I have managed to find online how to overlay a normal curve to a histogram in R, but I would like to retain the normal "frequency" y-axis of a histogram. See two code segments below, and notice how in the second, the y-axis is replaced with "density". How can I keep that y-axis as "frequency", as it is in the first plot.
我已经在网上找到了如何将一个正常的曲线覆盖到R的直方图上，但是我想保留直方图的正常的“频率”y轴。请参阅下面的两个代码段，并注意在第二段中，y轴被“密度”取代。如何保持y轴为“频率”，就像在第一个图中一样。
AS A BONUS: I'd like to mark the SD regions (up to 3 SD) on the density curve as well. How can I do this? I tried abline, but the line extends to the top of the graph and looks ugly.
作为奖励:我想在密度曲线上标出SD区域(最多3个SD)。我该怎么做呢?我试过abline，但这条线一直延伸到图形的顶端，看起来很难看。
g = d$mydata
hist(g)
g = d$mydata
m<-mean(g)
std<-sqrt(var(g))
hist(g, density=20, breaks=20, prob=TRUE,
xlab="x-variable", ylim=c(0, 2),
main="normal curve over histogram")
curve(dnorm(x, mean=m, sd=std),
col="darkblue", lwd=2, add=TRUE, yaxt="n")
See how in the image above, the y-axis is "density". I'd like to get that to be "frequency".
在上图中，y轴是“密度”。我想把它变成“频率”。
2 个解决方案
#1
39
Here's a nice easy way I found:
这是我找到的一个很简单的方法:
h <- hist(g, breaks = 10, density = 10,
col = "lightgray", xlab = "Accuracy", main = "Overall")
xfit <- seq(min(g), max(g), length = 40)
yfit <- dnorm(xfit, mean = mean(g), sd = sd(g))
yfit <- yfit * diff(h$mids[1:2]) * length(g)
lines(xfit, yfit, col = "black", lwd = 2)
#2
22
You just need to find the right multiplier, which can be easily calculated from the hist object.
你只需要找到正确的乘数，它可以很容易地从hist对象计算出来。
myhist <- hist(mtcars$mpg)
multiplier <- myhist$counts / myhist$density
mydensity <- density(mtcars$mpg)
mydensity$y <- mydensity$y * multiplier[1]
plot(myhist)
lines(mydensity)
A more complete version, with a normal density and lines at each standard deviation away from the mean (including the mean):
一个更完整的版本，在每个标准偏离平均值(包括平均值)的情况下，有一个正常的密度和线:
myhist <- hist(mtcars$mpg)
multiplier <- myhist$counts / myhist$density
mydensity <- density(mtcars$mpg)
mydensity$y <- mydensity$y * multiplier[1]
plot(myhist)
lines(mydensity)
myx <- seq(min(mtcars$mpg), max(mtcars$mpg), length.out= 100)
mymean <- mean(mtcars$mpg)
mysd <- sd(mtcars$mpg)
normal <- dnorm(x = myx, mean = mymean, sd = mysd)
lines(myx, normal * multiplier[1], col = "blue", lwd = 2)
sd_x <- seq(mymean - 3 * mysd, mymean + 3 * mysd, by = mysd)
sd_y <- dnorm(x = sd_x, mean = mymean, sd = mysd) * multiplier[1]
segments(x0 = sd_x, y0= 0, x1 = sd_x, y1 = sd_y, col = "firebrick4", lwd = 2)
#1
39
Here's a nice easy way I found:
这是我找到的一个很简单的方法:
h <- hist(g, breaks = 10, density = 10,
col = "lightgray", xlab = "Accuracy", main = "Overall")
xfit <- seq(min(g), max(g), length = 40)
yfit <- dnorm(xfit, mean = mean(g), sd = sd(g))
yfit <- yfit * diff(h$mids[1:2]) * length(g)
lines(xfit, yfit, col = "black", lwd = 2)
#2
22
You just need to find the right multiplier, which can be easily calculated from the hist object.
你只需要找到正确的乘数，它可以很容易地从hist对象计算出来。
myhist <- hist(mtcars$mpg)
multiplier <- myhist$counts / myhist$density
mydensity <- density(mtcars$mpg)
mydensity$y <- mydensity$y * multiplier[1]
plot(myhist)
lines(mydensity)
A more complete version, with a normal density and lines at each standard deviation away from the mean (including the mean):
一个更完整的版本，在每个标准偏离平均值(包括平均值)的情况下，有一个正常的密度和线:
myhist <- hist(mtcars$mpg)
multiplier <- myhist$counts / myhist$density
mydensity <- density(mtcars$mpg)
mydensity$y <- mydensity$y * multiplier[1]
plot(myhist)
lines(mydensity)
myx <- seq(min(mtcars$mpg), max(mtcars$mpg), length.out= 100)
mymean <- mean(mtcars$mpg)
mysd <- sd(mtcars$mpg)
normal <- dnorm(x = myx, mean = mymean, sd = mysd)
lines(myx, normal * multiplier[1], col = "blue", lwd = 2)
sd_x <- seq(mymean - 3 * mysd, mymean + 3 * mysd, by = mysd)
sd_y <- dnorm(x = sd_x, mean = mymean, sd = mysd) * multiplier[1]
segments(x0 = sd_x, y0= 0, x1 = sd_x, y1 = sd_y, col = "firebrick4", lwd = 2)