We again define the salary offers:
offer <- c(241, 590, 533, 425, 261)Here is a vector of logicals, with TRUE indicating that the element is greater than 400, and FALSE indicating that it’s not.
offer > 400## [1] FALSE TRUE TRUE TRUE FALSEWe can also get this vector:
offer < 550## [1] TRUE FALSE TRUE TRUE TRUEWe can combine those vector using &:
offer > 400 & offer < 550## [1] FALSE FALSE TRUE TRUE FALSEHere, TRUE corresponds to elements that are both greater than 400 and smaller than 550. We can now retrieve the elements that are between 400 and 550 using
offer[offer > 400 & offer < 550]## [1] 533 425Recall that FALSE means “drop” and TRUE means keep, so we just keep the elements between 400 and 550 (not including 400 and 500).
What about the elements of offer outside the range of 400..550? One way to go is
offer[!(offer > 400 & offer < 550)]## [1] 241 590 261The ! operator turns TRUE into FALSE and vice versa, so we keep the elements we dropped before, and drop the elements we kept.
Another way to obtain the same result is
offer[offer <= 400 | offer >= 550]## [1] 241 590 261Note that we are now using <= rather than <. That because we want offer <= 400 | offer >= 550 to be true when the element is equal to exactly 400 to be consistent with offer[!(offer > 400 & offer < 550)]. (Think this through if it’s not obvious.)
Let’s go back to offer. Actually, those were the average offers to various specialties of doctors:
offer <- c(241, 590, 533, 425, 261)
spec <- c("family doc", "cardiologist", "orthopedic", "dermatologist", "psychiatrist")So for example the offers to cardiologist were $590k per year in 2018. Suppose we want to automatically figure out which specialty makes the most money.
So, if we knew it were specialty number 2, we’d go spec[2]. But how do we figure out the 2?
Let’s try another tack. We’d like to compute spec[c(F, T, F, F, F)]. Can we compute the logical vector? Here’s an idea:
offer == max(offer)## [1] FALSE TRUE FALSE FALSE FALSECombining those ideas, we’ll get
spec[offer == max(offer)]## [1] "cardiologist"So how do we get the 2? We can first generate the vector
1:length(offer) # generate c(1, 2, 3, ..., length(offer))## [1] 1 2 3 4 5And now, we can simply go
(1:length(offer))[offer == max(offer)]## [1] 2The following is really unnecessarily complicated, but we could have
spec[(1:length(offer))[offer == max(offer)]]## [1] "cardiologist"Data frames are R’s way of storing tables (note that the salary data we had was also actually a table).
You can define a data frame using the following syntax:
offers <- data.frame(amount = c(241, 590, 533, 425, 261),
spec = c("family doc", "cardiologist", "orthopedic", "dermatologist", "psychiatrist"))Note the amount and spec are the names of columns in the new data frame. We define data frames column-by-column.
The value of offers will be displayed as follows in the console:
offers## amount spec
## 1 241 family doc
## 2 590 cardiologist
## 3 533 orthopedic
## 4 425 dermatologist
## 5 261 psychiatristLet’s load a data frame (you must previously have successfully run install.packages("babynames")).
library(babynames) #Load the data frame babynames into RHere are the first several rows of the data frame
head(babynames) # Display the first 5 rows of a data frame## # A tibble: 6 x 5
## year sex name n prop
## <dbl> <chr> <chr> <int> <dbl>
## 1 1880 F Mary 7065 0.0724
## 2 1880 F Anna 2604 0.0267
## 3 1880 F Emma 2003 0.0205
## 4 1880 F Elizabeth 1939 0.0199
## 5 1880 F Minnie 1746 0.0179
## 6 1880 F Margaret 1578 0.0162We can access, for example, row 2 and column “year” of the table like so:
babynames[2, "year"]## # A tibble: 1 x 1
## year
## <dbl>
## 1 1880If we want to access all of row 2, we can omit the second part:
babynames[2, ]## # A tibble: 1 x 5
## year sex name n prop
## <dbl> <chr> <chr> <int> <dbl>
## 1 1880 F Anna 2604 0.0267We can access rows 2 through 6 like so:
babynames[2:6, ]## # A tibble: 5 x 5
## year sex name n prop
## <dbl> <chr> <chr> <int> <dbl>
## 1 1880 F Anna 2604 0.0267
## 2 1880 F Emma 2003 0.0205
## 3 1880 F Elizabeth 1939 0.0199
## 4 1880 F Minnie 1746 0.0179
## 5 1880 F Margaret 1578 0.0162We can only take the columns "n" and "year", like this:
babynames[2:6, c("n", "year")]## # A tibble: 5 x 2
## n year
## <int> <dbl>
## 1 2604 1880
## 2 2003 1880
## 3 1939 1880
## 4 1746 1880
## 5 1578 1880Finally, if we want a particular column as a vector (rather than as a data frame), we can do the following (make sure to keep track of the quotes)
babynames[5:20, ]$name## [1] "Minnie" "Margaret" "Ida" "Alice" "Bertha" "Sarah"
## [7] "Annie" "Clara" "Ella" "Florence" "Cora" "Martha"
## [13] "Laura" "Nellie" "Grace" "Carrie"$ vs. [, "colname"]Note that sometimes, df[, "colname"] will yield a vector (when you are operating on one kind of data frames), and sometimes, it will a data frame. df$colname will always yield a vector. So if you want a vector, use the $ operator.
Let’s now write code that finds the most common name in 1999
babies.1999 <- babynames[babynames$year == 1999 & babynames$sex == "F", ]
max.name.count <- max(babies.1999$n)
(babies.1999$name)[max.name.count == babies.1999$n]## [1] "Emily"Let’s make this into a more general function:
most.common.name <- function(babynames, year, sex){
b.year.sex <- babynames[babynames$year == year & babynames$sex == sex, ]
(b.year.sex$name)[b.year.sex$n == max(b.year.sex$n)]
}tidyverseTo use pipes, first you need to install and load the tidyverse library. Use the following:
install.packages("tidyverse")
library(tidyverse)library(tidyverse)Let’s define two functions:
f <- function(x){
x^2
}
g <- function(y){
y + 1
}We can compute f(g(5)) and g(f(5)). Note that those are not the same: \(f(g(5)) = f(6) = 36\) but \(g(f(5)) = g(25) = 26\).
Now, we’ll compute the same qunatities using pipes:
5 %>% f %>% g ## [1] 26This is the same as computing g(f(10)). The way to think about it is this: we start with 10, then apply f to 10 and obtain f(10), and then apply g to 10 %>% f (i.e., f(10)) and obtain f(g(10)).
To compute f(g(10)), we can use:
5 %>% g %>% f ## [1] 36This is the same as
5 %>% g() %>% f()## [1] 36If the only thing that we’re sending to f is g(5), the parentheses are optional in this notation.
This is useful mostly because it’s easier to read something like x %>% f1 %>% f2 %>% f3 %>% f4 than to read f4(f3(f2(f1(x)))).
tidyverse/dplyr to wrangle datafilterThe filter function is used to select rows from a data frame. The following are all equivalent. They are different ways of selecting rows for which the year is 1880 and sex is "F".
filter(babynames, year == 1880, sex == "F")
idx <- (babynames$year == 1880) & (babynames$sex == "F") babynames[idx, ]
babynames[(babynames$year == 1880) & (babynames$sex == "F"), ]
babynames %>% filter(year == 1880 & sex == "F")
babynames %>% filter(year == 1880, sex == "F")
The last way is probably preferable, though different people might have different preferences.
selectWe can use select as follows:
b.y.n <- babynames %>% select(year, name)
b.y.n[1:20, ]## # A tibble: 20 x 2
## year name
## <dbl> <chr>
## 1 1880 Mary
## 2 1880 Anna
## 3 1880 Emma
## 4 1880 Elizabeth
## 5 1880 Minnie
## 6 1880 Margaret
## 7 1880 Ida
## 8 1880 Alice
## 9 1880 Bertha
## 10 1880 Sarah
## 11 1880 Annie
## 12 1880 Clara
## 13 1880 Ella
## 14 1880 Florence
## 15 1880 Cora
## 16 1880 Martha
## 17 1880 Laura
## 18 1880 Nellie
## 19 1880 Grace
## 20 1880 CarrieWe only kept the year and name columns here.
summarize and group_byFirst, let’s use summarize to compute the average number of babies per name:
babynames %>% summarize(b.pername = mean(n))## # A tibble: 1 x 1
## b.pername
## <dbl>
## 1 181.Inside of summarize, we can refer to columns in the data frame we are processing by name, without the $ operator. (And you shouldn’t be using the $ operator.) This is basically the same as not using summarize and computing mean(babynames$n) – not that interesting.
The real power of summarize is in using group_by – we can group rows and compute a function such as mean for every group of rows separately. For example, we could compute the average number of babies per name for different sexes separately.
babynames %>% group_by(sex) %>%
summarize(b.pername = mean(n))## # A tibble: 2 x 2
## sex b.pername
## <chr> <dbl>
## 1 F 151.
## 2 M 223.We see that the number of babies per name for "F" is much smaller than for "M" – female names are more diverse with fewer babies per name, and male names are less diverse with more babies per name.