Introduction to R - model answers
D G Rossiter
2019-12-31
Exercise 1
The number of seconds in a 365-day year:
365*24*60*60## [1] 31536000Exercise 2
A workspace object which contains the number of seconds in 365-day year, and the results.
(s.in.yr <- 365*24*60*60)## [1] 31536000Exercise 3
??log10
?log10The function name for base-10 logarithms, and the base-10 logarithm of 10, 100, and 1000.
log10(10); log10(100); log10(1000)## [1] 1## [1] 2## [1] 3Exercise 4
help(rbinom)There are three arguments, all of which must all be specified: n: number of observations; size: number of trials; prob: probability of success on each trial. The value returned is a vector of length n with the number of successes in each trial.
Exercise 5
(v <- rbinom(128, 24, 0.2))## [1] 6 2 4 4 7 6 6 6 5 4 8 6 3 2 3 5 3 4 6 3 4 7 5
## [24] 3 2 3 7 3 5 4 7 3 6 9 4 5 9 6 9 4 2 1 6 5 5 4
## [47] 10 5 4 6 4 5 8 9 4 1 3 8 2 7 7 4 4 6 10 2 3 3 2
## [70] 4 5 4 3 5 4 5 4 5 6 3 7 9 5 6 5 6 7 8 4 3 4 3
## [93] 4 6 5 2 8 6 5 4 5 5 6 5 2 7 5 5 1 6 7 4 2 5 9
## [116] 3 5 3 5 3 1 4 5 4 8 11 2 2The vector of the number of successes in 24 trials with probability of success 0.2 (20%), this simulation carried out 128 times.
Exercise 6
table(v <- rbinom(128, 24, 0.2))##
## 1 2 3 4 5 6 7 8 9 10 11 12
## 7 11 22 23 31 15 8 5 3 1 1 1The range is from 1 to 12; the modal value is 5
Exercise 7
A vector representing latitudes in degrees from \(0^\circ\) (equator) to \(+90^\circ\) (north pole), in intervales of \(5^\circ\):
(angles <- seq(0, 90, by=5))## [1] 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90Their cosines and its maximum value:
deg.rad <- pi/180
angles*deg.rad## [1] 0.00000000 0.08726646 0.17453293 0.26179939 0.34906585 0.43633231
## [7] 0.52359878 0.61086524 0.69813170 0.78539816 0.87266463 0.95993109
## [13] 1.04719755 1.13446401 1.22173048 1.30899694 1.39626340 1.48352986
## [19] 1.57079633round(angles.cos <- cos(angles*deg.rad),4)## [1] 1.0000 0.9962 0.9848 0.9659 0.9397 0.9063 0.8660 0.8192 0.7660 0.7071
## [11] 0.6428 0.5736 0.5000 0.4226 0.3420 0.2588 0.1736 0.0872 0.0000max(angles.cos)## [1] 1Exercise 8
Installing gstat:
install.packages("gstat", dependencies=TRUE)Or, use the Install toolbar button of Packages tab of the Help RStudio pane.
Loading gstat into the workspace:
library(gstat)Or, click next to the package’s name in Packages tab of the Help RStudio pane.
Help for the variogram function:
help(variogram, package="gstat")Description: “Calculates the sample variogram from data, or in case of a linear model is given, for the residuals, with options for directional, robust, and pooled variogram, and for irregular distance intervals”
Exercise 9
class(pi)## [1] "numeric"class(letters)## [1] "character"Exercise 10
help(variogram)The variogram function returns an object of class gstatVariogram.
Exercise 11
Datasets in the gstat package:
data(package="gstat")Exercise 12
library(gstat)
data(oxford)
summary(oxford)## PROFILE XCOORD YCOORD ELEV PROFCLASS
## Min. : 1.00 Min. :100 Min. : 100 Min. :540.0 Cr:19
## 1st Qu.: 32.25 1st Qu.:200 1st Qu.: 600 1st Qu.:558.0 Ct:36
## Median : 63.50 Median :350 Median :1100 Median :573.0 Ia:71
## Mean : 63.50 Mean :350 Mean :1100 Mean :573.6
## 3rd Qu.: 94.75 3rd Qu.:500 3rd Qu.:1600 3rd Qu.:584.5
## Max. :126.00 Max. :600 Max. :2100 Max. :632.0
## MAPCLASS VAL1 CHR1 LIME1 VAL2
## Cr:31 Min. :2.000 Min. :1.000 Min. :0.000 Min. :4.00
## Ct:36 1st Qu.:3.000 1st Qu.:2.000 1st Qu.:1.000 1st Qu.:4.00
## Ia:59 Median :4.000 Median :2.000 Median :4.000 Median :8.00
## Mean :3.508 Mean :2.468 Mean :2.643 Mean :6.23
## 3rd Qu.:4.000 3rd Qu.:3.000 3rd Qu.:4.000 3rd Qu.:8.00
## Max. :4.000 Max. :4.000 Max. :4.000 Max. :8.00
## CHR2 LIME2 DEPTHCM DEP2LIME
## Min. :2 Min. :0.000 Min. :10.00 Min. :20.00
## 1st Qu.:2 1st Qu.:4.000 1st Qu.:25.00 1st Qu.:20.00
## Median :2 Median :5.000 Median :36.00 Median :20.00
## Mean :3 Mean :3.889 Mean :46.25 Mean :30.32
## 3rd Qu.:4 3rd Qu.:5.000 3rd Qu.:64.75 3rd Qu.:40.00
## Max. :6 Max. :5.000 Max. :91.00 Max. :90.00
## PCLAY1 PCLAY2 MG1 OM1
## Min. :10.00 Min. :10.00 Min. : 19.00 Min. : 2.600
## 1st Qu.:20.00 1st Qu.:10.00 1st Qu.: 44.00 1st Qu.: 4.100
## Median :24.50 Median :10.00 Median : 72.00 Median : 5.350
## Mean :24.44 Mean :14.76 Mean : 93.53 Mean : 5.995
## 3rd Qu.:28.00 3rd Qu.:20.00 3rd Qu.:123.25 3rd Qu.: 7.175
## Max. :37.00 Max. :40.00 Max. :308.00 Max. :13.100
## CEC1 PH1 PHOS1 POT1
## Min. : 7.00 Min. :4.200 Min. : 1.700 Min. : 83.0
## 1st Qu.:12.00 1st Qu.:7.200 1st Qu.: 6.200 1st Qu.:127.0
## Median :15.00 Median :7.500 Median : 8.500 Median :164.0
## Mean :18.88 Mean :7.152 Mean : 8.752 Mean :181.7
## 3rd Qu.:25.25 3rd Qu.:7.600 3rd Qu.:10.500 3rd Qu.:194.8
## Max. :43.00 Max. :7.700 Max. :25.000 Max. :847.0str(oxford)## 'data.frame': 126 obs. of 22 variables:
## $ PROFILE : num 1 2 3 4 5 6 7 8 9 10 ...
## $ XCOORD : num 100 100 100 100 100 100 100 100 100 100 ...
## $ YCOORD : num 2100 2000 1900 1800 1700 1600 1500 1400 1300 1200 ...
## $ ELEV : num 598 597 610 615 610 595 580 590 598 588 ...
## $ PROFCLASS: Factor w/ 3 levels "Cr","Ct","Ia": 2 2 2 3 3 2 3 2 3 3 ...
## $ MAPCLASS : Factor w/ 3 levels "Cr","Ct","Ia": 2 3 3 3 3 2 2 3 3 3 ...
## $ VAL1 : num 3 3 4 4 3 3 4 4 4 3 ...
## $ CHR1 : num 3 3 3 3 3 2 2 3 3 3 ...
## $ LIME1 : num 4 4 4 4 4 0 2 1 0 4 ...
## $ VAL2 : num 4 4 5 8 8 4 8 4 8 8 ...
## $ CHR2 : num 4 4 4 2 2 4 2 4 2 2 ...
## $ LIME2 : num 4 4 4 5 5 4 5 4 5 5 ...
## $ DEPTHCM : num 61 91 46 20 20 91 30 61 38 25 ...
## $ DEP2LIME : num 20 20 20 20 20 20 20 20 40 20 ...
## $ PCLAY1 : num 15 25 20 20 18 25 25 35 35 12 ...
## $ PCLAY2 : num 10 10 20 10 10 20 10 20 10 10 ...
## $ MG1 : num 63 58 55 60 88 168 99 59 233 87 ...
## $ OM1 : num 5.7 5.6 5.8 6.2 8.4 6.4 7.1 3.8 5 9.2 ...
## $ CEC1 : num 20 22 17 23 27 27 21 14 27 20 ...
## $ PH1 : num 7.7 7.7 7.5 7.6 7.6 7 7.5 7.6 6.6 7.5 ...
## $ PHOS1 : num 13 9.2 10.5 8.8 13 9.3 10 9 15 12.6 ...
## $ POT1 : num 196 157 115 172 238 164 312 184 123 282 ...Exercise 13
data(women)
dim(women)## [1] 15 2colnames(women)## [1] "height" "weight"row.names(women)## [1] "1" "2" "3" "4" "5" "6" "7" "8" "9" "10" "11" "12" "13" "14"
## [15] "15"women[1,"height"]## [1] 58There are 15 observations (cases) and 2 attributes (fields) for each case. The column (field) names are height, weight and the row (cases) names are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. The first woman is 58 inches tall.
Exercise 14
names(which(sapply(oxford, is.factor)))## [1] "PROFCLASS" "MAPCLASS"Exercise 15
trees$hg <- trees$Height/trees$Girth
# thin trees have a height/girth ratio more than 1 s.d. above the mean
(thin.trees <- subset(trees, hg > (mean(trees$hg) + sd(trees$hg))))## Girth Height Volume hg
## 1 8.3 70 10.3 8.433735
## 2 8.6 65 10.3 7.558140
## 5 10.7 81 18.8 7.570093
## 6 10.8 83 19.7 7.685185
## 9 11.1 80 22.6 7.207207Exercise 16
data(diamonds, package="ggplot2")
hist(diamonds$price)
Exercise 17
model.hg <- lm(Height ~ Girth, data=trees)
# equivalent to: model <- lm(trees$Volume ~ trees$Height)
summary(model.hg)##
## Call:
## lm(formula = Height ~ Girth, data = trees)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.5816 -2.7686 0.3163 2.4728 9.9456
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 62.0313 4.3833 14.152 1.49e-14 ***
## Girth 1.0544 0.3222 3.272 0.00276 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.538 on 29 degrees of freedom
## Multiple R-squared: 0.2697, Adjusted R-squared: 0.2445
## F-statistic: 10.71 on 1 and 29 DF, p-value: 0.002758Only 24.4% of the variance in tree heights can be explained by its girth.
Exercise 18
model.vhg <- lm(Volume ~ Height + Girth, data=trees)
summary(model.vhg)##
## Call:
## lm(formula = Volume ~ Height + Girth, data = trees)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.4065 -2.6493 -0.2876 2.2003 8.4847
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -57.9877 8.6382 -6.713 2.75e-07 ***
## Height 0.3393 0.1302 2.607 0.0145 *
## Girth 4.7082 0.2643 17.816 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.882 on 28 degrees of freedom
## Multiple R-squared: 0.948, Adjusted R-squared: 0.9442
## F-statistic: 255 on 2 and 28 DF, p-value: < 2.2e-16Exercise 19
limit.01 <- function(v) {
changed <- 0
ix <- which(v < 0); v[ix] <- 0
changed <- changed+length(ix)
ix <- which(v > 1); v[ix] <- 1
changed <- changed+length(ix)
print(paste("Number of elements limited to 0..1:", changed))
return(v)
}Test of this function:
(test.v <- seq(-0.2, 1.2, by=0.1))## [1] -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1
## [15] 1.2limit.01(test.v)## [1] "Number of elements limited to 0..1: 5"## [1] 0.0 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.0 1.0Bonus Exercise
library(dplyr)
data(trees)
names(trees)## [1] "Girth" "Height" "Volume"trees %>%
filter(Volume > median(Volume)) %>%
mutate(thickness=round(Girth/Height,3)) %>%
arrange(thickness)## Girth Height Volume thickness
## 1 12.9 85 33.8 0.152
## 2 13.3 86 27.4 0.155
## 3 14.2 80 31.7 0.178
## 4 14.0 78 34.5 0.179
## 5 13.7 71 25.7 0.193
## 6 14.5 74 36.3 0.196
## 7 16.3 77 42.6 0.212
## 8 17.5 82 55.7 0.213
## 9 17.3 81 55.4 0.214
## 10 13.8 64 24.9 0.216
## 11 16.0 72 38.3 0.222
## 12 17.9 80 58.3 0.224
## 13 18.0 80 51.5 0.225
## 14 18.0 80 51.0 0.225
## 15 20.6 87 77.0 0.237