Related
How add the contours under the graph using the R as in plot 2?
I've searched a lot on the internet and found no example of how to do it in R! Is there any function or package to add the outline along with the chart?
#Function density probability
library(pbivnorm)
bsb <- function(t1,t2){
a1 <- sqrt(phi1/2)*(sqrt(((phi1+1)*t1)/(phi1*mu1))-sqrt(((phi1*mu1)/((phi1+1)*t1))))
a2 <- sqrt(phi2/2)*(sqrt(((phi2+1)*t2)/(phi2*mu2))-sqrt(((phi2*mu2)/((phi2+1)*t2))))
Phi2 <- pbivnorm(a1, a2, rho, recycle = TRUE)
b1 <- ((phi1+1)/(2*phi1*mu1))*sqrt(phi1/2)*(((phi1*mu1)/((phi1+1)*t1))^(1/2)+((phi1*mu1)/((phi1+1)*t1))^(3/2))
b2 <- ((phi2+1)/(2*phi2*mu2))*sqrt(phi2/2)*(((phi2*mu2)/((phi2+1)*t2))^(1/2)+((phi2*mu2)/((phi2+1)*t2))^(3/2))
fdp <- Phi2*b1*b2
return(fdp)
}
t1 <- seq(0.001,5,length=100)
t2 <- seq(0.001,5,length=100)
#Parameters
mu1=5
phi1=2
mu2=5
phi2=2
rho=0.9
z<-outer(t1,t2,bsb) # calculate density values
persp(t1, t2, z, # 3-D plot
main="Bivariate Birnbaum-Saunders",
col="lightgray",
theta=40, phi=10,
r=10,
d=0.9,
expand=0.5,
ltheta=90, lphi=80,
shade=0.9,
ticktype="detailed",
nticks=5)
As #alistaire pointed out, it actually requires a single line to get the plotly version, see for documentation to edit details of the plot (https://plot.ly/r/3d-surface-plots/)
test<-outer(t1,t2,bsb) # your output matrix
p <- plot_ly(z = ~test) %>% add_surface()
p
Resolved:
source("https://raw.githubusercontent.com/walmes/wzRfun/master/R/panel.3d.contour.R")
library(lattice)
library(manipulate)
library(colorRamps)
#Function density probability
library(pbivnorm)
bsb <- function(t1,t2){
a1 <- sqrt(phi1/2)*(sqrt(((phi1+1)*t1)/(phi1*mu1))-sqrt(((phi1*mu1)/((phi1+1)*t1))))
a2 <- sqrt(phi2/2)*(sqrt(((phi2+1)*t2)/(phi2*mu2))-sqrt(((phi2*mu2)/((phi2+1)*t2))))
Phi2 <- pbivnorm(a1, a2, rho, recycle = TRUE)
b1 <- ((phi1+1)/(2*phi1*mu1))*sqrt(phi1/2)*(((phi1*mu1)/((phi1+1)*t1))^(1/2)+((phi1*mu1)/((phi1+1)*t1))^(3/2))
b2 <- ((phi2+1)/(2*phi2*mu2))*sqrt(phi2/2)*(((phi2*mu2)/((phi2+1)*t2))^(1/2)+((phi2*mu2)/((phi2+1)*t2))^(3/2))
fdp <- Phi2*b1*b2
return(fdp)
}
#Parameters
mu1=5
phi1=2
mu2=5
phi2=2
rho=0.9
grid <- expand.grid(t1 = seq(0.001,8, by = 0.1),
t2 = seq(0.001,8, by = 0.1))
grid$z <- bsb(grid$t1,grid$t2)
manipulate({
## Makes the three-dimensional chart
colr <- colorRampPalette(c(c1, c2, c3), space="rgb")
arrows <- arr
wireframe(z ~ t1 + t2,
data = grid,
scales = list(arrows = FALSE),
zlim = extendrange(grid$z, f = 0.25),
panel.3d.wireframe = "panel.3d.contour",
nlevels = 8,
col = "gray40",
type = c("bottom"),
col.regions = colr(101),
drape = TRUE, colorkey=FALSE,
screen=list(z=z.angle, x=x.angle),
axis.line = list(col = "transparent"),
clip = list(panel = "off"),
par.settings = list(box.3d = list(col=c(1,NA,NA,1,1,NA,NA,NA,NA))))
},
## Controls the value of angles and colors
z.angle=slider(0, 360, step=10, initial=40),
x.angle=slider(-180, 0, step=5, initial=-80),
arr=checkbox(FALSE, "show.arrows"),
c1=picker("transparent","black","red","yellow","orange","green","blue","pink","violet"),
c2=picker("transparent","black","red","yellow","orange","green","blue","pink","violet"),
c3=picker("transparent","black","red","yellow","orange","green","blue","pink","violet")
)
If I have 10 values, each of which has a fitted value F, and an upper and lower confidence interval U and L:
set.seed(0815)
F <- runif(10, 1, 2)
L <- runif(10, 0, 1)
U <- runif(10, 2, 3)
How can I show these 10 fitted values and their confidence intervals in the same plot like the one below in R?
Here is a plotrix solution:
set.seed(0815)
x <- 1:10
F <- runif(10,1,2)
L <- runif(10,0,1)
U <- runif(10,2,3)
require(plotrix)
plotCI(x, F, ui=U, li=L)
And here is a ggplot solution:
set.seed(0815)
df <- data.frame(x =1:10,
F =runif(10,1,2),
L =runif(10,0,1),
U =runif(10,2,3))
require(ggplot2)
ggplot(df, aes(x = x, y = F)) +
geom_point(size = 4) +
geom_errorbar(aes(ymax = U, ymin = L))
UPDATE:
Here is a base solution to your edits:
set.seed(1234)
x <- rnorm(20)
df <- data.frame(x = x,
y = x + rnorm(20))
plot(y ~ x, data = df)
# model
mod <- lm(y ~ x, data = df)
# predicts + interval
newx <- seq(min(df$x), max(df$x), length.out=100)
preds <- predict(mod, newdata = data.frame(x=newx),
interval = 'confidence')
# plot
plot(y ~ x, data = df, type = 'n')
# add fill
polygon(c(rev(newx), newx), c(rev(preds[ ,3]), preds[ ,2]), col = 'grey80', border = NA)
# model
abline(mod)
# intervals
lines(newx, preds[ ,3], lty = 'dashed', col = 'red')
lines(newx, preds[ ,2], lty = 'dashed', col = 'red')
Here is a solution using functions plot(), polygon() and lines().
set.seed(1234)
df <- data.frame(x =1:10,
F =runif(10,1,2),
L =runif(10,0,1),
U =runif(10,2,3))
plot(df$x, df$F, ylim = c(0,4), type = "l")
#make polygon where coordinates start with lower limit and
# then upper limit in reverse order
polygon(c(df$x,rev(df$x)),c(df$L,rev(df$U)),col = "grey75", border = FALSE)
lines(df$x, df$F, lwd = 2)
#add red lines on borders of polygon
lines(df$x, df$U, col="red",lty=2)
lines(df$x, df$L, col="red",lty=2)
Now use example data provided by OP in another question:
Lower <- c(0.418116841, 0.391011834, 0.393297710,
0.366144073,0.569956636,0.224775521,0.599166016,0.512269587,
0.531378573, 0.311448219, 0.392045751,0.153614913, 0.366684097,
0.161100849,0.700274810,0.629714150, 0.661641288, 0.533404093,
0.412427559, 0.432905333, 0.525306427,0.224292061,
0.28893064,0.099543648, 0.342995605,0.086973739,0.289030388,
0.081230826,0.164505624, -0.031290586,0.148383474,0.070517523,0.009686605,
-0.052703529,0.475924192,0.253382210, 0.354011010,0.130295355,0.102253218,
0.446598823,0.548330752,0.393985810,0.481691632,0.111811248,0.339626541,
0.267831909,0.133460254,0.347996621,0.412472322,0.133671128,0.178969601,0.484070587,
0.335833224,0.037258467, 0.141312363,0.361392799,0.129791998,
0.283759439,0.333893418,0.569533076,0.385258093,0.356201955,0.481816148,
0.531282473,0.273126565,0.267815691,0.138127486,0.008865700,0.018118398,0.080143484,
0.117861634,0.073697418,0.230002398,0.105855042,0.262367348,0.217799352,0.289108011,
0.161271889,0.219663224,0.306117717,0.538088622,0.320711912,0.264395149,0.396061543,
0.397350946,0.151726970,0.048650180,0.131914718,0.076629840,0.425849394,
0.068692279,0.155144797,0.137939059,0.301912657,-0.071415593,-0.030141781,0.119450922,
0.312927614,0.231345972)
Upper.limit <- c(0.6446223,0.6177311, 0.6034427, 0.5726503,
0.7644718, 0.4585430, 0.8205418, 0.7154043,0.7370033,
0.5285199, 0.5973728, 0.3764209, 0.5818298,
0.3960867,0.8972357, 0.8370151, 0.8359921, 0.7449118,
0.6152879, 0.6200704, 0.7041068, 0.4541011, 0.5222653,
0.3472364, 0.5956551, 0.3068065, 0.5112895, 0.3081448,
0.3745473, 0.1931089, 0.3890704, 0.3031025, 0.2472591,
0.1976092, 0.6906118, 0.4736644, 0.5770463, 0.3528607,
0.3307651, 0.6681629, 0.7476231, 0.5959025, 0.7128883,
0.3451623, 0.5609742, 0.4739216, 0.3694883, 0.5609220,
0.6343219, 0.3647751, 0.4247147, 0.6996334, 0.5562876,
0.2586490, 0.3750040, 0.5922248, 0.3626322, 0.5243285,
0.5548211, 0.7409648, 0.5820070, 0.5530232, 0.6863703,
0.7206998, 0.4952387, 0.4993264, 0.3527727, 0.2203694,
0.2583149, 0.3035342, 0.3462009, 0.3003602, 0.4506054,
0.3359478, 0.4834151, 0.4391330, 0.5273411, 0.3947622,
0.4133769, 0.5288060, 0.7492071, 0.5381701, 0.4825456,
0.6121942, 0.6192227, 0.3784870, 0.2574025, 0.3704140,
0.2945623, 0.6532694, 0.2697202, 0.3652230, 0.3696383,
0.5268808, 0.1545602, 0.2221450, 0.3553377, 0.5204076,
0.3550094)
Fitted.values<- c(0.53136955, 0.50437146, 0.49837019,
0.46939721, 0.66721423, 0.34165926, 0.70985388, 0.61383696,
0.63419092, 0.41998407, 0.49470927, 0.26501789, 0.47425695,
0.27859380, 0.79875525, 0.73336461, 0.74881668, 0.63915795,
0.51385774, 0.52648789, 0.61470661, 0.33919656, 0.40559797,
0.22339000, 0.46932536, 0.19689011, 0.40015996, 0.19468781,
0.26952645, 0.08090917, 0.26872696, 0.18680999, 0.12847285,
0.07245286, 0.58326799, 0.36352329, 0.46552867, 0.24157804,
0.21650915, 0.55738088, 0.64797691, 0.49494416, 0.59728999,
0.22848680, 0.45030036, 0.37087676, 0.25147426, 0.45445930,
0.52339711, 0.24922310, 0.30184215, 0.59185198, 0.44606040,
0.14795374, 0.25815819, 0.47680880, 0.24621212, 0.40404398,
0.44435727, 0.65524894, 0.48363255, 0.45461258, 0.58409323,
0.62599114, 0.38418264, 0.38357103, 0.24545011, 0.11461756,
0.13821664, 0.19183886, 0.23203127, 0.18702881, 0.34030391,
0.22090140, 0.37289121, 0.32846615, 0.40822456, 0.27801706,
0.31652008, 0.41746184, 0.64364785, 0.42944100, 0.37347037,
0.50412786, 0.50828681, 0.26510696, 0.15302635, 0.25116438,
0.18559609, 0.53955941, 0.16920626, 0.26018389, 0.25378867,
0.41439675, 0.04157232, 0.09600163, 0.23739430, 0.41666762,
0.29317767)
Assemble into a data frame (no x provided, so using indices)
df2 <- data.frame(x=seq(length(Fitted.values)),
fit=Fitted.values,lwr=Lower,upr=Upper.limit)
plot(fit~x,data=df2,ylim=range(c(df2$lwr,df2$upr)))
#make polygon where coordinates start with lower limit and then upper limit in reverse order
with(df2,polygon(c(x,rev(x)),c(lwr,rev(upr)),col = "grey75", border = FALSE))
matlines(df2[,1],df2[,-1],
lwd=c(2,1,1),
lty=1,
col=c("black","red","red"))
Here is part of my program related to plotting confidence interval.
1. Generate the test data
ads = 1
require(stats); require(graphics)
library(splines)
x_raw <- seq(1,10,0.1)
y <- cos(x_raw)+rnorm(len_data,0,0.1)
y[30] <- 1.4 # outlier point
len_data = length(x_raw)
N <- len_data
summary(fm1 <- lm(y~bs(x_raw, df=5), model = TRUE, x =T, y = T))
ht <-seq(1,10,length.out = len_data)
plot(x = x_raw, y = y,type = 'p')
y_e <- predict(fm1, data.frame(height = ht))
lines(x= ht, y = y_e)
Result
2. Fitting the raw data using B-spline smoother method
sigma_e <- sqrt(sum((y-y_e)^2)/N)
print(sigma_e)
H<-fm1$x
A <-solve(t(H) %*% H)
y_e_minus <- rep(0,N)
y_e_plus <- rep(0,N)
y_e_minus[N]
for (i in 1:N)
{
tmp <-t(matrix(H[i,])) %*% A %*% matrix(H[i,])
tmp <- 1.96*sqrt(tmp)
y_e_minus[i] <- y_e[i] - tmp
y_e_plus[i] <- y_e[i] + tmp
}
plot(x = x_raw, y = y,type = 'p')
polygon(c(ht,rev(ht)),c(y_e_minus,rev(y_e_plus)),col = rgb(1, 0, 0,0.5), border = NA)
#plot(x = x_raw, y = y,type = 'p')
lines(x= ht, y = y_e_plus, lty = 'dashed', col = 'red')
lines(x= ht, y = y_e)
lines(x= ht, y = y_e_minus, lty = 'dashed', col = 'red')
Result
Some addition to the previous answers. It is nice to regulate the density of the polygon to avoid obscuring the data points.
library(MASS)
attach(Boston)
lm.fit2 = lm(medv~poly(lstat,2))
plot(lstat,medv)
new.lstat = seq(min(lstat), max(lstat), length.out=100)
preds <- predict(lm.fit2, newdata = data.frame(lstat=new.lstat), interval = 'prediction')
lines(sort(lstat), fitted(lm.fit2)[order(lstat)], col='red', lwd=3)
polygon(c(rev(new.lstat), new.lstat), c(rev(preds[ ,3]), preds[ ,2]), density=10, col = 'blue', border = NA)
lines(new.lstat, preds[ ,3], lty = 'dashed', col = 'red')
lines(new.lstat, preds[ ,2], lty = 'dashed', col = 'red')
Please note that you see the prediction interval on the picture, which is several times wider than the confidence interval. You can read here the detailed explanation of those two types of interval estimates.
If I have 10 values, each of which has a fitted value F, and an upper and lower confidence interval U and L:
set.seed(0815)
F <- runif(10, 1, 2)
L <- runif(10, 0, 1)
U <- runif(10, 2, 3)
How can I show these 10 fitted values and their confidence intervals in the same plot like the one below in R?
Here is a plotrix solution:
set.seed(0815)
x <- 1:10
F <- runif(10,1,2)
L <- runif(10,0,1)
U <- runif(10,2,3)
require(plotrix)
plotCI(x, F, ui=U, li=L)
And here is a ggplot solution:
set.seed(0815)
df <- data.frame(x =1:10,
F =runif(10,1,2),
L =runif(10,0,1),
U =runif(10,2,3))
require(ggplot2)
ggplot(df, aes(x = x, y = F)) +
geom_point(size = 4) +
geom_errorbar(aes(ymax = U, ymin = L))
UPDATE:
Here is a base solution to your edits:
set.seed(1234)
x <- rnorm(20)
df <- data.frame(x = x,
y = x + rnorm(20))
plot(y ~ x, data = df)
# model
mod <- lm(y ~ x, data = df)
# predicts + interval
newx <- seq(min(df$x), max(df$x), length.out=100)
preds <- predict(mod, newdata = data.frame(x=newx),
interval = 'confidence')
# plot
plot(y ~ x, data = df, type = 'n')
# add fill
polygon(c(rev(newx), newx), c(rev(preds[ ,3]), preds[ ,2]), col = 'grey80', border = NA)
# model
abline(mod)
# intervals
lines(newx, preds[ ,3], lty = 'dashed', col = 'red')
lines(newx, preds[ ,2], lty = 'dashed', col = 'red')
Here is a solution using functions plot(), polygon() and lines().
set.seed(1234)
df <- data.frame(x =1:10,
F =runif(10,1,2),
L =runif(10,0,1),
U =runif(10,2,3))
plot(df$x, df$F, ylim = c(0,4), type = "l")
#make polygon where coordinates start with lower limit and
# then upper limit in reverse order
polygon(c(df$x,rev(df$x)),c(df$L,rev(df$U)),col = "grey75", border = FALSE)
lines(df$x, df$F, lwd = 2)
#add red lines on borders of polygon
lines(df$x, df$U, col="red",lty=2)
lines(df$x, df$L, col="red",lty=2)
Now use example data provided by OP in another question:
Lower <- c(0.418116841, 0.391011834, 0.393297710,
0.366144073,0.569956636,0.224775521,0.599166016,0.512269587,
0.531378573, 0.311448219, 0.392045751,0.153614913, 0.366684097,
0.161100849,0.700274810,0.629714150, 0.661641288, 0.533404093,
0.412427559, 0.432905333, 0.525306427,0.224292061,
0.28893064,0.099543648, 0.342995605,0.086973739,0.289030388,
0.081230826,0.164505624, -0.031290586,0.148383474,0.070517523,0.009686605,
-0.052703529,0.475924192,0.253382210, 0.354011010,0.130295355,0.102253218,
0.446598823,0.548330752,0.393985810,0.481691632,0.111811248,0.339626541,
0.267831909,0.133460254,0.347996621,0.412472322,0.133671128,0.178969601,0.484070587,
0.335833224,0.037258467, 0.141312363,0.361392799,0.129791998,
0.283759439,0.333893418,0.569533076,0.385258093,0.356201955,0.481816148,
0.531282473,0.273126565,0.267815691,0.138127486,0.008865700,0.018118398,0.080143484,
0.117861634,0.073697418,0.230002398,0.105855042,0.262367348,0.217799352,0.289108011,
0.161271889,0.219663224,0.306117717,0.538088622,0.320711912,0.264395149,0.396061543,
0.397350946,0.151726970,0.048650180,0.131914718,0.076629840,0.425849394,
0.068692279,0.155144797,0.137939059,0.301912657,-0.071415593,-0.030141781,0.119450922,
0.312927614,0.231345972)
Upper.limit <- c(0.6446223,0.6177311, 0.6034427, 0.5726503,
0.7644718, 0.4585430, 0.8205418, 0.7154043,0.7370033,
0.5285199, 0.5973728, 0.3764209, 0.5818298,
0.3960867,0.8972357, 0.8370151, 0.8359921, 0.7449118,
0.6152879, 0.6200704, 0.7041068, 0.4541011, 0.5222653,
0.3472364, 0.5956551, 0.3068065, 0.5112895, 0.3081448,
0.3745473, 0.1931089, 0.3890704, 0.3031025, 0.2472591,
0.1976092, 0.6906118, 0.4736644, 0.5770463, 0.3528607,
0.3307651, 0.6681629, 0.7476231, 0.5959025, 0.7128883,
0.3451623, 0.5609742, 0.4739216, 0.3694883, 0.5609220,
0.6343219, 0.3647751, 0.4247147, 0.6996334, 0.5562876,
0.2586490, 0.3750040, 0.5922248, 0.3626322, 0.5243285,
0.5548211, 0.7409648, 0.5820070, 0.5530232, 0.6863703,
0.7206998, 0.4952387, 0.4993264, 0.3527727, 0.2203694,
0.2583149, 0.3035342, 0.3462009, 0.3003602, 0.4506054,
0.3359478, 0.4834151, 0.4391330, 0.5273411, 0.3947622,
0.4133769, 0.5288060, 0.7492071, 0.5381701, 0.4825456,
0.6121942, 0.6192227, 0.3784870, 0.2574025, 0.3704140,
0.2945623, 0.6532694, 0.2697202, 0.3652230, 0.3696383,
0.5268808, 0.1545602, 0.2221450, 0.3553377, 0.5204076,
0.3550094)
Fitted.values<- c(0.53136955, 0.50437146, 0.49837019,
0.46939721, 0.66721423, 0.34165926, 0.70985388, 0.61383696,
0.63419092, 0.41998407, 0.49470927, 0.26501789, 0.47425695,
0.27859380, 0.79875525, 0.73336461, 0.74881668, 0.63915795,
0.51385774, 0.52648789, 0.61470661, 0.33919656, 0.40559797,
0.22339000, 0.46932536, 0.19689011, 0.40015996, 0.19468781,
0.26952645, 0.08090917, 0.26872696, 0.18680999, 0.12847285,
0.07245286, 0.58326799, 0.36352329, 0.46552867, 0.24157804,
0.21650915, 0.55738088, 0.64797691, 0.49494416, 0.59728999,
0.22848680, 0.45030036, 0.37087676, 0.25147426, 0.45445930,
0.52339711, 0.24922310, 0.30184215, 0.59185198, 0.44606040,
0.14795374, 0.25815819, 0.47680880, 0.24621212, 0.40404398,
0.44435727, 0.65524894, 0.48363255, 0.45461258, 0.58409323,
0.62599114, 0.38418264, 0.38357103, 0.24545011, 0.11461756,
0.13821664, 0.19183886, 0.23203127, 0.18702881, 0.34030391,
0.22090140, 0.37289121, 0.32846615, 0.40822456, 0.27801706,
0.31652008, 0.41746184, 0.64364785, 0.42944100, 0.37347037,
0.50412786, 0.50828681, 0.26510696, 0.15302635, 0.25116438,
0.18559609, 0.53955941, 0.16920626, 0.26018389, 0.25378867,
0.41439675, 0.04157232, 0.09600163, 0.23739430, 0.41666762,
0.29317767)
Assemble into a data frame (no x provided, so using indices)
df2 <- data.frame(x=seq(length(Fitted.values)),
fit=Fitted.values,lwr=Lower,upr=Upper.limit)
plot(fit~x,data=df2,ylim=range(c(df2$lwr,df2$upr)))
#make polygon where coordinates start with lower limit and then upper limit in reverse order
with(df2,polygon(c(x,rev(x)),c(lwr,rev(upr)),col = "grey75", border = FALSE))
matlines(df2[,1],df2[,-1],
lwd=c(2,1,1),
lty=1,
col=c("black","red","red"))
Here is part of my program related to plotting confidence interval.
1. Generate the test data
ads = 1
require(stats); require(graphics)
library(splines)
x_raw <- seq(1,10,0.1)
y <- cos(x_raw)+rnorm(len_data,0,0.1)
y[30] <- 1.4 # outlier point
len_data = length(x_raw)
N <- len_data
summary(fm1 <- lm(y~bs(x_raw, df=5), model = TRUE, x =T, y = T))
ht <-seq(1,10,length.out = len_data)
plot(x = x_raw, y = y,type = 'p')
y_e <- predict(fm1, data.frame(height = ht))
lines(x= ht, y = y_e)
Result
2. Fitting the raw data using B-spline smoother method
sigma_e <- sqrt(sum((y-y_e)^2)/N)
print(sigma_e)
H<-fm1$x
A <-solve(t(H) %*% H)
y_e_minus <- rep(0,N)
y_e_plus <- rep(0,N)
y_e_minus[N]
for (i in 1:N)
{
tmp <-t(matrix(H[i,])) %*% A %*% matrix(H[i,])
tmp <- 1.96*sqrt(tmp)
y_e_minus[i] <- y_e[i] - tmp
y_e_plus[i] <- y_e[i] + tmp
}
plot(x = x_raw, y = y,type = 'p')
polygon(c(ht,rev(ht)),c(y_e_minus,rev(y_e_plus)),col = rgb(1, 0, 0,0.5), border = NA)
#plot(x = x_raw, y = y,type = 'p')
lines(x= ht, y = y_e_plus, lty = 'dashed', col = 'red')
lines(x= ht, y = y_e)
lines(x= ht, y = y_e_minus, lty = 'dashed', col = 'red')
Result
Some addition to the previous answers. It is nice to regulate the density of the polygon to avoid obscuring the data points.
library(MASS)
attach(Boston)
lm.fit2 = lm(medv~poly(lstat,2))
plot(lstat,medv)
new.lstat = seq(min(lstat), max(lstat), length.out=100)
preds <- predict(lm.fit2, newdata = data.frame(lstat=new.lstat), interval = 'prediction')
lines(sort(lstat), fitted(lm.fit2)[order(lstat)], col='red', lwd=3)
polygon(c(rev(new.lstat), new.lstat), c(rev(preds[ ,3]), preds[ ,2]), density=10, col = 'blue', border = NA)
lines(new.lstat, preds[ ,3], lty = 'dashed', col = 'red')
lines(new.lstat, preds[ ,2], lty = 'dashed', col = 'red')
Please note that you see the prediction interval on the picture, which is several times wider than the confidence interval. You can read here the detailed explanation of those two types of interval estimates.
I would like to plot a line that connects the outer points of the plot
plot(rnorm(1000), rnorm(1000), xlim=c(-5,5),ylim=c(-5,5))
and thus "bags" all points of the plot
Function to be used here is chull. Line 4 is there to close the circle. For more examples, see here.
x <- data.frame(x = rnorm(100), y = rnorm(100))
plot(x)
chx <- chull(x)
chx <- rbind(x = x[chx, ], x[chx[1], ])
lines(chx)
a <- rnorm(1000)
b <- rnorm(1000)
Plot_ConvexHull<-function(xcoord, ycoord, lcolor){
hpts <- chull(x = xcoord, y = ycoord)
hpts <- c(hpts, hpts[1])
lines(xcoord[hpts], ycoord[hpts], col = lcolor)
}
(xrange <- range(c(a)))
(yrange <- range(c(b)))
par(tck = 0.02, mgp = c(1.7, 0.3, 0))
plot(a, b, type = "p", pch = 1, col = "black", xlim = c(xrange), ylim = c(yrange))
Plot_ConvexHull(xcoord = a, ycoord = b, lcolor = "black")
I want to plot a regression line with (a = 0 and b = 1) and add the individual point deviations from this along with identifying the data point with name.
set.seed(123)
namelab <- paste ("ET", 1:10, sep = "")
xvar <- 1:10
yvar <- rnorm(10, 5, 5)
myd <- data.frame(namelab, xvar, yvar)
plot(xvar, yvar)
abline (a= 0, b = 1, col = "red", lty = 2)
Just manual sketch of my intention, I just labelled a single point just for example. The line drawn need a slim.
dev.new(width=4, height=4)
plot(xvar, yvar, asp=1)
a = 0
b = 1
abline (a, b, col = "red", lty = 2)
myd$xint = with(myd, (b*yvar + xvar - b*a) / (b^2 + 1))
myd$yint = with(myd, (b*yvar + b*xvar + a) / (b^2 + 1))
with(myd, segments(xvar, yvar, xint, yint))
with(myd, text(xvar, yvar, labels=namelab, pos=3, cex=0.5))
...and if you did want vertical as opposed to perpendicular offsets, here is a pretty straightforward option:
set.seed(123)
namelab <- paste ("ET", 1:10, sep = "")
xvar <- 1:10
yvar <- rnorm(10, 5, 5)
plot(xvar, yvar)
abline (a= 0, b = 1, col = "red", lty = 2)
segments(xvar,yvar,xvar,xvar)
text(xvar,yvar,namelab,pos=3)
For this to work for any value of a and b, you would use:
segments(xvar,yvar,xvar,((xvar*b)+a))