如何以置信区间绘制数据？

Question

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?

Answer 1

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')

Answer 2

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"))

Answer 3

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

Answer 4

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.