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I am trying to add lm model coefs of two parallel modelling results onto the same ggplot plot. Here is my working example:
library(ggplot2)
set.seed(100)
dat <- data.frame(
x <- rnorm(100, 1),
y <- rnorm(100, 10),
lev <- gl(n = 2, k = 50, labels = letters[1:2])
)
mod1 <- lm(y~x, dat = dat[lev %in% "a", ])
r1 <- paste("R^2==", round(summary(mod1)[[9]], 3))
p1<- paste("p==", round(summary(mod1)[[4]][2, 4], 3), sep= "")
lab1 <- paste(r1, p1, sep =",")
mod2 <- lm(y~x, dat = dat[lev %in% "b", ])
r2 <- paste("R^2==", round(summary(mod2)[[9]], 3))
p2 <- paste("p==", round(summary(mod2)[[4]][2, 4], 3), sep= "")
lab2 <- paste(r2, p2, sep =",")
ggplot(dat, aes(x = x, y = y, col = lev)) + geom_jitter() + geom_smooth(method = "lm") + annotate("text", x = 2, y = 12, label = lab1, parse = T) + annotate("text", x = 10, y = 8, label = lab2, parse = T)
Here is the promot shows:
Error in parse(text = text[[i]]) : <text>:1:12: unexpected ','
1: R^2== 0.008,
Now the problem is that I could label either R2 or p value seperately, but not both of them together. How could I do to put the two results into one single line on the figure?
BTW, any other efficienty way of doing the same thing as my code? I have nine subplots that I want to put into one full plot, and I don't want to add them one by one.
++++++++++++++++++++++++++ Some update ++++++++++++++++++++++++++++++++++
Following #G. Grothendieck 's kind suggestion and idea, I tried to wrap the most repeatative part of the codes into a function, so I could finish all the plot with a few lines. Now the problem is that, whatever I changed the input variables, the output plot are basically the same, except the axis labels. Can anyone explain why? The following is the working code I used:
library(ggplot2)
library(ggpubr)
set.seed(100)
dat <- data.frame(
x = rnorm(100, 1),
y = rnorm(100, 10),
z = rnorm(100, 25),
lev = gl(n = 2, k = 50, labels = letters[1:2])
)
test <- function(dat, x, y){
fmt <- "%s: Adj ~ R^2 == %.3f * ',' ~ {p == %.3f}"
mod1 <- lm(y ~ x, dat, subset = lev == "a")
sum1 <- summary(mod1)
lab1 <- sprintf(fmt, "a", sum1$adj.r.squared, coef(sum1)[2, 4])
mod2 <- lm(y ~ x, dat, subset = lev == "b")
sum2 <- summary(mod2)
lab2 <- sprintf(fmt, "b", sum2$adj.r.squared, coef(sum2)[2, 4])
colors <- 1:2
p <- ggplot(dat, aes(x = x, y = y, col = lev)) +
geom_jitter() +
geom_smooth(method = "lm") +
annotate("text", x = 2, y = c(12, 8), label = c(lab1, lab2),
parse = TRUE, hjust = 0, color = colors) +
scale_color_manual(values = colors)
return(p)
}
ggarrange(test(dat, x, z), test(dat, y, z))
There are several problems here:
x, y and lev are arguments to data.frame so they must be specified using = rather than <-
make use of the subset= argument in lm
use sprintf instead of paste to simplify the specification of labels
label the text strings a and b and make them the same color as the corresponding lines to identify which is which
the formula syntax needs to be corrected. See fmt below.
it would be clearer to use component names and accessor functions of the summary objects where available
use TRUE rather than T because the latter can be overridden if there is a variable called T but TRUE can never be overridden.
use hjust=0 and adjust the x= and y= in annotate to align the two text strings
combine the annotate statements
place the individual terms of the ggplot statement on separate lines for improved readability
This gives:
library(ggplot2)
set.seed(100)
dat <- data.frame(
x = rnorm(100, 1),
y = rnorm(100, 10),
lev = gl(n = 2, k = 50, labels = letters[1:2])
)
fmt <- "%s: Adj ~ R^2 == %.3f * ',' ~ {p == %.3f}"
mod1 <- lm(y ~ x, dat, subset = lev == "a")
sum1 <- summary(mod1)
lab1 <- sprintf(fmt, "a", sum1$adj.r.squared, coef(sum1)[2, 4])
mod2 <- lm(y ~ x, dat, subset = lev == "b")
sum2 <- summary(mod2)
lab2 <- sprintf(fmt, "b", sum2$adj.r.squared, coef(sum2)[2, 4])
colors <- 1:2
ggplot(dat, aes(x = x, y = y, col = lev)) +
geom_jitter() +
geom_smooth(method = "lm") +
annotate("text", x = 2, y = c(12, 8), label = c(lab1, lab2),
parse = TRUE, hjust = 0, color = colors) +
scale_color_manual(values = colors)
Unless I'm misunderstanding your question, the problem's with the parse = T arguments to your annotate calls. I don't think your strings need to be parsed. Try parse = F instead, or just drop the parameter, as the default value seems to be FALSE anyway
I have a large dataset of gene expression from ~10,000 patient samples (TCGA), and I'm plotting a predicted expression value (x) and the actual observed value (y) of a certain gene signature. For my downstream analysis, I need to draw a precise line through the plot and calculate different parameters in samples above/below the line.
No matter how I draw a line through the data (geom_smooth(method = 'lm', 'glm', 'gam', or 'loess')), the line always seems imperfect - it doesn't cut through the data to my liking (red line is lm in figure).
After playing around for a while, I realized that the 2d kernel density lines (geom_density2d) actually do a good job of showing the slope/trends of my data, so I manually drew a line that kind of cuts through the density lines (black line in figure).
My question: how can I automatically draw a line that cuts through the kernel density lines, as for the black line in the figure? (Rather than manually playing with different intercepts and slopes till something looks good).
The best approach I can think of is to somehow calculate intercept and slope of the longest diameter for each of the kernel lines, take an average of all those intercepts and slopes and plot that line, but that's a bit out of my league. Maybe someone here has experience with this and can help?
A more hacky approach may be getting the x,y coords of each kernel density line from ggplot_build, and going from there, but it feels too hacky (and is also out of my league).
Thanks!
EDIT: Changed a few details to make the figure/analysis easier. (Density lines are smoother now).
Reprex:
library(MASS)
set.seed(123)
samples <- 10000
r <- 0.9
data <- mvrnorm(n=samples, mu=c(0, 0), Sigma=matrix(c(2, r, r, 2), nrow=2))
x <- data[, 1] # standard normal (mu=0, sd=1)
y <- data[, 2] # standard normal (mu=0, sd=1)
test.df <- data.frame(x = x, y = y)
lm(y ~ x, test.df)
ggplot(test.df, aes(x, y)) +
geom_point(color = 'grey') +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = c(2,2)) + ### EDIT: h = c(2,2)
geom_smooth(method = "glm", se = F, lwd = 1, color = 'red') +
geom_abline(intercept = 0, slope = 0.7, lwd = 1, col = 'black') ## EDIT: slope to 0.7
Figure:
I generally agree with #Hack-R.
However, it was kind of a fun problem and looking into ggplot_build is not such a big deal.
require(dplyr)
require(ggplot2)
p <- ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = c(2,2))
#basic version of your plot
p_built <- ggplot_build(p)
p_data <- p_built$data[[1]]
p_maxring <- p_data[p_data[['level']] == min(p_data[['level']]),] %>%
select(x,y) # extracts the x/y coordinates of the points on the largest ellipse from your 2d-density contour
Now this answer helped me to find the points on this ellipse which are furthest apart.
coord_mean <- c(x = mean(p_maxring$x), y = mean(p_maxring$y))
p_maxring <- p_maxring %>%
mutate (mean_dev = sqrt((x - mean(x))^2 + (y - mean(y))^2)) #extra column specifying the distance of each point to the mean of those points
coord_farthest <- c('x' = p_maxring$x[which.max(p_maxring$mean_dev)], 'y' = p_maxring$y[which.max(p_maxring$mean_dev)])
# gives the coordinates of the point farthest away from the mean point
farthest_from_farthest <- sqrt((p_maxring$x - coord_farthest['x'])^2 + (p_maxring$y - coord_farthest['y'])^2)
#now this looks which of the points is the farthest from the point farthest from the mean point :D
coord_fff <- c('x' = p_maxring$x[which.max(farthest_from_farthest)], 'y' = p_maxring$y[which.max(farthest_from_farthest)])
ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = c(2,2)) +
# geom_segment using the coordinates of the points farthest apart
geom_segment((aes(x = coord_farthest['x'], y = coord_farthest['y'],
xend = coord_fff['x'], yend = coord_fff['y']))) +
geom_smooth(method = "glm", se = F, lwd = 1, color = 'red') +
# as per your request with your geom_smooth line
coord_equal()
coord_equal is super important, because otherwise you will get super weird results - it messed up my brain too. Because if the coordinates are not set equal, the line will seemingly not pass through the point furthest apart from the mean...
I leave it to you to build this into a function in order to automate it. Also, I'll leave it to you to calculate the y-intercept and slope from the two points
Tjebo's approach was kind of good initially, but after a close look, I found that it found the longest distance between two points on an ellipse. While this is close to what I wanted, it failed with either an irregular shape of the ellipse, or the sparsity of points in the ellipse. This is because it measured the longest distance between two points; whereas what I really wanted is the longest diameter of an ellipse; i.e.: the semi-major axis. See image below for examples/details.
Briefly:
To find/draw density contours of specific density/percentage:
R - How to find points within specific Contour
To get the longest diameter ("semi-major axis") of an ellipse:
https://stackoverflow.com/a/18278767/3579613
For function that returns intercept and slope (as in OP), see last piece of code.
The two pieces of code and images below compare two Tjebo's approach vs. my new approach based on the above posts.
#### Reprex from OP
require(dplyr)
require(ggplot2)
require(MASS)
set.seed(123)
samples <- 10000
r <- 0.9
data <- mvrnorm(n=samples, mu=c(0, 0), Sigma=matrix(c(2, r, r, 2), nrow=2))
x <- data[, 1] # standard normal (mu=0, sd=1)
y <- data[, 2] # standard normal (mu=0, sd=1)
test.df <- data.frame(x = x, y = y)
#### From Tjebo
p <- ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = 2)
p_built <- ggplot_build(p)
p_data <- p_built$data[[1]]
p_maxring <- p_data[p_data[['level']] == min(p_data[['level']]),][,2:3]
coord_mean <- c(x = mean(p_maxring$x), y = mean(p_maxring$y))
p_maxring <- p_maxring %>%
mutate (mean_dev = sqrt((x - mean(x))^2 + (y - mean(y))^2)) #extra column specifying the distance of each point to the mean of those points
p_maxring = p_maxring[round(seq(1, nrow(p_maxring), nrow(p_maxring)/23)),] #### Make a small ellipse to illustrate flaws of approach
coord_farthest <- c('x' = p_maxring$x[which.max(p_maxring$mean_dev)], 'y' = p_maxring$y[which.max(p_maxring$mean_dev)])
# gives the coordinates of the point farthest away from the mean point
farthest_from_farthest <- sqrt((p_maxring$x - coord_farthest['x'])^2 + (p_maxring$y - coord_farthest['y'])^2)
#now this looks which of the points is the farthest from the point farthest from the mean point :D
coord_fff <- c('x' = p_maxring$x[which.max(farthest_from_farthest)], 'y' = p_maxring$y[which.max(farthest_from_farthest)])
farthest_2_points = data.frame(t(cbind(coord_farthest, coord_fff)))
plot(p_maxring[,1:2], asp=1)
lines(farthest_2_points, col = 'blue', lwd = 2)
#### From answer in another post
d = cbind(p_maxring[,1], p_maxring[,2])
r = ellipsoidhull(d)
exy = predict(r) ## the ellipsoid boundary
lines(exy)
me = colMeans((exy))
dist2center = sqrt(rowSums((t(t(exy)-me))^2))
max(dist2center) ## major axis
lines(exy[dist2center == max(dist2center),], col = 'red', lwd = 2)
#### The plot here is made from the data in the reprex in OP, but with h = 0.5
library(MASS)
set.seed(123)
samples <- 10000
r <- 0.9
data <- mvrnorm(n=samples, mu=c(0, 0), Sigma=matrix(c(2, r, r, 2), nrow=2))
x <- data[, 1] # standard normal (mu=0, sd=1)
y <- data[, 2] # standard normal (mu=0, sd=1)
test.df <- data.frame(x = x, y = y)
## MAKE BLUE LINE
p <- ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = 0.5) ## NOTE h = 0.5
p_built <- ggplot_build(p)
p_data <- p_built$data[[1]]
p_maxring <- p_data[p_data[['level']] == min(p_data[['level']]),][,2:3]
coord_mean <- c(x = mean(p_maxring$x), y = mean(p_maxring$y))
p_maxring <- p_maxring %>%
mutate (mean_dev = sqrt((x - mean(x))^2 + (y - mean(y))^2))
coord_farthest <- c('x' = p_maxring$x[which.max(p_maxring$mean_dev)], 'y' = p_maxring$y[which.max(p_maxring$mean_dev)])
farthest_from_farthest <- sqrt((p_maxring$x - coord_farthest['x'])^2 + (p_maxring$y - coord_farthest['y'])^2)
coord_fff <- c('x' = p_maxring$x[which.max(farthest_from_farthest)], 'y' = p_maxring$y[which.max(farthest_from_farthest)])
## MAKE RED LINE
## h = 0.5
## Given the highly irregular shape of the contours, I will use only the largest contour line (0.95) for draing the line.
## Thus, average = 1. See function below for details.
ln = long.diam("x", "y", test.df, h = 0.5, average = 1) ## NOTE h = 0.5
## PLOT
ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = 0.5) + ## NOTE h = 0.5
geom_segment((aes(x = coord_farthest['x'], y = coord_farthest['y'],
xend = coord_fff['x'], yend = coord_fff['y'])), col = 'blue', lwd = 2) +
geom_abline(intercept = ln[1], slope = ln[2], color = 'red', lwd = 2) +
coord_equal()
Finally, I came up with the following function to deal with all this. Sorry for the lack of comments/clarity
#### This will return the intercept and slope of the longest diameter (semi-major axis).
####If Average = TRUE, it will average the int and slope across different density contours.
long.diam = function(x, y, df, probs = c(0.95, 0.5, 0.1), average = T, h = 2) {
fun.df = data.frame(cbind(df[,x], df[,y]))
colnames(fun.df) = c("x", "y")
dens = kde2d(fun.df$x, fun.df$y, n = 200, h = h)
dx <- diff(dens$x[1:2])
dy <- diff(dens$y[1:2])
sz <- sort(dens$z)
c1 <- cumsum(sz) * dx * dy
levels <- sapply(probs, function(x) {
approx(c1, sz, xout = 1 - x)$y
})
names(levels) = paste0("L", str_sub(formatC(probs, 2, format = 'f'), -2))
#plot(fun.df$x,fun.df$y, asp = 1)
#contour(dens, levels = levels, labels=probs, add=T, col = c('red', 'blue', 'green'), lwd = 2)
#contour(dens, add = T, col = 'red', lwd = 2)
#abline(lm(fun.df$y~fun.df$x))
ls <- contourLines(dens, levels = levels)
names(ls) = names(levels)
lines.info = list()
for (i in 1:length(ls)) {
d = cbind(ls[[i]]$x, ls[[i]]$y)
exy = predict(ellipsoidhull(d))## the ellipsoid boundary
colnames(exy) = c("x", "y")
me = colMeans((exy)) ## center of the ellipse
dist2center = sqrt(rowSums((t(t(exy)-me))^2))
#plot(exy,type='l',asp=1)
#points(d,col='blue')
#lines(exy[order(dist2center)[1:2],])
#lines(exy[rev(order(dist2center))[1:2],])
max.dist = data.frame(exy[rev(order(dist2center))[1:2],])
line.fit = lm(max.dist$y ~ max.dist$x)
lines.info[[i]] = c(as.numeric(line.fit$coefficients[1]), as.numeric(line.fit$coefficients[2]))
}
names(lines.info) = names(ls)
#plot(fun.df$x,fun.df$y, asp = 1)
#contour(dens, levels = levels, labels=probs, add=T, col = c('red', 'blue', 'green'), lwd = 2)
#abline(lines.info[[1]], col = 'red', lwd = 2)
#abline(lines.info[[2]], col = 'blue', lwd = 2)
#abline(lines.info[[3]], col = 'green', lwd = 2)
#abline(apply(simplify2array(lines.info), 1, mean), col = 'black', lwd = 4)
if (isTRUE(average)) {
apply(simplify2array(lines.info), 1, mean)
} else {
lines.info[[average]]
}
}
Finally, here's the final implementation of the different answers:
library(MASS)
set.seed(123)
samples = 10000
r = 0.9
data = mvrnorm(n=samples, mu=c(0, 0), Sigma=matrix(c(2, r, r, 2), nrow=2))
x = data[, 1] # standard normal (mu=0, sd=1)
y = data[, 2] # standard normal (mu=0, sd=1)
#plot(x, y)
test.df = data.frame(x = x, y = y)
#### Find furthest two points of contour
## BLUE
p <- ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 2, contour = T, h = 2)
p_built <- ggplot_build(p)
p_data <- p_built$data[[1]]
p_maxring <- p_data[p_data[['level']] == min(p_data[['level']]),][,2:3]
coord_mean <- c(x = mean(p_maxring$x), y = mean(p_maxring$y))
p_maxring <- p_maxring %>%
mutate (mean_dev = sqrt((x - mean(x))^2 + (y - mean(y))^2))
coord_farthest <- c('x' = p_maxring$x[which.max(p_maxring$mean_dev)], 'y' = p_maxring$y[which.max(p_maxring$mean_dev)])
farthest_from_farthest <- sqrt((p_maxring$x - coord_farthest['x'])^2 + (p_maxring$y - coord_farthest['y'])^2)
coord_fff <- c('x' = p_maxring$x[which.max(farthest_from_farthest)], 'y' = p_maxring$y[which.max(farthest_from_farthest)])
#### Find the average intercept and slope of 3 contour lines (0.95, 0.5, 0.1), as in my long.diam function above.
## RED
ln = long.diam("x", "y", test.df)
#### Plot everything. Black line is GLM
ggplot(test.df, aes(x, y)) +
geom_point(color = 'grey') +
geom_density2d(color = 'red', lwd = 1, contour = T, h = 2) +
geom_smooth(method = "glm", se = F, lwd = 1, color = 'black') +
geom_abline(intercept = ln[1], slope = ln[2], col = 'red', lwd = 1) +
geom_segment((aes(x = coord_farthest['x'], y = coord_farthest['y'],
xend = coord_fff['x'], yend = coord_fff['y'])), col = 'blue', lwd = 1) +
coord_equal()
library(ROCR);
lig <- unique(read.table("ligands.txt")[,1]);
dec <- unique(read.table("decoys.txt")[,1]);
uniqRes <- read.table("file1.txt",header=T);
colnames(uniqRes)[1]="LigandName";
uniqRes$IsActive <- as.numeric(uniqRes$LigandName %in% lig)
predTOTALuq <- prediction(uniqRes$TOTAL*-1, uniqRes$IsActive)
perfTOTALuq <- performance(predTOTALuq, 'tpr','fpr')
jpeg("hivpr_Rinter_ROC.jpg")
plot(perfTOTALuq,main="hivpr - ROC Curves",col="blue")
abline(0,1,col="grey")
dev.off()
here is the code for plotting single curve by taking data from single file.
i want to plot 3 curves in same plot by taking data from three different files i.e. file 1, file 2, file 3
please help me to do so
you can add abline or curve directly.
df1 <- data.frame(x = 1:10, y = 1:10)
df2 <- data.frame(x = 1:13, y = 2:14)
df3 <- data.frame(x = 6:10, y = 2:6)
lx <- range(c(df1$x, df2$x, df3$x))
ly <- range(c(df1$y, df2$y, df3$y))
plot(df1, main = "hivpr - ROC Curves", xlim = lx, ylim = ly, col = "blue")
abline(0, 1, col = "blue")
points(df2, col = 'red3')
points(df3, col = 'yellow')
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.