Raytracing many objects and exporting shaded ground as a raster - r
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).
Related
Radar plot in R Shiny
I'm really new to R Shiny (starting playing with it today!), but this code isn't working for me... R keeps saying "the data must be given as dataframe." which, as far as I can tell, it is a dataframe (and it says it is when I check with is.data.frame). # Load packages ---- library(shiny) library(fmsb) # Load data ---- industry <- read.csv("data/industry.csv") # User interface ---- ui <- fluidPage( titlePanel("L&D Capabilities 2023"), sidebarLayout( sidebarPanel( helpText("Check which L&D capabilities your industry has in-house in 2023."), selectInput("var", label = "Choose a variable to display", choices = c("Central government", "Local government", "IT and Telecoms", "Professional services, law and accountancy", "Finance, banking and insurance", "Health", "Social care/housing association", "Other charity/voluntary sector", "Retail", "Engineering", "Manufacturing", "Pharmaceutical", "Transport", "Utilities", "Hospitality", "Education (HE, FE)", "Art, media and design", "Other", "Consulting"), selected = "Central government"), ), mainPanel(plotOutput("radarPlot")) ) ) # Server logic ---- server <- function(input, output) { output$radarPlot <- renderPlot({ data <- switch(input$var, "Central government" = industry$Centralgov, "Local government" = industry$Localgov, "IT and Telecoms" = industry$IT, "Professional services, law and accountancy" = industry$PS, "Finance, banking and insurance" = industry$Finance, "Health" = industry$Health, "Social care/housing association" = industry$Social, "Other charity/voluntary sector" = industry$Charity, "Retail" = industry$Retail, "Engineering" = industry$Engineering, "Manufacturing" = industry$Manufacturing, "Pharmaceutical" = industry$Pharmaceutical, "Transport" = industry$Transport, "Utilities" = industry$Utilities, "Hospitality" = industry$Hospitality, "Education (HE, FE)" = industry$Education, "Consulting" = industry$Consulting, "Art, media and design" = industry$Art, "Other" = counties$Other) radarchart(data) }) } # Run app ---- shinyApp(ui, server) Any ideas what's going on? Or what I'm missing? Many thanks! EDIT: here's my data > dput(industry) structure(list(Max = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), Min = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Centralgov = c(0.6, 0.18, 0.27, 0.27, 0.27, 0.27, 0.36, 0.3, 0.55, 0.45, 0.1, 0, 0.1, 0.27, 0.64, 0.09, 0.09, 0.18, 0.27, 0, 0.09, 0.18, 0.25, 0.29, 0.14), Localgov = c(0.36, 0.5, 0.36, 0.5, 0.42, 0.42, 0.09, 0.27, 0.36, 0.55, 0.3, 0.36, 0.55, 0.45, 0.73, 0.36, 0.18, 0.45, 0.64, 0.36, 0.27, 0.18, 0.3, 0.2, 0.6), IT = c(0.73, 0.33, 0.47, 0.51, 0.38, 0.18, 0.34, 0.38, 0.62, 0.41, 0.19, 0.38, 0.49, 0.41, 0.62, 0.32, 0.22, 0.38, 0.58, 0.51, 0.33, 0.34, 0.41, 0.15, 0.37 ), PS = c(0.73, 0.4, 0.56, 0.6, 0.48, 0.48, 0.29, 0.24, 0.63, 0.56, 0.29, 0.41, 0.27, 0.36, 0.71, 0.28, 0.16, 0.48, 0.4, 0.52, 0.36, 0.38, 0.29, 0.25, 0.13), Finance = c(0.9, 0.44, 0.66, 0.66, 0.61, 0.52, 0.44, 0.5, 0.86, 0.62, 0.32, 0.39, 0.48, 0.59, 0.86, 0.3, 0.27, 0.5, 0.52, 0.52, 0.57, 0.51, 0.56, 0.33, 0.29), Health = c(0.88, 0.33, 0.47, 0.65, 0.28, 0.37, 0.33, 0.29, 0.78, 0.47, 0.18, 0.13, 0.47, 0.5, 0.78, 0.26, 0.16, 0.41, 0.58, 0.5, 0.38, 0.39, 0.33, 0.13, 0.29), Social = c(0.7, 0.25, 0.5, 0.33, 0.3, 0.2, 0.1, 0.4, 0.5, 0.2, 0, 0.22, 0, 0.2, 0.4, 0.1, 0.3, 0.1, 0.3, 0.3, 0.33, 0.3, 0.33, 0, 0.11), Charity = c(0.8, 0.55, 0.62, 0.44, 0.5, 0.31, 0.08, 0.33, 0.58, 0.5, 0.4, 0.36, 0.33, 0.38, 0.82, 0.15, 0.08, 0.36, 0.22, 0.42, 0.2, 0.42, 0.18, 0.22, 0.11), Retail = c(0.62, 0.38, 0.46, 0.27, 0.25, 0.09, 0.08, 0.31, 0.82, 0.46, 0.25, 0.27, 0.25, 0.54, 0.69, 0.08, 0.17, 0.31, 0.67, 0.5, 0.33, 0.5, 0.38, 0.18, 0.08), Engineering = c(0.6, 0, 0.4, 0.25, 0.17, 0.17, 0, 0, 0.33, 0.5, 0.25, 0.33, 0.6, 0.17, 0.33, 0, 0, 0.33, 0.33, 0.17, 0.17, 0.5, 0.2, 0, 0), Manufacturing = c(0.56, 0.22, 0.35, 0.42, 0.42, 0.4, 0.24, 0.2, 0.56, 0.41, 0.24, 0.11, 0.21, 0.3, 0.63, 0.1, 0, 0.25, 0.42, 0.58, 0.21, 0.35, 0.25, 0.33, 0.06), Pharmaceutical = c(0.43, 0.25, 0, 0.71, 0.63, 0.25, 0.13, 0.13, 0.63, 0.43, 0, 0, 0, 0, 0.38, 0.25, 0.13, 0.38, 0.38, 0.5, 0, 0, 0.33, 0, 0.17), Transport = c(0.77, 0.62, 0.79, 0.57, 0.71, 0.64, 0.14, 0.5, 0.79, 0.46, 0.38, 0.21, 0.36, 0.38, 0.64, 0.43, 0.29, 0.21, 0.57, 0.64, 0.29, 0.54, 0.57, 0.36, 0.15), Utilities = c(1, 0.6, 0.4, 0.33, 0.2, 0.2, 0.6, 0.6, 0.8, 0.6, 0.25, 0.2, 0.8, 0.4, 1, 0.4, 0.4, 0.6, 0.4, 0.6, 0.2, 0.2, 0.2, 0.2, 0), Hospitality = c(0.67, 0, 0.67, 0.4, 0.67, 0.33, 0.33, 0.83, 0.83, 0.2, 0.67, 0.17, 0.2, 0.33, 0.83, 0.33, 0.33, 0, 0.67, 1, 0.5, 0.33, 0.33, 0.6, 0.33), Education = c(0.87, 0.33, 0.47, 0.53, 0.41, 0.38, 0.5, 0.47, 0.65, 0.41, 0.2, 0.31, 0.47, 0.65, 0.53, 0.24, 0.29, 0.38, 0.56, 0.41, 0.31, 0.19, 0.38, 0.27, 0.35), Consulting = c(0.67, 0.5, 0.67, 1, 0.33, 0.33, 0.17, 0.5, 1, 0.6, 0.33, 0.6, 0.4, 0.67, 0.5, 0.17, 0.17, 0.4, 0.6, 0.5, 0.5, 0.33, 0.4, 0.25, 0.25), Art = c(1, 0.2, 0.6, 0.5, 0.4, 0.2, 0.2, 0.6, 0.6, 0.4, 0.2, 0.2, 0.6, 0.5, 0.5, 0.2, 0.2, 0.2, 0.2, 0.6, 0.4, 0.4, 0.4, 0.2, 0.25), Other = c(0.67, 0.57, 0.71, 0.29, 0.57, 0.43, 0.14, 0.5, 0.67, 0.29, 0.57, 0.29, 0.43, 0.57, 0.71, 0.29, 0.43, 0.29, 0.43, 0.57, 0.71, 0.43, 0.5, 0.6, 0.4)), class = "data.frame", row.names = c("In-person classroom delivery", "Strategy and governance", "Stakeholder engagement", "Instructional design", "Crafting learning journeys / blended solutions", "Supporting ongoing workplace performance", "Facilitating social and collaborative learning", "Understanding learner behaviour", "Virtual classroom / webinar delivery", "Digital content development", "Performance consulting", "Business acumen", "Marketing and communications", "Coaching and mentoring", "Learning management / administration", "Analytics / data management", "Evaluating impact", "Technology/infrastructure", "Project management", "Leveraging L&D expertise", "Knowledge management", "Negotiation, persuasion, and influence", "Learning experience design", "Community engagement", "Research capabilities"))
Like I mentioned in my comment, radarchart requires at least three variables. In this solution, I set the plot to only render when there are at least three options selected. I've changed the creation of the dropdown to allow multiple selections (so that three can be selected. ui <- fluidPage( titlePanel("Title"), sidebarLayout(sidebarPanel( helpText("help goes here"), selectInput( "var", label = "Choose at least three options to plot", multiple = T, choices = setNames( names(industry)[3:21], c("Central government", "Local government", "IT and Telecoms", "Professional services, law and accountancy", "Finance, banking and insurance", "Health", "Social care/housing association", "Other charity/voluntary sector", "Retail", "Engineering","Manufacturing", "Pharmaceutical", "Transport", "Utilities", "Hospitality", "Education (HE, FE)", "Consulting", "Art, media and design", "Other")), selected = "Central government") ), mainPanel(plotOutput("radarPlot"))) ) server <- function(input, output, session) { output$radarPlot <- renderPlot({ if(length(input$var) >= 3) { radarchart(industry[, input$var]) } }) } # Run app ---- shinyApp(ui, server) However, this is not technically accurate since the first row is supposed to be the max value. The second row is supposed to be the min value. (Which is how your data frame is orientated if you transpose it.) Here's another version of the server. In this version, I've modified the data so that the first and second rows are the max and min. # add max and min to the top of every column industry2 <- t(industry) %>% as.data.frame() %>% mutate(max = 1, min = 0) %>% select(max, min, everything()) %>% t() %>% as.data.frame() server2 <- function(input, output, session) { output$radarPlot <- renderPlot({ if(length(input$var) >= 3) { radarchart(industry[, input$var]) } }) } # Run app ---- shinyApp(ui, server2) This renders an identical plot to what I pictured for the first version (so I am not sure why the radarchart function calls for this information.) I'm not sure how attached you are to fmsb::radarchart, but there are a lot of other options: ggplot2, plotly, highcharter, echarts, etc. Here's an alternative using the Plotly library. In this option, I've set multiple in selectInput to false (based on what you were originally trying to plot). I've set the range so the graphs are comparable. A few other changes are annotated in the comments in the code. library(plotly) library(shiny) ui2 <- fluidPage( titlePanel("With Plotly"), sidebarLayout(sidebarPanel( helpText("help goes here"), selectInput( "var", label = "Choose a field to plot", multiple = F, # <<- FALSE now (can remove argument, default is false) choices = setNames( names(industry)[3:21], c("Central government", "Local government", "IT and Telecoms", "Professional services, law and accountancy", "Finance, banking and insurance", "Health", "Social care/housing association", "Other charity/voluntary sector", "Retail", "Engineering","Manufacturing", "Pharmaceutical", "Transport", "Utilities", "Hospitality", "Education (HE, FE)", "Consulting", "Art, media and design", "Other")), selected = "Central government") ), mainPanel(plotlyOutput("radarPlot"))) # <<-- plotlyOutput instead of plotOutput ) server3 <- function(input, output, session) { output$radarPlot <- renderPlotly({ # <- render plotly instead of plot plot_ly(r = industry[, input$var], theta = rownames(industry), fill = "toself", type = "scatterpolar", mode = "markers") %>% layout( polar = list( angularaxis = list(showticklabels = F), radialaxis = list(range = c(0, 1)) )) }) } # Run app ---- shinyApp(ui2, server3) I hid the labels, because they overlapped. However, when you hover, you can see the row names.
Calculate derivative of Cumulative Distribution (CDF) to get Probability Density (PDF)
The following code calculates the Cumulative Distribution function (CDF) for vector VP. I would like to use the CDF to get the Probability Density function (PDF). In other words, I need to calculate the derivative of CDF. How can I do that in R? VP <- c(0.36, 0.3, 0.36, 0.47, 0, 0.05, 0.4, 0, 0, 0.15, 0.89, 0.03, 0.45, 0.21, 0, 0.18, 0.04, 0.53, 0, 0.68, 0.06, 0.09, 0.58, 0.03, 0.23, 0.27, 0, 0.12, 0.12, 0, 0.32, 0.07, 0.04, 0.07, 0.39, 0, 0.25, 0.28, 0.42, 0.55, 0.04, 0.07, 0.18, 0.17, 0.06, 0.39, 0.65, 0.15, 0.1, 0.32, 0.52, 0.55, 0.71, 0.93, 0, 0.36) set.seed(0) CF <- round(sapply(1:1000, function(i) sample(VP, length(VP), replace=TRUE)),2) Breaks <- c(max(CF,1.0), 1.0, 0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1, 0) CDF <- round(sapply(Breaks, function(b) sum(CF<=b)/length(CF)),2)
diff is the discrete difference operator, so I think you're looking for diff(CDF)/diff(Breaks) note that this vector will be one shorter than the original CDF and Breaks vectors you might have to do something about reversing your CDF and Breaks vectors to get sensible results ...
You can also try the empirical cdf function: CDF <- ecdf(VP) and the histogram function can also provide a sample density function PDF <- hist(VP, freq=F) Have a look at PDF$counts and PDF$breaks.
Error with hist function (need finite 'ylim' values) (breaks related)
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)
Make table/matrix of probability densities and associated 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)
Execute function multiple times and create matrix with results
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)