, 0.92, 0.35, 0.25, 0.34, 0.25, 0.34, 0.16, -0.36, 0, 0.28, 0.28, -0.28, -0.34, -0.23, 0.78, -0.41, 0.65, -0.32, -0.54, -0.36, 0.92, 0.25, 0.47, -0.1, 0.78, -0.54, 0.63, 0.65, -0.28, 0.25, 0.07, 0.35, 0.62, -0.28, -0.36, -0.54, 0.47, 0.47, 1.43, 0.63, -0.28, 0.03, 0.92, 0.92), .Dim = c(26L, 5L))
Breaks <- c(max(CFh,1.0), 1.0, 0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1, 0,
-0.1, -0.2, -0.3, -0.4, -0.5, -0.6, -0.7, -0.8, -0.9, -1.0, min(CFh,-1.0))
h <- hist(CFh, plot=TRUE, breaks=Breaks)
You have repeated values in your Breaks vector. This causes a problem with binning. Make sure the values are unique
h <- hist(CFh, plot=TRUE, breaks=unique(Breaks))
Continuing on from the above solution,
using unique breaks,
you could also consider a simpler way to create breaks as a sequence
Breaks <- unique(c(max(CFh,1.0), seq(1, -1, by=-0.1), min(CFh,-1.0)))
h <- hist(CFh, plot=TRUE, breaks=Breaks)
The following code takes vector V1, and bootstraps 10000 times a randomized rnomal sample made out of V1, creating with the results a matrix with 10000 columns. It then creates a histogram for that matrix.
V1 <- c(0.18, 0.2, 0.24, 0.35, -0.22, -0.17, 0.28, -0.28, -0.14, 0.03, 0.87, -0.2, 0.06, -0.1, -0.72, 0.18, 0.01, 0.31, -0.36, 0.61, -0.16, -0.07, -0.13, 0.01, -0.09, 0.26, -0.14, 0.08, -0.62, -0.2, 0.3, -0.21, -0.11, 0.05, 0.06, -0.28, -0.27, 0.17, 0.42, -0.05, -0.15, 0.05, -0.07, -0.22, -0.34, 0.16, 0.34, 0.1, -0.12, 0.24, 0.45, 0.37, 0.61, 0.9, -0.25, 0.02)
xxx <- round(sapply(1:10000, function(i) rnorm(length(V1), mean=sample
(V1, length(V1), replace=TRUE), sd(V1))),2)
h <- hist(xxx, plot=T)
I would like to create a printout of its probability density function in matrix or table format, i.e. get a matrix with 1.0, 0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1, 0, -0,1 -0.2, -0.3, -0.4, -0.5, -0.6, -0.7, -0.8, -0.9, -1.0 as breaks, and the associated probability densities on the next column.
My problems are two. First, specifying the breaks I want fails. Secondly, making a matrix with h$breaks and h$density also fails. Any insights would be much appreciated. Thank you.
#This works, but I want to specify the breaks
h <- hist(xxx, plot=T, breaks=20)
#This gives error "some 'x' not counted; maybe 'breaks' do not span range of 'x'"
h <- hist(xxx, plot=T, breaks=c(1.0, 0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1, 0, -0,1 -0.2, -0.3, -0.4, -0.5, -0.6, -0.7, -0.8, -0.9, -1.0))
#This gives error number of rows of result is not a multiple of vector length
ddd<- cbind(h$breaks, h$density)
At first you have a type error, you confused “.” and “,”.
At second the breaks must span the whole range of observations like this:
h <- hist(xxx, plot=F, breaks=c(max(xxx),1.0, 0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1, 0, -0.1, -0.2, -0.3, -0.4, -0.5, -0.6, -0.7, -0.8, -0.9, -1.0,min(xxx)))
And at last, you want to pick “mids” instead of breaks:
ddd<- cbind(h$mids, h$density)
The following code takes vector V1 and creates one bootstrapped sample called BV1. I would like to run that i times and place all the BVi vectors in matrix MV. Using a function of the apply family rather than a for loop if possible.
V1 <- c(0.18, 0.2, 0.24, 0.35, -0.22, -0.17, 0.28, -0.28, -0.14, 0.03, 0.87, -0.2, 0.06, -0.1, -0.72, 0.18, 0.01, 0.31, -0.36, 0.61, -0.16, -0.07, -0.13, 0.01, -0.09, 0.26, -0.14, 0.08, -0.62, -0.2, 0.3, -0.21, -0.11, 0.05, 0.06, -0.28, -0.27, 0.17, 0.42, -0.05, -0.15, 0.05, -0.07, -0.22, -0.34, 0.16, 0.34, 0.1, -0.12, 0.24, 0.45, 0.37, 0.61, 0.9, -0.25, 0.02)
BV1 <- sample(V1, length(V1), replace=TRUE)
I will then use that matrix to calculate a distribution of the bootstrapped summary statistics. Thanks for your help.
We can use replicate to repeat the sample 'n' times and output as a matrix.
replicate(4, sample(V1, length(V1), replace=TRUE))
If we look at replicate
function (n, expr, simplify = "array")
sapply(integer(n), eval.parent(substitute(function(...) expr)),
simplify = simplify)
it uses sapply (so the OP's need for apply family of functions is covered)