Sub-setting Data. Tzu L. Phang

Transcription

1 Sub-setting Data Tzu L. Phang

2 Subsetting in R Let s start with a (dummy) vectors. x <- c(5.4, 6.2, 7.1, 4.8, 7.5) names(x) <- c('a', 'b', 'c', 'd', 'e') x

3 Accessing elements using their indices Using indices x[1] x[4] Using a series of indices x[c(1, 3)] Or slices of the vector: x[1:4]

4 Warning But, if outside... return nothing... x[6] If we ask for the 0th element, we get an empty vector: x[0]

5 Skipping and removing elements Negative number: return all except that: x[-2] We can skip multiple elements: x[c(-1, -5)] # or x[-c(1,5)]

6 Challenge 1 Given the following code: x <- c(5.4, 6.2, 7.1, 4.8, 7.5) names(x) <- c("a", "b", "c", "d", "e") print(x) Come up with at least 3 different commands that will produce the following output: x[2:4]

7 Solution to challenge 1 x[2:4] x[-c(1, 5)] x[c("b", "c", "d")] x[c(2, 3, 4)]

8 Subsetting by name extracting by name: x[c("a", "c")] To skip (or remove) a single named element: x[-which(names(x) == "a")]

9 Skipping Multiple Named Indices Skipping multiple named indices: x[-which(names(x) %in% c("a", "c"))] The %in% goes through each element of its left argument, in this case the names of x, and asks, Does this element occur in the second argument?.

10 Subsetting through other logical operations Subset through logical operations: x[c(true, TRUE, FALSE, FALSE)] Logical vector is also recycled: x[c(true, FALSE)]

11 Combining logical conditions Combine multiple logical criteria. & &&, the logical AND operator: returns TRUE if both the left and right are TRUE., the logical OR operator: returns TRUE, if either the left or right (or both) are TRUE.!, the logical NOT operator: converts TRUE to FALSE and FALSE to TRUE. It can negate a single logical condition (eg!true becomes FALSE), or a whole vector of conditions(eg!c(true, FALSE) becomes c(false, TRUE)). all, Which returns TRUE if every element of the vector is TRUE any, Which returns TRUE if one or more elements of the vector are TRUE

12 Challenge 3 Given the following code: x <- c(5.4, 6.2, 7.1, 4.8, 7.5) names(x) <- c("a", "b", "c", "d", "e") print(x) Write a subsetting command to return the values in x that are greater than 4 and less than 7.

13 Solution to challenge 3 x_subset <- x[x < 7 & x > 4] print(x_subset)

14 Handling special values There are a number of special functions you can use to filter out this data: is.na will return all positions in a vector, matrix, or data.frame containing NA. is.nan will do for NaN is.infinite will do for Inf is.finite will return all positions in a vector, matrix, or data.frame that do not contain NA, NaN or Inf. na.omit will filter out all missing values from a vector

15 Factor subsetting Factor subsetting works the same way as vector subsetting. f <- factor(c("a", "a", "b", "c", "c", "d")) f[f == "a"] f[f %in% c("b", "c")] f[1:3] An important note is that skipping elements will not remove the level even if no more of that category exists in the factor: f[-3]

16 Matrix subsetting Matrices are also subsetted using the [ function. It takes two arguments: the first applying to the rows the second to its columns: set.seed(1) m <- matrix(rnorm(6*4), ncol=4, nrow=6) m[3:4, c(3,1)] You can leave the first or second arguments blank to retrieve all the rows or columns respectively: m[, c(3,4)] If we only access one row or column, R will automatically convert the result to a vector: m[3,]

17 Matrix subsetting... cont... If you want to keep the output as a matrix, you need to specify a third argument; drop = FALSE: m[3,, drop=false] Unlike vectors, if we try to access a row or column outside of the matrix, R will throw an error: m[, c(3,6)]

18 Matric: assign by column Matrices are laid out in column-major format by default. matrix(1:6, nrow=2, ncol=3) If you wish to populate the matrix by row, use byrow=true: matrix(1:6, nrow=2, ncol=3, byrow=true)

19 Challenge 4 Given the following code: m <- matrix(1:18, nrow=3, ncol=6) print(m) ## [,1] [,2] [,3] [,4] [,5] [,6] ## [1,] ## [2,] ## [3,] Which of the following commands will extract the values 11 and 14? A. m[2,4,2,5] B. m[2:5] C. m[4:5,2] D. m[2,c(4,5)]

20 Solution to challenge 4 D

21 List subsetting There are three functions used to subset lists. [: will subset a list [[: will subset the element of the list $: will subset the element by name xlist <- list(a = "Software Carpentry", b = 1:10, data = head(iris)) xlist[1] xlist[[1]] xlist[['a']] xlist$a

22 Subset list limitation You can t extract more than one element at once: xlist[[1:2]] Nor use it to skip elements: > xlist[[-1]]

23 Challenge 5 Given the following list: xlist <- list(a = "Software Carpentry", b = 1:10, data = head(iris)) Using your knowledge of both list and vector subsetting, extract the number 2 from xlist. Hint: the number 2 is contained within the b item in the list.

24 Solution to challenge 5 xlist$b[2] xlist[[2]][2] xlist[["b"]][2]

25 Challenge 6 Given a linear model: mod <- aov(pop ~ lifeexp, data=gapminder) Extract the residual degrees of freedom (hint: attributes() will help you)

26 Solution to challenge 6 attributes(mod) ## `df.residual` is one of the names of `mod` mod$df.residual

27 Gapminder: Life Expectancy GapMinder Video: let s watch... Lets read in the gapminder dataset that we downloaded previously: gapminder <- read.csv("data/gapminder-fiveyeardata.csv")

28 Data frames: practice with gapminder dataset Fix each of the following common data frame subsetting errors: 1. Extract observations collected for the year 1957 gapminder[gapminder$year == 1957,] 2. Extract all columns except 1 through to 4 gapminder[,-c(1:4)] 3. Extract the rows where the life expectancy is longer the 80 years gapminder[gapminder$lifeexp > 80,]

29 Data frames: practice with gapminder dataset... cont Extract the first row, and the fourth and fifth columns (lifeexp and gdppercap). gapminder[1, c(4, 5)] 5. Advanced: extract rows that contain information for the years 2002 and 2007 gapminder[gapminder$year == 2002 gapminder$year == 2007,] gapminder[gapminder$year %in% c(2002, 2007),]