How do I add subscripts to labels in ggplot? - r
I'm doing an analysis on air pollutants using Bayesian Kernel Machine Regression, using the bkmr package in R.
https://jenfb.github.io/bkmr/overview.html
The link is to Jennifer Bobb's instructions on how to use this package. I don't think it is relevant to the issue though. What I want to do is have PM2.5, O3, and NO2 show up in my charts with the 2.5, 3, and 2 as subscripts. I'm trying to use this function and getting no luck:
colnames(dat) <- c("LTE4", "$O[3]", "$PM[2.5]", "$NO[2]", "Diethyl", "Dimethyl", "age", "tmpf", "relh", "sex", "agany", "agself", "asthma")
When I do this what happens I just see these labels show up in the plots with with the $ and [] instead of subscripted numbers. Any ideas?
This is the full code I am using:
### January BKMR Analysis ###
## Hierarchical Variable Selection ##
## Updated June 6, 2022 ##
# Reading in necessary packages
library(tidyverse)
library(bkmr)
trio_semipro <- readRDS("C:/Users/Matt/OneDrive/Documents/Fresno Thesis/Thesis Code/trio_semipro.rds")
trio_semipro
dim(trio_semipro)
head(trio_semipro)
trio_semipro$log_lte4 <- log(trio_semipro$Final)
# Separating out dataframes for winter and summer to run separate models for each season
trio_semipro_w <- trio_semipro %>%
filter(visit_month == 1)
trio_semipro_s <- trio_semipro %>%
filter(visit_month == 2)
# Summer and Winter Dataframes
trio_semipro_w
trio_semipro_s
head(trio_semipro_w)
#view(trio_semipro_w)
dat = cbind(trio_semipro_w$log_lte4, trio_semipro_w$O3,
trio_semipro_w$PM25, trio_semipro_w$NO2, trio_semipro_w$diethyl, trio_semipro_w$dimethyl,
trio_semipro_w$age, trio_semipro_w$tmpf, trio_semipro_w$relh, trio_semipro_w$sex, trio_semipro_w$agriculture_anyone,
trio_semipro_w$agriculture_self, trio_semipro_w$asthma)
head(dat)
colnames(dat) = c("LTE4", "$O[3]", "$PM[2.5]", "$NO[2]", "Diethyl", "Dimethyl", "age", "tmpf", "relh", "sex", "agany", "agself", "asthma")
dat = as.data.frame(dat)
dat$sex
# recode the binary variable to be 0, 1 and NA
dat$agself = dat$agself-1
dat$agself[which(dat$agself==2)]=NA
dat$agself
# recode sex variable
dat$sex = dat$sex -1
# recode agany variable
dat$agany = dat$agany - 1
dat$agany[which(dat$agany==2)]=NA
#recode asthma variable
dat$asthma = dat$asthma - 1
dat$asthma[which(dat$asthma==2)]=NA
dat$asthma
dat$sex
dat$agany
# good
head(dat)
complete_dat = dat[-which(apply(dat, 1, anyNA)),]
dim(complete_dat)
# Fit BKMR
zscaled <- apply(complete_dat[,(2:6)], 2, scale)
yscaled <- scale(complete_dat$lte4)
xscaled <- cbind(scale(complete_dat[,7:9]), complete_dat[,10:13])
fit_bkmr = kmbayes(y=yscaled, Z= zscaled, X = xscaled,
iter = 20000, varsel = TRUE, groups=c(1,1,1,2,2), verbose=FALSE)
plot(fit_bkmr$sigsq.eps, type = "l")
TracePlot(fit = fit_bkmr, par = "beta", comp = 4)
TracePlot(fit = fit_bkmr, par = "sigsq.eps")
TracePlot(fit = fit_bkmr, par = "r", comp = 1)
# Estimating posterior inclusion probabilities
ExtractPIPs(fit_bkmr)
# Estimating h
y <- yscaled
Z <- zscaled
X <- xscaled
med_vals <- apply(Z, 2, median)
Znew <- matrix(med_vals, nrow = 1)
# Summarize model output
pred.resp.univar <- PredictorResponseUnivar(fit = fit_bkmr)
library(ggplot2) # Using ggplot to plot cross sections of h
ggplot(pred.resp.univar, aes(z, est, ymin = est - 1.96*se, ymax = est + 1.96*se)) +
geom_smooth(stat = "identity") +
geom_hline(yintercept = 0, lty = 5, col = "red2", alpha = 0.4) +
facet_wrap(~ variable, nrow = 1) +
ylab("h(z)")
# visualze the bivarate exposure-response function for two predictors, where
# all of the other predictors are fixed at a particular percentile.
pred.resp.bivar <- PredictorResponseBivar(fit = fit_bkmr, min.plot.dist = 1)
ggplot(pred.resp.bivar, aes(z1, z2, fill = est)) +
geom_raster() +
facet_grid(variable2 ~ variable1) +
scale_fill_gradientn(colours=c("#0000FFFF","#FFFFFFFF","#FF0000FF")) +
xlab("expos1") +
ylab("expos2") +
ggtitle("h(expos1, expos2)")
Related
How to select appropriate sin() terms to fit a time series using R
I want to fit a time series with sin() function because it has a form of some periods (crests and troughs). However, for now I only guessed it, e.g., 1 month, two months, ..., 1 year, 2 year. Is there some function in R to estimate the multiple periods in a data series? Below is an example which I want to fit it using the combination of sin() functions. The expression in lm() is a try after several guesses (red line in the Figure below). How can I find the sin() terms with appropriate periods? t <- 1:365 y <- c(-1,-1.3,-1.6,-1.8,-2.1,-2.3,-2.5,-2.7,-2.9,-3,-2,-1.1,-0.3,0.5,1.1,1.6,2.1,2.5,2.8,3.1,3.4,3.7,4.2,4.6,5,5.3,5.7,5.9,6.2,5.8,5.4,5,4.6,4.2,3.9,3.6,3.4,3.1,2.9,2.8,2.6,2.5,2.3,1.9,1.5,1.1,0.8,0.5,0.2,0,-0.1,-0.3,-0.4,-0.5,-0.5,-0.6,-0.7,-0.8,-0.9,-0.8,-0.6,-0.3,-0.1,0.1,0.4,0.6,0.9,1.1,1.3,1.5,1.7,2.1,2.4,2.7,3,3.3,3.5,3.8,4.3,4.7,5.1,5.5,5.9,6.2,6.4,6.6,6.7,6.8,6.8,6.9,7,6.9,6.8,6.7, 6.5,6.4,6.4,6.3,6.2,6,5.9,5.7,5.6,5.5,5.4,5.4,5.1,4.9,4.8,4.6,4.5,4.4,4.3,3.9,3.6,3.3,3,2.8,2.6,2.4,2.6,2.5,2.4,2.3,2.3,2.2,2.2,2.3,2.4,2.4,2.5,2.5,2.6,2.6,2.4,2.1,1.9,1.8,1.6,1.4,1.3,1,0.7,0.5,0.2,0,-0.2,-0.4,-0.2,-0.1,0.1,0.1,0.1,0.1,0.1,0.1,0,0,-0.1,-0.1,-0.2,-0.2,-0.3,-0.3,-0.4,-0.5,-0.5,-0.6,-0.7,-0.7,-0.8,-0.8,-0.8,-0.9,-0.9,-0.9,-1.3,-1.6,-1.9,-2.1,-2.3,-2.6,-2.9,-2.9,-2.9,-2.9, -2.9,-3,-3,-3,-2.8,-2.7,-2.5,-2.4,-2.3,-2.2,-2.1,-2,-2,-1.9,-1.9,-1.8,-1.8,-1.8,-1.9,-1.9,-2,-2.1,-2.2,-2.2,-2.3,-2.4,-2.5,-2.6,-2.7,-2.8,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.8,-2.8,-2.7,-2.7,-2.6,-2.6,-2.8,-3,-3.1,-3.3,-3.4,-3.5,-3.6,-3.5,-3.4,-3.3,-3.3,-3.2,-3,-2.9,-2.8,-2.8,-2.7,-2.6,-2.6,-2.6,-2.5,-2.6,-2.7,-2.8,-2.8,-2.9,-3,-3,-3,-3,-2.9,-2.9,-2.9,-2.9,-2.9,-2.8, -2.7,-2.6,-2.5,-2.4,-2.3,-2.3,-2.1,-1.9,-1.8,-1.7,-1.5,-1.4,-1.3,-1.5,-1.7,-1.8,-1.9,-2,-2.1,-2.2,-2.4,-2.5,-2.6,-2.7,-2.8,-2.8,-2.9,-3.1,-3.2,-3.3,-3.4,-3.5,-3.5,-3.6,-3.6,-3.5,-3.4,-3.3,-3.2,-3.1,-3,-2.7,-2.3,-2,-1.8,-1.5,-1.3,-1.1,-0.9,-0.7,-0.6,-0.5,-0.3,-0.2,-0.1,-0.3,-0.5,-0.6,-0.7,-0.8,-0.9,-1,-1.1,-1.1,-1.2,-1.2,-1.2,-1.2,-1.2,-0.8,-0.4,-0.1,0.2,0.5,0.8,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0.6,0.3,0,-0.2,-0.5,-0.7,-0.8) dt <- data.frame(t = t, y = y) plot(x = dt$t, y = dt$y) lm <- lm(y ~ sin(2*3.1416/365*t)+cos(2*3.1416/365*t)+ sin(2*2*3.1416/365*t)+cos(2*2*3.1416/365*t)+ sin(2*4*3.1416/365*t)+cos(2*4*3.1416/365*t)+ sin(2*5*3.1416/365*t)+cos(2*5*3.1416/365*t)+ sin(2*6*3.1416/365*t)+cos(2*6*3.1416/365*t)+ sin(2*0.5*3.1416/365*t)+cos(2*0.5*3.1416/365*t), data = dt) summary(lm)$adj.r.squared plot(dt$y); lines(predict(lm), type = "l", col = "red")
Package forecast has the fourier function (see here), which allows you to model fourier series terms based on time series objects. For example: library(forecast) dt$y <- ts(dt$y, frequency = 365) lm<- lm(y ~ fourier(y, K=6), dt) plot(dt$t, dt$y); lines(predict(lm), type = "l", col = "red")
Following my comment to the question, In catastrophic-failure's answer replace Mod by Re as in SleuthEye's answer. Then call nff(y, 20, col = "red"). I realized that there is another correction to function nff to be made: substitute length(x) or xlen for the magical number 73. Here is the function corrected. nff = function(x = NULL, n = NULL, up = 10L, plot = TRUE, add = FALSE, main = NULL, ...){ #The direct transformation #The first frequency is DC, the rest are duplicated dff = fft(x) #The time xlen <- length(x) t = seq_along(x) #Upsampled time nt = seq(from = 1L, to = xlen + 1L - 1/up, by = 1/up) #New spectrum ndff = array(data = 0, dim = c(length(nt), 1L)) ndff[1] = dff[1] #Always, it's the DC component if(n != 0){ ndff[2:(n+1)] <- dff[2:(n+1)] #The positive frequencies always come first #The negative ones are trickier ndff[(length(ndff) - n + 1):length(ndff)] <- dff[(xlen - n + 1L):xlen] } #The inverses indff = fft(ndff/xlen, inverse = TRUE) idff = fft(dff/xlen, inverse = TRUE) if(plot){ if(!add){ plot(x = t, y = x, pch = 16L, xlab = "Time", ylab = "Measurement", main = ifelse(is.null(main), paste(n, "harmonics"), main)) lines(y = Re(idff), x = t, col = adjustcolor(1L, alpha = 0.5)) } lines(y = Re(indff), x = nt, ...) } ret = data.frame(time = nt, y = Mod(indff)) return(ret) } y <- c(-1,-1.3,-1.6,-1.8,-2.1,-2.3,-2.5,-2.7,-2.9,-3,-2,-1.1,-0.3,0.5,1.1,1.6,2.1,2.5,2.8,3.1,3.4,3.7,4.2,4.6,5,5.3,5.7,5.9,6.2,5.8,5.4,5,4.6,4.2,3.9,3.6,3.4,3.1,2.9,2.8,2.6,2.5,2.3,1.9,1.5,1.1,0.8,0.5,0.2,0,-0.1,-0.3,-0.4,-0.5,-0.5,-0.6,-0.7,-0.8,-0.9,-0.8,-0.6,-0.3,-0.1,0.1,0.4,0.6,0.9,1.1,1.3,1.5,1.7,2.1,2.4,2.7,3,3.3,3.5,3.8,4.3,4.7,5.1,5.5,5.9,6.2,6.4,6.6,6.7,6.8,6.8,6.9,7,6.9,6.8,6.7, 6.5,6.4,6.4,6.3,6.2,6,5.9,5.7,5.6,5.5,5.4,5.4,5.1,4.9,4.8,4.6,4.5,4.4,4.3,3.9,3.6,3.3,3,2.8,2.6,2.4,2.6,2.5,2.4,2.3,2.3,2.2,2.2,2.3,2.4,2.4,2.5,2.5,2.6,2.6,2.4,2.1,1.9,1.8,1.6,1.4,1.3,1,0.7,0.5,0.2,0,-0.2,-0.4,-0.2,-0.1,0.1,0.1,0.1,0.1,0.1,0.1,0,0,-0.1,-0.1,-0.2,-0.2,-0.3,-0.3,-0.4,-0.5,-0.5,-0.6,-0.7,-0.7,-0.8,-0.8,-0.8,-0.9,-0.9,-0.9,-1.3,-1.6,-1.9,-2.1,-2.3,-2.6,-2.9,-2.9,-2.9,-2.9, -2.9,-3,-3,-3,-2.8,-2.7,-2.5,-2.4,-2.3,-2.2,-2.1,-2,-2,-1.9,-1.9,-1.8,-1.8,-1.8,-1.9,-1.9,-2,-2.1,-2.2,-2.2,-2.3,-2.4,-2.5,-2.6,-2.7,-2.8,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.9,-2.8,-2.8,-2.7,-2.7,-2.6,-2.6,-2.8,-3,-3.1,-3.3,-3.4,-3.5,-3.6,-3.5,-3.4,-3.3,-3.3,-3.2,-3,-2.9,-2.8,-2.8,-2.7,-2.6,-2.6,-2.6,-2.5,-2.6,-2.7,-2.8,-2.8,-2.9,-3,-3,-3,-3,-2.9,-2.9,-2.9,-2.9,-2.9,-2.8, -2.7,-2.6,-2.5,-2.4,-2.3,-2.3,-2.1,-1.9,-1.8,-1.7,-1.5,-1.4,-1.3,-1.5,-1.7,-1.8,-1.9,-2,-2.1,-2.2,-2.4,-2.5,-2.6,-2.7,-2.8,-2.8,-2.9,-3.1,-3.2,-3.3,-3.4,-3.5,-3.5,-3.6,-3.6,-3.5,-3.4,-3.3,-3.2,-3.1,-3,-2.7,-2.3,-2,-1.8,-1.5,-1.3,-1.1,-0.9,-0.7,-0.6,-0.5,-0.3,-0.2,-0.1,-0.3,-0.5,-0.6,-0.7,-0.8,-0.9,-1,-1.1,-1.1,-1.2,-1.2,-1.2,-1.2,-1.2,-0.8,-0.4,-0.1,0.2,0.5,0.8,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0.6,0.3,0,-0.2,-0.5,-0.7,-0.8) res <- nff(y, 20, col = "red") str(res) #> 'data.frame': 3650 obs. of 2 variables: #> $ time: num 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 ... #> $ y : num 1.27 1.31 1.34 1.37 1.4 ... Created on 2022-10-17 with reprex v2.0.2
The functions sinusoid and mvrm from package BNSP allow one to specify the number of harmonics and if that number is too high, the algorithm can remove some of the unnecessary terms and avoid overfitting. # Specify the model model <- y ~ sinusoid(t, harmonics = 20, amplitude = 1, period = 365) # Fit the model m1 <- mvrm(formula = model, data = dt, sweeps = 5000, burn = 3000, thin = 2, seed = 1, StorageDir = getwd()) # ggplot plotOptionsM <- list(geom_point(data = dt, aes(x = t, y = y))) plot(x = m1, term = 1, plotOptions = plotOptionsM, intercept = TRUE, quantiles = c(0.005, 0.995), grid = 100) In this particular example, among the 20 harmonics, the 19 appear to be important.
Monte Carlo simulations for VAR models
I've been trying to estimate VAR models using Monte Carlo Simulation. I have 3 endogenous variables. I need some guidance regarding this. First of all, I want to add an outlier as a percentage of the sample size. Second (second simulation for same model), I want to add multivariate contaminated normal distribution like 0.9N (0, I) + 0.1((0,0,0)',(100, 100, 100)) instead of outlier. Could you tell me how to do these? Thank you. RR <- function(n, out){ # n is number of observations k <- 3 # Number of endogenous variables p <- 2 # Number of lags # add outlier n[1]<- n[1]+out # Generate coefficient matrices B1 <- matrix(c(.1, .3, .4, .1, -.2, -.3, .03, .1, .1), k) # Coefficient matrix of lag 1 B2 <- matrix(c(0, .2, .1, .07, -.4, -.1, .5, 0, -.1), k) # Coefficient matrix of lag 2 M <- cbind(B1, B2) # Companion form of the coefficient matrices # Generate series DT <- matrix(0, k, n + 2*p) # Raw series with zeros for (i in (p + 1):(n + 2*p)){ # Generate series with e ~ N(0,1) DT[, i] <- B1%*%DT[, i-1] + B2%*%DT[, i-2] + rnorm(k, 0, 1) } DT <- ts(t(DT[, -(1:p)])) # Convert to time series format #names <- c("V1", "V2", "V3") # Rename variables colnames(DT) <- c("Y1", "Y2", "Y3") #plot.ts(DT) # Plot the series # estimate VECM vecm1 <- VECM(DT, lag = 2, r = 2, include = "const", estim ="ML") vecm2 <- VECM(DT, lag = 2, r = 1, include = "const", estim ="ML") # mse mse1 <- mean(vecm1$residuals^2) mse2 <- mean(vecm2$residuals^2) #param_list <- unname(param_list) return(list("mse1" = mse1, "mse2" = mse2, "mse3" = mse3)) } # defined the parameter grids(define the parameters ranges we want to run our function with) n_grid = c(50, 80, 200, 400) out_grid = c(0 ,5, 10) # collect parameter grids in a list (to enter it into the Monte Carlo function) prml = list("n" = n_grid, "out" = out_grid) # run simulation RRS <- MonteCarlo(func = RR, nrep = 1000, param_list = prml) summary(RRS) # make table: rows = "n" cols = "out" MakeTable(output = RRS, rows = rows, cols = cols)
How can I improve the quality/graphics of my R plot for a Naive Bayes classifier visual
I tried a Naive Bayes classifier to see if I can predict if a person, given their age and estimated salary, would purchase a particular vehicle or not. The plot I got in the visualisation section looks not very smooth and clean, with white lines running across my plot. I'm assuiming the graphics/resolution is the problem but I am not sure. This is a snippet of what the dataset looks like Age EstimatedSalary Purchased 19 19000 0 35 20000 0 26 43000 0 27 57000 0 19 76000 0 27 58000 0 Here is the code # Loading the data set data <- read.csv(" *A csv sheet on people's age, salaries and whether or not they will purchase a certain vehicle* ") data <- data[, 3:5] attach(data) # Encoding the dependent variable data$Purchased <- factor(data$Purchased, levels = c(0, 1)) attach(data) # Splitting the dataset library(caTools) set.seed(404) split <- sample.split(Purchased, SplitRatio = 0.75) train_set <- subset(data, split == T) test_set <- subset(data, split == F) # Feature scaling train_set[-3] <- scale(train_set[-3]) test_set[-3] <- scale(test_set[-3]) # Training the model library(e1071) classifier <- naiveBayes(x = train_set[-3], y = train_set$Purchased) # Predicting test results y_pred <- predict(classifier, newdata = test_set[-3]) # Construct the confusion matrix (cm <- table(test_set[, 3], y_pred)) Below is the code that I used to visualise the results # Visualising the results library(ElemStatLearn) set <- test_set x1 <- seq(min(set[, 1]) - 1, max(set[, 1]) + 1, by = 0.01) x2 <- seq(min(set[, 2]) - 1, max(set[, 2]) + 1, by = 0.01) grid_set <- expand.grid(x1, x2) colnames(grid_set) <- c("Age", "EstimatedSalary") y_grid <- predict(classifier, newdata = grid_set) plot(set[, -3], main = "Naive Bayes: Test set", xlab = "Age", ylab = "EstimatedSalary", xlim = range(x1), ylim = range(x2)) contour(x1, x2, matrix(as.numeric(y_grid), length(x1), length(x2)), add = T) points(grid_set, pch = ".", col = ifelse(y_grid == 1, "Springgreen3", "tomato")) points(set, pch = 21, bg = ifelse(set[, 3] == 1, "green4", "red3")) Naive Bayes classifier plot on the test set predictions Would like to know the reason for the white lines running up and down the plot and why it does not look smooth?
So I figured out what was giving me the weird lines and the low quality resolution. Adding the "cex = n" parameter to the "points()" function in the graph with n = 5 solved this. Revised block of code set <- test_set x1 <- seq(min(set[, 1]) - 1, max(set[, 1]) + 1, by = 0.01) x2 <- seq(min(set[, 2]) - 1, max(set[, 2]) + 1, by = 0.01) grid_set <- expand.grid(x1, x2) colnames(grid_set) <- c("Age", "EstimatedSalary") y_grid <- predict(classifier, newdata = grid_set) plot(set[, -3], main = "Naive Bayes: Test set", xlab = "Age", ylab = "EstimatedSalary", xlim = range(x1), ylim = range(x2)) contour(x1, x2, matrix(as.numeric(y_grid), length(x1), length(x2)), add = T) points(grid_set, pch = ".", col = ifelse(y_grid == 1, "Springgreen3", "tomato"), cex = 5) points(set, pch = 21, bg = ifelse(set[, 3] == 1, "green4", "red3")) The revised line of code in the above block points(grid_set, pch = ".", col = ifelse(y_grid == 1, "Springgreen3", "tomato"), cex = 5) However the case, I would still like to know the reason behind how this happened because the explanation available in R about the functions and the parameters were not that clear to me. Would appreciate any help given!
How to extract the Prediction Intervals of a Gaussian Process Regression via caret kernlab package?
I am trying to use a Gaussian Process Regression (GPR) model to predict hourly streamflow discharges in a river. I've got good results applying the caret::kernlab train () function (thanks Kuhn!). Since the uncertainty idea is one of the main inherent ones advantages of the GPR, I would like to know if anyone could help me to access the results related to the prediction inteval of the test dataset. I'll put an extract of the code I've been working. Since my real data are huge (and sincerely, I don't know how to put it here), I'll example with the data(airquality). The main goal in this particular example is to predict airquality$Ozone, using as predictos the lag-variables of airquality$Temperature. rm(list = ls()) data(airquality) airquality = na.omit(as.data.frame(airquality)); str(airquality) library(tidyverse) library(magrittr) airquality$Ozone %>% plot(type = 'l') lines(airquality$Temp, col = 2) legend("topleft", legend = c("Ozone", "Temperature"), col=c(1, 2), lty = 1:1, cex = 0.7, text.font = 4, inset = 0.01, box.lty=0, lwd = 1) attach(airquality) df_lags <- airquality %>% mutate(Temp_lag1 = lag(n = 1L, Temp)) %>% na.omit() ESM_train = data.frame(df_lags[1:81, ]) # Training Observed 75% dataset ESM_test = data.frame(df_lags[82:nrow(df_lags), ]) # Testing Observed 25% dataset grid_gaussprRadial = expand.grid(.sigma = c(0.001, 0.01, 0.05, 0.1, 0.5, 1, 2)) # Sigma parameters searching for GPR # TRAIN MODEL ############################ # Tuning set library(caret) set.seed(111) cvCtrl <- trainControl( method ="repeatedcv", repeats = 1, number = 20, allowParallel = TRUE, verboseIter = TRUE, savePredictions = "final") # Train (aprox. 4 seconds time-simulation) attach(ESM_train) set.seed(111) system.time(Model_train <- caret::train(Ozone ~ Temp + Temp_lag1, trControl = cvCtrl, data = ESM_train, metric = "MAE", # Using MAE since I intend minimum values are my focus preProcess = c("center", "scale"), method = "gaussprRadial", # Setting RBF kernel function tuneGrid = grid_gaussprRadial, maxit = 1000, linout = 1)) # Regression type plot(Model_train) Model_train ESM_results_train <- Model_train$resample %>% mutate(Model = "") # K-fold Training measures # Select the interested TRAIN data and arrange them as dataframe Ozone_Obs_Tr = Model_train$pred$obs Ozone_sim = Model_train$pred$pred Resid = Ozone_Obs_Tr - Ozone_sim train_results = data.frame(Ozone_Obs_Tr, Ozone_sim, Resid) # Plot Obs x Simulated train results library(ggplot2) ggplot(data = train_results, aes(x = Ozone_Obs_Tr, y = Ozone_sim)) + geom_point() + geom_abline(intercept = 0, slope = 1, color = "black") # TEST MODEL ############################ # From "ESM_test" dataframe, we predict ESM Ozone time series, adding it in "ESM_forecasted" dataframe ESM_forecasted = ESM_test %>% mutate(Ozone_Pred = predict(Model_train, newdata = ESM_test, variance.model = TRUE)) str(ESM_forecasted) # Select the interested TEST data and arrange them as a dataframe Ozone_Obs = ESM_forecasted$Ozone Ozone_Pred = ESM_forecasted$Ozone_Pred # Plot Obs x Predicted TEST results ggplot(data = ESM_forecasted, aes(x = Ozone_Obs, y = Ozone_Pred)) + geom_point() + geom_abline(intercept = 0, slope = 1, color = "black") # Model performance ##### library(hydroGOF) gof_TR = gof(Ozone_sim, Ozone_Obs_Tr) gof_TEST = gof(Ozone_Pred,Ozone_Obs) Performances = data.frame( Train = gof_TR, Test = gof_TEST ); Performances # Plot the TEST prediction attach(ESM_forecasted) plot(Ozone_Obs, type = "l", xlab = "", ylab = "", ylim = range(Ozone_Obs, Ozone_Pred)) lines(Ozone_Pred , col = "coral2", lty = 2, lwd = 2) legend("top", legend = c("Ozone Obs Test", "Ozone Pred Test"), col=c(1, "coral2"), lty = 1:2, cex = 0.7, text.font = 4, inset = 0.01, box.lty=0, lwd = 2) These last lines generate the following plot: The next, and last, step would be to extract the prediction intervals, which is based on a gaussian distribution around each prediction point, to plot it together with this last plot. The caret::kernlab train() appliance returned better prediction than, for instance, just kernlab::gaussprRadial(), or even tgp::bgp() packages. For both of them I could find the prediction interval. For example, to pick up the prediction intervals via tgp::bgp(), it could be done typing: Upper_Bound <- Ozone_Pred$ZZ.q2 #Ozone_Pred - 2 * sigma^2 Lower_Bound <- Ozone_Pred$ZZ.q1 #Ozone_Pred + 2 * sigma^2 Therefore, via caret::kernlab train(), I hope the required standard deviations could be found typing something as Model_train$... or maybe, with Ozone_Pred$... Moreover, at link: https://stats.stackexchange.com/questions/414079/can-mad-median-absolute-deviation-or-mae-mean-absolute-error-be-used-to-calc, Stephan Kolassa author explained that we could estimate the prediction intervals through MAE, or even RMSE. But I didn't understand if this is my point, since the MAE I got is just the comparison between Obs x Predicted Ozone data, in this example. Please, this solution is very important to me! I think I am near to obtain my main results, but I don't know anymore how to try. Thanks a lot, friends!
I don't really know how the caret framework works, but getting a prediction interval for a GP regression with a Gaussian likelihood is easy enough to do manually. First we just need a function for the squared exponential kernel, also called the radial basis function kernel, which is what you were using. sf here is the scale factor (unused in the kernlab implementation), and ell is the length scale, called sigma in the kernlab implementation: covSEiso <- function(x1, x2 = x1, sf = 1.0, ell = 1.0) { sf <- sf^2 ell <- -0.5 * (1 / (ell^2)) n <- nrow(x1) m <- nrow(x2) d <- ncol(x1) result <- matrix(0, nrow = n, ncol = m) for ( j in 1:m ) { for ( i in 1:n ) { result[i, j] <- sf * exp(ell * sum((x1[i, ] - x2[j, ])^2)) } } return(result) } I'm not sure what your code says about which length scale to use; below I will use a length scale of 25 and scale factor of 50 (obtained via GPML's hyperparameter optimization routines). Then we use the covSEiso() function above to get the relevant covariances, and the rest is application of basic Gaussian identities. I would refer you to Chapter 2 of Rasmussen and Williams (2006) (graciously provided for free online). data(airquality) library(tidyverse) library(magrittr) df_lags <- airquality %>% mutate(Temp_lag1 = lag(n = 1L, Temp)) %>% na.omit() ESM_train <- data.frame(df_lags[1:81, ]) # Training Data 75% dataset ESM_test <- data.frame(df_lags[82:nrow(df_lags), ]) # Testing Data 25% dataset ## For convenience I'll define separately the training and test inputs X <- ESM_train[ , c("Temp", "Temp_lag1")] Xstar <- ESM_test[ , c("Temp", "Temp_lag1")] ## Get the kernel manually K <- covSEiso(X, ell = 25, sf = 50) ## We also need covariance between the test cases Kstar <- covSEiso(Xstar, X, ell = 25, sf = 50) Ktest <- covSEiso(Xstar, ell = 25, sf = 50) ## Now the 95% credible region for the posterior is predictive_mean <- Kstar %*% solve(K + diag(nrow(K))) %*% ESM_train$Ozone predictive_var <- Ktest - (Kstar %*% solve(K + diag(nrow(K))) %*% t(Kstar)) ## Then for the prediction interval we only need to add the observation noise z <- sqrt(diag(predictive_var)) + 25 interval_high <- predictive_mean + 2 * z interval_low <- predictive_mean - 2 * z Then we can check out the prediction intervals This all is pretty easy to do via my gplmr package (available on GitHub) which can call GPML from R if you have Octave installed: data(airquality) library(tidyverse) library(magrittr) library(gpmlr) df_lags <- airquality %>% mutate(Temp_lag1 = lag(n = 1L, Temp)) %>% na.omit() ESM_train <- data.frame(df_lags[1:81, ]) # Training Data 75% dataset ESM_test <- data.frame(df_lags[82:nrow(df_lags), ]) # Testing Data 25% dataset X <- as.matrix(ESM_train[ , c("Temp", "Temp_lag1")]) y <- ESM_train$Ozone Xs <- as.matrix(ESM_test[ , c("Temp", "Temp_lag1")]) ys <- ESM_test$Ozone hyp0 <- list(mean = numeric(), cov = c(0, 0), lik = 0) hyp <- set_hyperparameters(hyp0, "infExact", "meanZero", "covSEiso","likGauss", X, y) gp_res <- gp(hyp, "infExact", "meanZero", "covSEiso", "likGauss", X, y, Xs, ys) predictive_mean <- gp_res$YMU interval_high <- gp_res$YMU + 2 * sqrt(gp_res$YS2) interval_low <- gp_res$YMU - 2 * sqrt(gp_res$YS2) Then just plot the predictions, as above: plot(NULL, xlab = "", ylab = "", xaxt = "n", yaxt = "n", xlim = range(ESM_test$Temp), ylim = range(c(interval_high, interval_low))) axis(1, tick = FALSE, line = -0.75) axis(2, tick = FALSE, line = -0.75) mtext("Temp", 1, 1.5) mtext("Ozone", 2, 1.5) idx <- order(ESM_test$Temp) polygon(c(ESM_test$Temp[idx], rev(ESM_test$Temp[idx])), c(interval_high[idx], rev(interval_low[idx])), border = NA, col = "#80808080") lines(ESM_test$Temp[idx], predictive_mean[idx]) points(ESM_test$Temp, ESM_test$Ozone, pch = 19) plot(NULL, xlab = "", ylab = "", xaxt = "n", yaxt = "n", xlim = range(ESM_test$Temp_lag1), ylim = range(c(interval_high, interval_low))) axis(1, tick = FALSE, line = -0.75) axis(2, tick = FALSE, line = -0.75) mtext("Temp_lag1", 1, 1.5) mtext("Ozone", 2, 1.5) idx <- order(ESM_test$Temp_lag1) polygon(c(ESM_test$Temp_lag1[idx], rev(ESM_test$Temp_lag1[idx])), c(interval_high[idx], rev(interval_low[idx])), border = NA, col = "#80808080") lines(ESM_test$Temp_lag1[idx], predictive_mean[idx]) points(ESM_test$Temp_lag1, ESM_test$Ozone, pch = 19)
Adapting the meansd moderator option in sjPlot interaction
I am using sjPlot, the sjp.int function, to plot an interaction of an lme. The options for the moderator values are means +/- sd, quartiles, all, max/min. Is there a way to plot the mean +/- 2sd? Typically it would be like this: model <- lme(outcome ~ var1+var2*time, random=~1|ID, data=mydata, na.action="na.omit") sjp.int(model, show.ci=T, mdrt.values="meansd") Many thanks Reproducible example: #create data mydata <- data.frame( SID=sample(1:150,400,replace=TRUE),age=sample(50:70,400,replace=TRUE), sex=sample(c("Male","Female"),200, replace=TRUE),time= seq(0.7, 6.2, length.out=400), Vol =rnorm(400),HCD =rnorm(400)) mydata$time <- as.numeric(mydata$time) #insert random NAs NAins <- NAinsert <- function(df, prop = .1){ n <- nrow(df) m <- ncol(df) num.to.na <- ceiling(prop*n*m) id <- sample(0:(m*n-1), num.to.na, replace = FALSE) rows <- id %/% m + 1 cols <- id %% m + 1 sapply(seq(num.to.na), function(x){ df[rows[x], cols[x]] <<- NA } ) return(df) } mydata2 <- NAins(mydata,0.1) #run the lme which gives error message model = lme(Vol ~ age+sex*time+time* HCD, random=~time|SID,na.action="na.omit",data=mydata2);summary(model) mydf <- ggpredict(model, terms=c("time","HCD [-2.5, -0.5, 2.0]")) #lmer works model2 = lmer(Vol ~ age+sex*time+time* HCD+(time|SID),control=lmerControl(check.nobs.vs.nlev = "ignore",check.nobs.vs.rankZ = "ignore", check.nobs.vs.nRE="ignore"), na.action="na.omit",data=mydata2);summary(model) mydf <- ggpredict(model2, terms=c("time","HCD [-2.5, -0.5, 2.0]")) #plotting gives problems (jittered lines) plot(mydf)
With sjPlot, it's currently not possible. However, I have written a package especially dedicated to compute and plot marginal effects: ggeffects. This package is a bit more flexible (for marginal effects plots). In the ggeffects-package, there's a ggpredict()-function, where you can compute marginal effects at specific values. Once you know the sd of your model term in question, you can specify these values in the function call to plot your interaction: library(ggeffects) # plot interaction for time and var2, for values # 10, 30 and 50 of var2 mydf <- ggpredict(model, terms = c("time", "var2 [10,30,50]")) plot(mydf) There are some examples in the package-vignette, see especially this section. Edit Here are the results, based on your reproducible example (note that GitHub-Version is currently required!): # requires at least the GitHub-Versiob 0.1.0.9000! library(ggeffects) library(nlme) library(lme4) library(glmmTMB) #create data mydata <- data.frame( SID = sample(1:150, 400, replace = TRUE), age = sample(50:70, 400, replace = TRUE), sex = sample(c("Male", "Female"), 200, replace = TRUE), time = seq(0.7, 6.2, length.out = 400), Vol = rnorm(400), HCD = rnorm(400) ) mydata$time <- as.numeric(mydata$time) #insert random NAs NAins <- NAinsert <- function(df, prop = .1) { n <- nrow(df) m <- ncol(df) num.to.na <- ceiling(prop * n * m) id <- sample(0:(m * n - 1), num.to.na, replace = FALSE) rows <- id %/% m + 1 cols <- id %% m + 1 sapply(seq(num.to.na), function(x) { df[rows[x], cols[x]] <<- NA }) return(df) } mydata2 <- NAins(mydata, 0.1) # run the lme, works now model = lme( Vol ~ age + sex * time + time * HCD, random = ~ time | SID, na.action = "na.omit", data = mydata2 ) summary(model) mydf <- ggpredict(model, terms = c("time", "HCD [-2.5, -0.5, 2.0]")) plot(mydf) lme-plot # lmer also works model2 <- lmer( Vol ~ age + sex * time + time * HCD + (time | SID), control = lmerControl( check.nobs.vs.nlev = "ignore", check.nobs.vs.rankZ = "ignore", check.nobs.vs.nRE = "ignore" ), na.action = "na.omit", data = mydata2 ) summary(model) mydf <- ggpredict(model2, terms = c("time", "HCD [-2.5, -0.5, 2.0]"), ci.lvl = NA) # plotting works, but only w/o CI plot(mydf) lmer-plot # lmer also works model3 <- glmmTMB( Vol ~ age + sex * time + time * HCD + (time | SID), data = mydata2 ) summary(model) mydf <- ggpredict(model3, terms = c("time", "HCD [-2.5, -0.5, 2.0]")) plot(mydf) plot(mydf, facets = T) glmmTMB-plots