Computer Lab Week 9: solutions
STAT221
In this computer lab we will use a dataset on the chemical composition of wines. This dataset can be downloaded here
# Read Data into R Warning: whole address must be in one line
wine.fl = “http://archive.ics.uci.edu/ml/machine-learning-databases/wine/wine.data” wine = read.csv(wine.fl, header = F)
# Names of the variables
wine.names = c(“Alcohol”, “Malic acid”, “Ash”, “Alcalinity of ash”, “Magnesium”,
“Total phenols”, “Flavanoids”, “Nonflavanoid phenols”, “Proanthocyanins”, “Color intensity”, “Hue”, “OD.Ratio”, “Proline”)
colnames(wine)[2:14] = wine.names colnames(wine)[1] = “Class”
Question 1.
Produce a scatterplot of OD.Ratio versus Proanthocyanins from the wine data set and superimpose onto it a line representing a box kernel smoother using the default bandwidth.
plot(wine$Proanthocyanins, wine$OD.Ratio) lines(ksmooth(wine$Proanthocyanins, wine$OD.Ratio))
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wine$Proanthocyanins
Comment: While the line looks ¡®choppy¡¯ at all points, it looks most choppy in areas where there is least data.
Question 2.
Look up the help page for ksmooth, and then play around with different values of the bandwidth to see if you can get a smoother or less choppy line, but still using the box kernel. Add these lines to the scatterp[lot using different colours for each line.]
plot(wine$Proanthocyanins, wine$OD.Ratio) lines(ksmooth(wine$Proanthocyanins, wine$OD.Ratio), col = “red”)
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wine$OD.Ratio
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lines(ksmooth(wine$Proanthocyanins, wine$OD.Ratio, bandwidth = 0.1), col = “blue”) lines(ksmooth(wine$Proanthocyanins, wine$OD.Ratio, bandwidth = 1), col = “green”)
Question 3.
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wine$Proanthocyanins
Now switch to using the ¡®¡¯normal¡± kernel in ksmooth. Again produce a scatterplot of OD.Ratio versus Proanthocyanins and superimpose onto it a line representing a normal kernel smoother using the default bandwidth.
plot(wine$Proanthocyanins, wine$OD.Ratio) lines(ksmooth(wine$Proanthocyanins, wine$OD.Ratio, “normal”))
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wine$Proanthocyanins
Comment: The line is now smooth, and we can see an almost linear relationship between the two variables up to a value of Proanthocyanins of around 2.0.
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wine$OD.Ratio wine$OD.Ratio
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Question 4.
Now plot a default lowess smoother and a default loess smoother, again superimposed on a scatterplot of OD.Ratio versus Proanthocyanins.
# the lowess smoother
plot(wine$Proanthocyanins, wine$OD.Ratio) lines(lowess(wine$Proanthocyanins, wine$OD.Ratio), col = “red”, lwd = 2)
# the loess smoother
Proanthocyanins = sort(wine$Proanthocyanins)
OD.Ratio = wine$OD.Ratio[order(wine$Proanthocyanins)] #plot(Proanthocyanins, OD.Ratio)
loess.fit = loess(OD.Ratio ~ Proanthocyanins) lines(Proanthocyanins, predict(loess.fit), col = “blue”, lwd = 2)
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wine$Proanthocyanins
Why are they different? – think about what the defaults are for each of these smoothers.
Play around with the default settings and see if you can get the two lines to be identical: the idea here is just to play around with the different arguments and learn something about how they work, by having a goal of trying to get the lines to be the same. So don¡¯t worry too much if you can¡¯t get them to be identical, just give it a go.
# the lowess smoother
plot(wine$Proanthocyanins, wine$OD.Ratio)
lines(lowess(wine$Proanthocyanins, wine$OD.Ratio, f = 0.3), col = “red”, lwd = 3)
# the loess smoother
Proanthocyanins = sort(wine$Proanthocyanins)
OD.Ratio = wine$OD.Ratio[order(wine$Proanthocyanins)] #plot(Proanthocyanins, OD.Ratio)
loess.fit = loess(OD.Ratio ~ Proanthocyanins,
span = 0.3, family = “symmetric”, degree = 1, surface = “direct”) lines(Proanthocyanins, predict(loess.fit), col = “blue”, lwd = 1)
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wine$OD.Ratio
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wine$Proanthocyanins
If you get the output below for your loess smoother, then you have forgotten to sort your data into order so that R can draw a smooth curve using lines . . .
plot(wine$Proanthocyanins, wine$OD.Ratio)
loess.fit = loess(wine$OD.Ratio ~ wine$Proanthocyanins) lines(wine$Proanthocyanins, predict(loess.fit), col = “red”)
Question 5.
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wine$Proanthocyanins
Examine the effect of changing the bandwidth by producing 10 different plots arranged in a grid with two columns and five rows (using mfrow). For each of these 10 plots, produce a scatterplot of OD.Ratio versus Proanthocyanins and superimpose onto each one a single line representing a normal kernel smoother using the bandwidths 0.1, 0.2, . . . , 1.0. Add a title to each plot that says what the bandwidth is of the smoother shown in that plot.
The plots are shown below to give you an idea of what you need to produce.
One approach is to loop around the code that produces a single plot, changing the bandwidth each time.
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wine$OD.Ratio wine$OD.Ratio
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Another approach might be to write a function that produces a single plot and which has the bandwidth as its argument, and then run that function for each of the 10 bandwidths. The R function lapply might be useful for doing this.
Finally, it might be useful to play around with the margins around the outside of each of the 10 plots to make them narrower. Have a look at the help for par using ?par, and then read about the graphical parameter called mar. You can adjust the default settings of mar to reduce the size of the margins. The default settings are par(mfrow=c(1,1), mar=c(5,4,4,2)).
kplotfn = function(bw){
plot(wine$Proanthocyanins, wine$OD.Ratio, main=bw, ylab=”OD.ratio”) lines(ksmooth(wine$Proanthocyanins, wine$OD.Ratio, “normal”, bandwidth=bw),
}
col=’red’, lwd=2)
bws = seq(.1, 1, by = .1) par(mfrow=c(5,2), mar=c(2,4,2,1)+.1) lapply(bws, kplotfn)
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par(mfrow=c(1,1))
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wine$Proanthocyanins
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wine$Proanthocyanins
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wine$Proanthocyanins
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wine$Proanthocyanins
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wine$Proanthocyanins
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wine$Proanthocyanins
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wine$Proanthocyanins
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wine$Proanthocyanins
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OD.ratio
OD.ratio
OD.ratio
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OD.ratio
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OD.ratio
OD.ratio
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OD.ratio
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