stringr practice

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Stringr package

Install and load stringr library

library(tidyverse)

package 㤼㸱tidyverse㤼㸲 was built under R version 3.2.5Error in loadNamespace(j <- i[[1L]], c(lib.loc, .libPaths()), versionCheck = vI[[j]]) : 
  there is no package called ‘lubridate’
Error: package or namespace load failed for ‘tidyverse’

Turning numbers into strings

format(a,digits=2)

[1] "1.2e+05" "2.5e-04"

formatC(a, format = "f")

[1] "123456.0000" "0.0002"     

# Scientific format
formatC(a,format = "e")

[1] "1.2346e+05" "2.5000e-04"

# or 
formatC(a,format = "g")

[1] "1.235e+05" "0.00025"  

Get to know a dataset

library(babynames)
head(babynames)

# Create a new column named name_female
name_female<-babynames %>% dplyr::select(name,sex) %>% dplyr::filter(sex=="F") %>% dplyr::mutate(Sex_female="Female")
tail(name_female)

name_male<-babynames %>% dplyr::select(name,sex) %>% dplyr::filter(sex!="F") %>% dplyr::mutate(Sex_female="Male")
head(name_male)

boy_length<-str_length(name_male$name)
head(boy_length)

[1] 4 7 5 7 6 5

# calculate the mean of name length
mean(boy_length)

[1] 5.993554

# extract a subset of characters, get the last letter of girl's name
babynames_sub<-str_sub(babynames$name[babynames$sex=="F"],-1,-1)
head(babynames_sub)

[1] "y" "a" "a" "h" "e" "t"

Finding pattern

# Find all pattern "zz" in name column
pattern_zz<-str_detect(babynames$name,pattern = "zz")
head(pattern_zz)

[1] FALSE FALSE FALSE FALSE FALSE FALSE

sum(pattern_zz)

[1] 2313

head(babynames[pattern_zz,])

# Find the babynames containing u
u_name<-str_subset(babynames$name,fixed("u"))
head(babynames[u_name,])

Repalce strings

a<-c("Ha Van & Tuyen","Ha Van & Tri","Ha Van & Tuong")
str_replace(a, pattern = "&", replacement = "")

[1] "Ha Van  Tuyen" "Ha Van  Tri"   "Ha Van  Tuong"

# If have many % in one sentence, using str_replace_all instead
b<-c("Ha % Van % Tuyen: % A Massey % student")
str_replace_all(b,pattern = "%",replacement = "")

[1] "Ha  Van  Tuyen:  A Massey  student"