R Programming for Biologists: Workshop 1

Assigning a character:

tmp <- "hello world!"
print(tmp)

[1] "hello world!"

Working with numbers:

a <- 20
b <- 5
c <- a**2 + b**2
c

[1] 425

Why do I use <- instead of =?

One of the main functions we'll be using is c.

test_vec <- c(5, 42, 44, 6)
test_vec

[1]  5 42 44  6

Getting a particular index from a vector:

test_vec[3]

[1] 44

Note that the first index is indicated by 1, not 0:

test_vec[0]

numeric(0)

logical_vec <- c(TRUE, F, T, 10 > 2, 10 == 1)
logical_vec

[1]  TRUE FALSE  TRUE  TRUE FALSE

Note no " characters:

class(c("TRUE", "FALSE"))

[1] "character"

Beware working with variables with Boolean names:

T <- FALSE
print(T)

[1] FALSE

These two lines have very different meanings:

      T == 24
      T = 24

Using comparison operators will return logical vectors

"CAT" == "DOG"

[1] FALSE

x <- 2
x < 3

[1] TRUE

x > 10

[1] FALSE

Defining a character vector:

tmp <- c("hey,", "this", "is", "multiple", "strings")
class(tmp)

[1] "character"

Converting to a character vector:

tmp2 <- c(10, 20, 40.0, 19)
tmp2 <- as.character(tmp2)
tmp2

[1] "10" "20" "40" "19"

class(tmp2)

[1] "character"

Character columns in tables will be read in as factors by default (set StringsAsFactors=False to avoid this).

Explicitly defining factors is better:

tmp <- c("hey,", "this", "is", "multiple", "strings")
tmp

[1] "hey,"     "this"     "is"       "multiple" "strings" 

tmp <- factor(tmp)
tmp

[1] hey,     this     is       multiple strings 
Levels: hey, is multiple strings this

tmp <- factor(c("treated_WT", "control_WT", "treated_KO", "control_WT", "control_KO", "control_KO", "treated_WT", "treated_KO"))
print(tmp)

[1] treated_WT control_WT treated_KO control_WT control_KO control_KO
[7] treated_WT treated_KO
Levels: control_KO control_WT treated_KO treated_WT

You can explicitly set the factor levels you want:

tmp2 <- factor(tmp, levels = c("control_WT", "treated_WT", "control_KO", "treated_KO"))
print(tmp2)

[1] treated_WT control_WT treated_KO control_WT control_KO control_KO
[7] treated_WT treated_KO
Levels: control_WT treated_WT control_KO treated_KO

What class is the object z below?

      z <- c(4154163, "Hi there gang!")

What class is the object funInSun below?

      print(funInSun)

[[1]]
[1] 1841   51

[[2]]
[1] "heya"

You can combine objects of different types in lists.

tmp <- list("prime"=c(1, 2, 5, 7, 11), "animals"=c("cow", "chicken"))
tmp

$prime
[1]  1  2  5  7 11

$animals
[1] "cow"     "chicken"

The key advantage of using lists is that you can use the function lapply, e.g.

tmp2 <- list("set1"=c(1, 2, 5, 7, 11), "set2"=c(3, 8, 10), "set3"=c(4, 8))
lapply(tmp2, sum)

$set1
[1] 26

$set2
[1] 21

$set3
[1] 12

Only for data of the same type. Take up less memory than dataframes.

test_matrix <- matrix(c(10, 32, 13, 54), nrow=2, ncol=2)
test_matrix

     [,1] [,2]
[1,]   10   13
[2,]   32   54

test_matrix2 <- matrix(c("dog", "cat", 13, 54), nrow=2, ncol=2)
test_matrix2

     [,1]  [,2]
[1,] "dog" "13"
[2,] "cat" "54"

Columns can be of different types. Take up more memory, but are easier to interact with.

test_df <- as.data.frame(matrix(c(10, 32, 13, 54), nrow=2, ncol=2))
test_df

  V1 V2
1 10 13
2 32 54

test_df2 <- data.frame(pet=c("dog", "cat"), livestock=c("cow", "sheep"), stringsAsFactors = FALSE)
test_df2

  pet livestock
1 dog       cow
2 cat     sheep

Select a single column:

test_df2$pets

NULL

Matrices and dataframes can both have row and column names.

test_df2 <- data.frame(pet=c("dog", "cat"), livestock=c("cow", "sheep"), stringsAsFactors = FALSE)
test_df2

  pet livestock
1 dog       cow
2 cat     sheep

rownames(test_df2) <- c("Bill", "Sandy")
colnames(test_df2) <- c("Pet", "Livestock")
test_df2

      Pet Livestock
Bill  dog       cow
Sandy cat     sheep

To get the row corresponding to Bill (note the column is left blank):

test_df2["Bill", ]

     Pet Livestock
Bill dog       cow

To get Sandy's livestock:

test_df2["Sandy", "Livestock"]

[1] "sheep"

However, we can't remove rows or columns by name (at least not without an extra step).

test_df2[-"Bill",]
Error in -"Bill" : invalid argument to unary operator

Return element at first row and second column:

test_df2[1, 2]

[1] "cow"

Remove the first row:

test_df2[-1,]

      Pet Livestock
Sandy cat     sheep

Reading:

in_table <- read.table("myfile.txt", header=TRUE, row.names=1, sep="\t", stringsAsFactors=FALSE)

There are many other options, but these ones are good to be aware of. Note that in this case the first column would be interpreted as the rownames.

Writing:

write.table(in_table_modified, file="newfile.txt", quote=FALSE, sep="\t", col.names=NA, row.names=TRUE)

Make sure to set quote=FALSE so that no quotes are in your output file.

• Load the dataset mtcars with the command

data(mtcars)

• What class is this object?
  
  
• Contrast the outputs of summary and str.
  
  
• Take a look at the first 10 rows of this object with:

head(mtcars, 10)

• Plot a scatterplot of miles per gallon against displacement in the mtcars table.
  
  
• Change the axes names and point type (with the pch option)

• Make a copy of the mtcars object, but only keep the rows called: “Valiant”, “Merc 230”, and “Lotus Europa”.
  
  
• Make a another copy of this object, but this time only with rows that contain “Merc” using grep.
  
  
• Get the mean miles per gallon for all Merc cars.

• Load the airquality dataset with:

data(airquality)

• How many NA values are in this table?
  
  
• When you sum a logical vector TRUE is interpreted as 1 and FALSE is interpreted as 0. with this in mind how could you use the rowSums function to identify rows that have any NA values?
  
  
• Make a new dataframe with rows with any NA values removed.