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Names and Indexing

Names and Indexing#

Names#

In R, it is not just variables that have names. We have seen that data frames can have column names, and it’s also possible to give them row names. In fact, any element in a vector or list can be given a name, and these names are accesible through a simple function.

# Naming a vector
x <- 1:5
names(x) # NULL

names(x) <- c("A","B","C","D","E")
names(x) # "A" "B" "C" "D" "E"
x

This is slightly different in a matrix or data.frame, where you can name the rows and columns.

# Naming rows and columns
df <- data.frame(1:3,4:6,7:9)
df
#     X1.3 X4.6 X7.9
# 1    1    4    7
# 2    2    5    8
# 3    3    6    9

rownames(df) # "1" "2" "3"
colnames(df) # "X1.3" "X4.6" "X7.9"

rownames(df) <- c("A","B","C")
colnames(df) <- c("X","Y","Z")
df
#   X Y Z
# A 1 4 7
# B 2 5 8
# C 3 6 9

Indexing#

Sometimes you want to refer to only part of a vector, matrix or data.frame – perhaps a single column or even single item. This is called slicing and requires an understanding of how R indexes the elements in objects.

For a vector, you can either reference an item by its position or name.

# Slicing a vector
x <- c("Chris","Field","Bioinformatician")
names(x) <- c("Name","Surname","Job")

x[1] #Name  "Chris"
x["Name"] #Name  "Chris"

For a matrix or data.frame, the same methods work for indexing the row or column of the object, or both. The convention is that first you give the row, then the column, separated by a comma, and if one is left blank it implies you want ‘all’ rows or columns. For this example we are going to load up a pre-made set of data that comes with R.

# Slicing a data.frame
data(swiss)

swiss[1,] # outputs all the elements in the first row of the swiss data.frame
swiss[,1] # outputs all the elements in the first column of the swiss data.frame
swiss[1,1] # outputs the element in the first row and first column of the swiss data.frame
swiss["Gruyere",] # outputs all the elements withtin the row named "Gruyere" of the swiss data.frame
swiss[,"Fertility"] # outputs all the elements withtin the column named "Fertility" of the swiss data.frame

Finally there are two additional ways to access items in a list, or columns (only) of a data frame.

# Accessing a list item
l <- list(names=c("Anna", "Ben", "Chris"), scores=c(23, 31, 34))
l$names #"Anna"  "Ben"   "Chris"
l[[1]] #"Anna"  "Ben"   "Chris"
l[["names"]] #"Anna"  "Ben"   "Chris"

# The difference between single and double brackets for a list
l[1] # Produces a list of one item
# $names
# [1] "Anna"  "Ben"   "Chris"

l[1:2] # Produces a list of two items
# $names
# [1] "Anna"  "Ben"   "Chris"
# $scores
# [1] 23 31 34

l[[1]] # Produces a vector
# "Anna"  "Ben"   "Chris"

l[[1:2]] # Produces a single item, the second entry in the first item in the list
# "Ben"

# Accessing a column of a data frame
swiss$Fertility
swiss[[1]]
swiss[["Fertility"]]

You can also slice multiple items by giving a vector of numbers or names. Remember that R automatically translates the code n:m into a range of integers from n to m.

# Slicing a range
x[1:2]
swiss[1:3,]
swiss[4:5,1:3]
swiss[c("Aigle","Vevey"),c("Fertility","Catholic")]

Exercise 0.4

  • Load the pre-made data set swiss

# Load the data
data(swiss)

  • Look at the row and column names, then try to rename the columns

# Look at the row and col names, try to rename
rownames(swiss)
colnames(swiss)
colnames(swiss) <- c("A","B","C","D","E","F")

  • What happens if you give fewer names than there are columns?

# What happens if I don't give enough names
colnames(swiss) <- c("A","B")
# The other columns are named NA, which is a problem

  • Create a vector containing the numbers 1 to 10 and then create a vector containing the first ten square numbers

# Create the vectors
n <- 1:10
sq <- n*n

  • Slice the vector to check the value of the 7th square number

# Slice the 7th square
sq[7]

  • Returning to the swiss data set, extract the data for just Sion

# Reload (because we renamed things) and extract data for Sion
data(swiss)
swiss["Sion",]

  • Now extract only the Catholic data for the first ten places, and than just for Sion

# Extract just Catholic data
swiss$Catholic
# or
swiss[,"Catholic"]
# or even, but this is less reliable if things move around
swiss[,5]

# Extract more specific data
swiss[1:10,"Catholic"]
swiss["Sion","Catholic"]

  • Finally use vectors to find the data on Examination and Education for Neuchatel and Sierre

# Extract very specific data
swiss[c("Neuchatel","Sierre"),c("Examination","Education")]
# Note that in the original table, Neuchatel appears after Sierre, but here they are reported in the order I gave

Logical Slicing#

We have seen that we can give R a vector of numbers or names and it will slice out the corresponding data from a vector or data frame. We can actually go further than that and use a vector of logical values, TRUE or FALSE to determine which elements we want to slice out. Furthermore, we can write the vector as a variable ahead of time if we like.

# Slice using premade vectors
places <- c("Neuchatel","Sierre")
cols_of_interest <- c("Examination","Education")

data(swiss)
swiss[places,cols_of_interest]

# Slice using a logical
cols_of_interest <- c(FALSE, FALSE, TRUE, TRUE, FALSE, FALSE)
swiss[places,cols_of_interest]

Now the really clever bit is that we can generate a vector of logical values using the data itself, with any of the comparison functions such as >, <, ==.

# Logical slicing
isCatholic <- swiss$Catholic > 50
swiss[isCatholic,]

# Logical slicing without saving the vector ahead of time
swiss[swiss$Fertility < 50,]

Exercise 0.5

  • Reload the swiss data set, in case you have edited it

# Reload the data
data(swiss)

  • Create a vector with the column names in alphabetical order and use it to ‘slice’ the table (we are really just rearranging!)

# Create a sorted vector of column names
names_sorted <- c("Agriculture", "Catholic", "Education", "Examination", "Fertility", "Infant.Mortality")
swiss_sorted <- swiss[,names_sorted]

  • Slice the table to see just the places with an Agriculture score less than 50

# Find places with low agriculture
low_ag <- swiss$Agriculture < 50
swiss[low_ag,]
# or directly
swiss[swiss$Agriculture < 50,]

  • Now, making sure to save the results into new variables, split the table into two based on whether a place has more or less than 50 in Catholic

# Split the table by catholic score
low_cath <- swiss[swiss$Catholic < 50,]
hi_cath <- swiss[swiss$Catholic >= 50,]

  • In each table, look at the Catholic column data only, what do you notice about it?

# Look at the values
low_cath$Catholic
hi_cath$Catholic
# With only a couple of exceptions, the values are either very low or very high - the distribution of scores is bimodal!
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