Homework 2
STAT 486/586 – Introduction to Statistical Computing
Yuhang (Tom) LIN
Due: February 8, 2022
R commands
### Create the score vector scores <- c(9.3, 8.7, 9.5, 10, 9, 8.9) ### Indicator for minimum and maximum scores I <- c(which.min(scores), which.max(scores)) ### Compute the average after removing the minimum and maximum (ave <- mean(scores[-I]))[1] 9.175### Method 1: Rep only ndays <- rep(c(366, rep(365, 3)), length.out=21) names(ndays) <- 2000:2020 ndays2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 366 365 365 365 366 365 365 365 366 365 365 365 366 365 365 365 2016 2017 2018 2019 2020 366 365 365 365 366### Method 2: Rep with the recycling rule ndays.recy <- rep(365, 21) + c(1, rep(0, 3))Warning in rep(365, 21) + c(1, rep(0, 3)): longer object length is not a multiple of shorter object lengthnames(ndays.recy) <- 2000:2020 ndays.recy2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 366 365 365 365 366 365 365 365 366 365 365 365 366 365 365 365 2016 2017 2018 2019 2020 366 365 365 365 366### Check if they give identical outputs identical(ndays, ndays.recy)[1] TRUE### Choose the range of i i <- 1: 100000 ### Compute the desired value x <- ((-1)^(i+1))/i ### First few entries x[1: 10][1] 1.0000000 -0.5000000 0.3333333 -0.2500000 0.2000000 -0.1666667 [7] 0.1428571 -0.1250000 0.1111111 -0.1000000### Compute the summation (result <- sum(x))[1] 0.6931422Increase the length of
ito see if the summation result varies a lot. It is a good approximation if results are similar. For example, we can increase the maximum ofifrom 100,000 to 1,000,000.### New range of i i.new <- 1: 1000000 ### Compute the desired value x.new <- ((-1)^(i.new+1))/i.new ### Compute the summation (result.new <- sum(x.new))[1] 0.6931467### Difference in result abs(result.new - result)[1] 4.499975e-06The difference is small, which indicates our approximations are good.
### Create height vector height <- c(Chris=1.8, Mary=1.65) ### Create weight vector weight <- c(70, 49) ### Compute BMI (bmi <- weight/(height^2))Chris Mary 21.60494 17.99816#### Create the vector x <- c(1, 1, 4, 6, 4, 7, 5, 9)### Sort the vector sort(x)[1] 1 1 4 4 5 6 7 9### Default rank rank(x)[1] 1.5 1.5 3.5 6.0 3.5 7.0 5.0 8.0Note that numbers with
0.5indicate there are ties.### Rank of minimum rank(x, ties.method="min")[1] 1 1 3 6 3 7 5 8### The number of unique elements in x length(unique(x))[1] 6### Method 1: Use diff function sum(diff(x) < 0)[1] 2### Method 2: Compare subsequences sum(x[-1] < x[-length(x)])[1] 2### Get the order of "distance" I <- order(abs(x - 5)) ### Four numbers closest to 5 (z <- x[I][1:4])[1] 5 4 6 4### Average mean(z)[1] 4.75### Numbers that differ from 5 by at most 2 (w <- x[abs(x - 5) <= 2])[1] 4 6 4 7 5### Average mean(w)[1] 5.2
trunc(.5 + -2:4)[1] -1 0 0 1 2 3 4round(.5 + -2:4)[1] -2 0 0 2 2 4 4Because we have
.5 + -2:4[1] -1.5 -0.5 0.5 1.5 2.5 3.5 4.5truncwould truncate the values toward 0, whileroundwould “round to even” (go to the even digit) when rounding off a 5.
Penguins
### Load library
library(palmerpenguins)### Inspect the data frame penguinsstr(penguins)tibble [344 x 8] (S3: tbl_df/tbl/data.frame) $ species : Factor w/ 3 levels "Adelie","Chinstrap",..: 1 1 1 1 1 1 1 1 1 1 ... $ island : Factor w/ 3 levels "Biscoe","Dream",..: 3 3 3 3 3 3 3 3 3 3 ... $ bill_length_mm : num [1:344] 39.1 39.5 40.3 NA 36.7 39.3 38.9 39.2 34.1 42 ... $ bill_depth_mm : num [1:344] 18.7 17.4 18 NA 19.3 20.6 17.8 19.6 18.1 20.2 ... $ flipper_length_mm: int [1:344] 181 186 195 NA 193 190 181 195 193 190 ... $ body_mass_g : int [1:344] 3750 3800 3250 NA 3450 3650 3625 4675 3475 4250 ... $ sex : Factor w/ 2 levels "female","male": 2 1 1 NA 1 2 1 2 NA NA ... $ year : int [1:344] 2007 2007 2007 2007 2007 2007 2007 2007 2007 2007 ...head(penguins)summary(penguins)species island bill_length_mm bill_depth_mm Adelie :152 Biscoe :168 Min. :32.10 Min. :13.10 Chinstrap: 68 Dream :124 1st Qu.:39.23 1st Qu.:15.60 Gentoo :124 Torgersen: 52 Median :44.45 Median :17.30 Mean :43.92 Mean :17.15 3rd Qu.:48.50 3rd Qu.:18.70 Max. :59.60 Max. :21.50 NA's :2 NA's :2 flipper_length_mm body_mass_g sex year Min. :172.0 Min. :2700 female:165 Min. :2007 1st Qu.:190.0 1st Qu.:3550 male :168 1st Qu.:2007 Median :197.0 Median :4050 NA's : 11 Median :2008 Mean :200.9 Mean :4202 Mean :2008 3rd Qu.:213.0 3rd Qu.:4750 3rd Qu.:2009 Max. :231.0 Max. :6300 Max. :2009 NA's :2 NA's :2### Get the decreasing indicator I <- order(penguins$body_mass_g, decreasing=TRUE) ### The species of the 5 largest penguins (large.spec <- penguins[I[1:5], "species"])### Tabulate the result in the previous part table(large.spec)large.spec Adelie Chinstrap Gentoo 0 0 5### Mean without NA values mean(penguins$body_mass_g, na.rm=TRUE)[1] 4201.754### Median without NA values median(penguins$body_mass_g, na.rm=TRUE)[1] 4050### Histogram hist(penguins$body_mass_g, xlab="Body mass", ylab="Count", main="Histogram for the body mass")
### Side-by-side boxplot boxplot(body_mass_g ~ sex, data=penguins, xlab="Sex", ylab="Body mass", main="Side-by-side boxplot for the body mass")
### Bar chart barplot(table(penguins$species), xlab="Species", ylab="Count", main="Barchart for the number of penguins with different species")
### Create a new level for NA values penguins$sex <- as.character(penguins$sex) penguins$sex[is.na(penguins$sex)] <- "not applicable" penguins$sex <- factor(penguins$sex) ### Group indicator for plotting I <- 1:length(levels(penguins$sex)) ### Scatterplot plot(penguins$flipper_length_mm, penguins$body_mass_g, col=as.numeric(penguins$sex), pch=as.numeric(penguins$sex), xlab="Flipper length", ylab="Body mass", main="Scatterplot for body mass versus flipper length") ### Create legend legend("topleft", legend=levels(penguins$sex), col=I, pch=I)