Exercise 1 Download the dataframe pirates.txt from http://nathanieldphillips.com/wp-content/uploads/2015/05/pirate_survey.txt. The data are stored in a comma-separated text file with headers. Load the dataframe into an object called pirates.
pirates <- read.table(
file= "http://nathanieldphillips.com/wp-content/uploads/2015/05/pirate_survey.txt",
header=T,
sep ="\t",
stringsAsFactors=F)
head(pirates)## id sex headband age college tattoos tchests.found parrots.lifetime
## 1 1 female yes 37 JSSFP 9 6 2
## 2 2 female yes 34 JSSFP 14 6 5
## 3 3 male no 29 CCCC 9 1 13
## 4 4 female yes 30 JSSFP 9 1 5
## 5 5 male yes 16 CCCC 12 4 1
## 6 6 male yes 27 CCCC 6 1 1
## favorite.pirate sword.type sword.speed
## 1 Hook cutlass 0.2572612
## 2 Lewis Scot cutlass 0.2935024
## 3 Jack Sparrow scimitar 0.8698797
## 4 Lewis Scot cutlass 0.1832235
## 5 Hook cutlass 0.3754281
## 6 Jack Sparrow cutlass 0.1542810Exercise 2 Let’s clean up the dataframe. Some of the values don’t seem to be appropriate. For example, when I look at the column sex, I see some bad values. For each of the columns, try to figure out which values are appropriate (hint: use table()), and recode all inappropriate values as NA.
table(pirates$sex)##
## female male other
## 482 477 41pirates$sex[!(pirates$sex %in% c("male", "female", "other/NA"))] <- NA
table(pirates$sex)##
## female male
## 482 477table(pirates$sex, useNA = "always")##
## female male <NA>
## 482 477 41table(pirates$headband)##
## no yes
## 97 903pirates$headband[!(pirates$headband %in% c("yes", "no"))] <- NA
table(pirates$headband, useNA = "always")##
## no yes <NA>
## 97 903 0summary(pirates$age)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 6.00 24.00 28.00 27.64 31.00 46.00pirates$age[!(pirates$age %in% c(15:120))] <- NA
summary(pirates$age, useNA = "always")## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 15.00 24.00 28.00 27.77 31.00 46.00 8table(pirates$age, useNA = "always")##
## 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
## 3 13 11 21 22 30 49 42 44 48 58 64 69 74 72
## 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44
## 76 56 34 46 42 27 30 16 15 11 9 3 1 1 3
## 45 46 <NA>
## 1 1 8table(pirates$college)##
## CCCC JSSFP
## 653 347table(pirates$tattoos)##
## 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
## 4 7 7 19 41 47 48 74 100 123 138 101 104 84 51 23 13 11
## 18 20
## 3 2summary(pirates$tattoos)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.000 8.000 10.000 9.569 12.000 20.000pirates$tattoos[!(pirates$tattoos %in% c(0:1000))] <- NA
summary(pirates$tattoos)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.000 8.000 10.000 9.569 12.000 20.000table(pirates$tattoos, useNA = "always")##
## 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
## 4 7 7 19 41 47 48 74 100 123 138 101 104 84 51
## 15 16 17 18 20 <NA>
## 23 13 11 3 2 0summary(pirates$tchests.found)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.000 2.000 5.000 7.414 11.000 51.000summary(pirates$parrots.lifetime)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.000 1.000 2.000 2.825 4.000 25.000pirates$parrots.lifetime[!(pirates$parrots.lifetime %in% c(1:22))] <- NA
summary(pirates$parrots.lifetime)## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 1.00 1.00 3.00 3.41 4.00 19.00 179table(pirates$parrots.lifetime, useNA = "always")##
## 1 2 3 4 5 6 7 8 9 10 11 12 13 14 17
## 230 176 130 84 54 42 34 21 17 9 7 3 7 2 3
## 19 <NA>
## 2 179table(pirates$favorite.pirate)##
## Anicetus Blackbeard Edward Low Hook Jack Sparrow
## 146 107 105 102 426
## Lewis Scot
## 114pirates$favorite.pirate[!(pirates$favorite.pirate %in% c("Anicetus", "Blackbeard", "Edward Low", "Hook", "Jack Sparrow", "Lewis Scot"))] <- NA
table(pirates$favorite.pirate, useNA = "always")##
## Anicetus Blackbeard Edward Low Hook Jack Sparrow
## 146 107 105 102 426
## Lewis Scot <NA>
## 114 0head(pirates)## id sex headband age college tattoos tchests.found parrots.lifetime
## 1 1 female yes 37 JSSFP 9 6 2
## 2 2 female yes 34 JSSFP 14 6 5
## 3 3 male no 29 CCCC 9 1 13
## 4 4 female yes 30 JSSFP 9 1 5
## 5 5 male yes 16 CCCC 12 4 1
## 6 6 male yes 27 CCCC 6 1 1
## favorite.pirate sword.type sword.speed
## 1 Hook cutlass 0.2572612
## 2 Lewis Scot cutlass 0.2935024
## 3 Jack Sparrow scimitar 0.8698797
## 4 Lewis Scot cutlass 0.1832235
## 5 Hook cutlass 0.3754281
## 6 Jack Sparrow cutlass 0.1542810table(pirates$sword.type)##
## banana cutlass sabre scimitar
## 42 827 65 66summary(pirates$sword.speed)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.001725 0.209800 0.504700 1.051000 1.123000 19.040000Exercise 3 A fellow pirate captain wants to know if there is a relationship between the sex of my pirates and the number of treasure chests they have found. Using aggregate() figure out the mean number of treasure chests found by males, females, and other/NA.
table(pirates$sex)##
## female male
## 482 477aggregated.data <- aggregate(tchests.found ~ sex, FUN = mean, na.rm = T, data = pirates)
aggregated.data## sex tchests.found
## 1 female 8.095436
## 2 male 6.761006Exercise 4 Each pirate only uses one kind of sword - and their speed with their preferred sword is represented in sword.speed Which sword types tend to have the fastest (i.e.; smallest) sword speed? Test this by calculating the median sword.speed for each sword type
head(pirates)## id sex headband age college tattoos tchests.found parrots.lifetime
## 1 1 female yes 37 JSSFP 9 6 2
## 2 2 female yes 34 JSSFP 14 6 5
## 3 3 male no 29 CCCC 9 1 13
## 4 4 female yes 30 JSSFP 9 1 5
## 5 5 male yes 16 CCCC 12 4 1
## 6 6 male yes 27 CCCC 6 1 1
## favorite.pirate sword.type sword.speed
## 1 Hook cutlass 0.2572612
## 2 Lewis Scot cutlass 0.2935024
## 3 Jack Sparrow scimitar 0.8698797
## 4 Lewis Scot cutlass 0.1832235
## 5 Hook cutlass 0.3754281
## 6 Jack Sparrow cutlass 0.1542810sword.mean <- aggregate(sword.speed ~ sword.type, FUN = median, na.rm = T, data = pirates)
sword.mean## sword.type sword.speed
## 1 banana 1.9427477
## 2 cutlass 0.3979775
## 3 sabre 1.5646515
## 4 scimitar 2.7334913Exercise 5 Is there a relationship between whether or not a pirate wears a headband and their speed with their sword? Test this in two ways.
First, calculate the median sword speed, separately for each level of headband use using aggregate(). What is your conclusion?Second, calculate the median sword speed for all combinations of both sex AND sword.type. That is, calculate the median sword.speed for headband-wearers who use cutlasses, headband-wearers who use sabres, … headband-nonwearers who use cutlasses, headband-nonwearers who use sabres… (hint: include two independent variables in the formula argument to aggregate()). Does your conclusion change? If so, what do you think is going on?
sword.headband <- aggregate(sword.speed ~ headband, FUN = median, na.rm = T, data = pirates)
sword.headband## headband sword.speed
## 1 no 1.1441930
## 2 yes 0.4584107sword.sex.type <- aggregate(sword.speed ~ sex + sword.type, FUN = median, na.rm = T, data = pirates)
sword.sex.type## sex sword.type sword.speed
## 1 female banana 2.0944283
## 2 male banana 1.3423544
## 3 female cutlass 0.4223560
## 4 male cutlass 0.3811121
## 5 female sabre 1.5193133
## 6 male sabre 1.8346521
## 7 female scimitar 2.8816917
## 8 male scimitar 2.2276789sword.speed.headband <- aggregate(sword.speed ~ headband + sword.type, FUN = median, na.rm = T, data = pirates)
sword.speed.headband## headband sword.type sword.speed
## 1 no banana 1.0902691
## 2 yes banana 6.7000876
## 3 no cutlass 0.3638043
## 4 yes cutlass 0.4020915
## 5 no sabre 1.2050550
## 6 yes sabre 1.9749380
## 7 no scimitar 2.0775947
## 8 yes scimitar 4.0558626sword.all <- aggregate(sword.speed ~ headband + sword.type + sex, FUN = median, na.rm = T, data = pirates)
sword.all## headband sword.type sex sword.speed
## 1 no banana female 1.0902691
## 2 yes banana female 6.4681032
## 3 no cutlass female 0.3065877
## 4 yes cutlass female 0.4339311
## 5 no sabre female 1.2998394
## 6 yes sabre female 1.9153348
## 7 no scimitar female 2.5258784
## 8 yes scimitar female 4.9011985
## 9 no banana male 1.0012506
## 10 yes banana male 6.7000876
## 11 no cutlass male 0.8152493
## 12 yes cutlass male 0.3784077
## 13 no sabre male 1.0841687
## 14 yes sabre male 2.3333113
## 15 no scimitar male 1.2479079
## 16 yes scimitar male 3.7374398conculision: The pirates without a headband are fastern the pirates with a headband. Apparently the female pirate with a handband and a bana sword ist the fastest.
Exercise 6: Does a pirate’s favorite pirate say anything about them? Do pirates whose favorite pirate is Hook have more tattoos on average or a faster sword speed than those whose favorite pirate is Blackbeard? Using dplyr, create the following aggregated dataframe which shows aggregated data depending on the pirates’ favorite pirate. Here are the four basic steps
First, define the dataframe THEN… (%>%)
Second, group the data by favorite.pirate, THEN… (%>%)
Third, use the summary function to tell R you will summarise the data
Fourth, define each of the new summary statistics (then close the summarize function)library("dplyr")##
## Attaching package: 'dplyr'
##
## The following object is masked from 'package:stats':
##
## filter
##
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionpirates.fav <- pirates %>%
group_by(favorite.pirate) %>% summarise (tattoos.mean = mean(tattoos, na.rm = T), sword.speed.median = median(sword.speed, na.rm = T))
pirates.fav## Source: local data frame [6 x 3]
##
## favorite.pirate tattoos.mean sword.speed.median
## 1 Anicetus 9.739726 0.5063631
## 2 Blackbeard 8.934579 0.5765946
## 3 Edward Low 9.285714 0.5187047
## 4 Hook 9.235294 0.5297751
## 5 Jack Sparrow 9.748826 0.5125712
## 6 Lewis Scot 9.833333 0.4175671