问题
Original question
I want to smooth my explanatory variable, something like Speed data of a vehicle, and then use this smoothed values. I searched a lot, and find nothing that directly is my answer.
I know how to calculate the kernel density estimation (density() or KernSmooth::bkde()) but I don't know then how to calculate the smoothed values of speed.
Re-edited question
Thanks to @ZheyuanLi, I am able to better explain what I have and what I want to do. So I have re-edited my question as below.
I have some speed measurement of a vehicle during a time, stored as a data frame vehicle:
t speed
1 0 0.0000000
2 1 0.0000000
3 2 0.0000000
4 3 0.0000000
5 4 0.0000000
. . .
. . .
1031 1030 4.8772222
1032 1031 4.4525000
1033 1032 3.2261111
1034 1033 1.8011111
1035 1034 0.2997222
1036 1035 0.2997222
Here is a scatter plot:
I want to smooth speed against t, and I want to use kernel smoothing for this purpose. According to @Zheyuan's advice, I should use ksmooth():
fit <- ksmooth(vehicle$t, vehicle$speed)
However, I found that the smoothed values are exactly the same as my original data:
sum(abs(fit$y - vehicle$speed)) # 0
Why is this happening? Thanks!
回答1:
Answer to old question
You need to distinguish "kernel density estimation" and "kernel smoothing".
Density estimation, only works with a single variable. It aims to estimate how spread out this variable is on its physical domain. For example, if we have 1000 normal samples:
x <- rnorm(1000, 0, 1)
We can assess its distribution by kernel density estimator:
k <- density(x)
plot(k); rug(x)
The rugs on the x-axis shows the locations of your x values, while the curve measures the density of those rugs.
Kernel smoother, is actually a regression problem, or scatter plot smoothing problem. You need two variables: one response variable y, and an explanatory variable x. Let's just use the x we have above for the explanatory variable. For response variable y, we generate some toy values from
y <- sin(x) + rnorm(1000, 0, 0.2)
Given the scatter plot between y and x:
we want to find a smooth function to approximate those scattered dots.
The Nadaraya-Watson kernel regression estimate, with R function ksmooth() will help you:
s <- ksmooth(x, y, kernel = "normal")
plot(x,y, main = "kernel smoother")
lines(s, lwd = 2, col = 2)
If you want to interpret everything in terms of prediction:
- kernel density estimation: given
x, predict density ofx; that is, we have an estimate of the probabilityP(grid[n] < x < grid[n+1]), wheregridis some gird points; - kernel smoothing: given
x, predicty; that is, we have an estimate of the functionf(x), which approximatesy.
In both cases, you have no smoothed value of explanatory variable x. So your question: "I want to smooth my explanatory variable" makes no sense.
Do you actually have a time series?
"Speed of a vehicle" sounds like you are monitoring the speed along time t. If so, get a scatter plot between speed and t, and use ksmooth().
Other smoothing approach like loess() and smooth.spline() are not of kernel smoothing class, but you can compare.
回答2:
Answer on re-edited question
The default bandwidth for ksmooth() is 0.5:
ksmooth(x, y, kernel = c("box", "normal"), bandwidth = 0.5,
range.x = range(x),
n.points = max(100L, length(x)), x.points)
For you time series data with lag 1, this means there will be no other speed data in the neighbourhood (i-0.5, i+0.5), for time t = i, except speed[i]. As a result, no local weighted average is done!
You need to choose a larger bandwidth. For example, if we hope to average over 20 values, we should set bandwidth = 10 (not 20 as it is two-sided). This is what we get:
fit <- ksmooth(vehicle$t, vehicle$speed, bandwidth = 10)
plot(vehicle, cex = 0.5)
lines(fit,col=2,lwd = 2)
Smoothness selection
One problem with ksmooth(), is that you must set bandwidth yourself. You can see that this parameter shapes the fitted curve greatly. Large bandwidth makes the curve smooth, but far away from data; while small bandwidth does the reverse.
Is there an optimal bandwidth? Is there a way to select the best one?
Yes, use sm.regression() from sm package, with cross-validation method for selecting bandwidth.
fit <- sm.regression(vehicle$t, vehicle$speed, method = "cv", eval.points = 0:1035)
## plot will be automatically generated!
You can check that fit$h is 18.7.
Other approach
Perhaps you think sm.regression() oversmooths your data? Well, use loess(), or my favourite: smooth.spline().
I had an answer:
- regarding
smooth.spline()at smooth.spline(): fitted model does not match user-specified degree of freedom; this one is very technical! - regarding
smooth.spline()at R smooth.spline(): smoothing spline is not smooth but overfitting my data; this one is practical modelling. - regarding
loess()at Problems displaying LOESS regression line and confidence interval; this one is about general use ofloess().
Here, I would demonstrate the use of smooth.spline():
fit <- smooth.spline(vehicle$t, vehicle$speed, all.knots = TRUE, control.spar = list(low = -2, hight = 2))
# Call:
# smooth.spline(x = vehicle$t, y = vehicle$speed, all.knots = TRUE,
# control.spar = list(low = -2, hight = 2))
# Smoothing Parameter spar= 0.2519922 lambda= 4.379673e-11 (14 iterations)
# Equivalent Degrees of Freedom (Df): 736.0882
# Penalized Criterion: 3.356859
# GCV: 0.03866391
plot(vehicle, cex = 0.5)
lines(fit$x, fit$y, col = 2, lwd = 2)
Or using its regression spline version:
fit <- smooth.spline(vehicle$t, vehicle$speed, nknots = 200)
plot(vehicle, cex = 0.5)
lines(fit$x, fit$y, col = 2, lwd = 2)
You really need to read my first link above, to understand why I use control.spar in the first case, while without it in the second case.
More powerful package
I would definitely recommend mgcv. I have several answers regarding mgcv, but I don't want to overwhelm you. So, I will not make extension here. Learn to use ksmooth(), smooth.spline() and loess() well. In future, when you meet more complicated problem, come back to stack overflow and ask for help!
来源：https://stackoverflow.com/questions/37952793/scatter-plot-kernel-smoothing-ksmooth-does-not-smooth-my-data-at-all