In this practical we are going to play around with the different types of elements in R.
vec1 with values 1 through 6 and one named vec2 with letters A through F. vec1 <- c(1, 2, 3, 4, 5, 6)
vec2 <- c("A", "B", "C", "D", "E", "F")To create a vector we used c(), which stands for ‘concatenation’. It is just a series of numbers or letters.
vec1 and one from vec2. The dimensions for both matrices are 3 rows by 2 columns. mat1 <- matrix(vec1, nrow = 3, ncol = 2)
mat2 <- matrix(vec2, nrow = 3, ncol = 2)To create a matrix we used matrix(). For a matrix we need to specify the dimensions (in this case 3 rows and 2 columns) and the input (in this case vec1 or vec2) needs to match these dimensions.
vec1## [1] 1 2 3 4 5 6vec2## [1] "A" "B" "C" "D" "E" "F"mat1## [,1] [,2]
## [1,] 1 4
## [2,] 2 5
## [3,] 3 6mat2## [,1] [,2]
## [1,] "A" "D"
## [2,] "B" "E"
## [3,] "C" "F"vec1 and mat1 contain numbers and vec2 and mat2 contain characters.
vec1 and vec2 with 6 rows and 2 columns. Inspect this matrix.mat3 <- matrix(c(vec1, vec2), 6, 2)
mat3## [,1] [,2]
## [1,] "1" "A"
## [2,] "2" "B"
## [3,] "3" "C"
## [4,] "4" "D"
## [5,] "5" "E"
## [6,] "6" "F"or
mat3b <- cbind(vec1, vec2)
is.matrix(mat3b)## [1] TRUEmat3b## vec1 vec2
## [1,] "1" "A"
## [2,] "2" "B"
## [3,] "3" "C"
## [4,] "4" "D"
## [5,] "5" "E"
## [6,] "6" "F"If one or more elements in the matrix represent characters, all other elements are also converted to characters. A matrix is just for either numeric or character elements. Notice that the second approach (the column bind approach from mat3b) returns a matrix where the column names are already set to the name of the bound objects.
To solve the problem of charactered numbers we can create a dataframe. A dataframe is essentially a matrix that allows for character elements. The use of a dataframe is often preferred over the use of a matrix in R, except for purposes where pure numerical calculations are done, such as in matrix algebra. However, most datasets do contain character information and a dataframe would normally be your preferred choice when working with your own collected datasets in R.
dat3 where vec1 and vec2 are both columns. Name the columns V1 and V2, respectively. Use function data.frame().dat3 <- data.frame(V1 = vec1, V2 = vec2)
dat3## V1 V2
## 1 1 A
## 2 2 B
## 3 3 C
## 4 4 D
## 5 5 E
## 6 6 Fdat3b where vec1 and vec2 are both columns. Name the columns V1 and V2, respectively. Use function as.data.frame() on the matrix obtained from Question 4. This is a tricky situation. At face value, everything may seem to be in order. But, be aware that the code
dat3b <- as.data.frame(mat3)
dat3b## V1 V2
## 1 1 A
## 2 2 B
## 3 3 C
## 4 4 D
## 5 5 E
## 6 6 Fdoes not work properly (at least not as intended) as the matrix nature of mat3 turned everything into a character value and you have lost the numerical nature of vec1. It may appear to be working, but if we check if column 1 is numerical, it turns out not to be the case.
is.numeric(dat3[, 1])## [1] TRUEis.numeric(dat3b[, 1])## [1] FALSEThe first column in matrix dat3b obtained from Question 5 is indeed not numeric. As a matter of fact, it is also not a character variable.
is.character(dat3b[, 1])## [1] FALSERather tricky; the function as.data.frame() has converted the first variable to a factor.
is.factor(dat3b[, 1])## [1] TRUEThis is due to us not specifying the variable correctly in the matrix we used to create the dataframe. Factors are categorical variables that are depicted by numbers. Character vectors are converted to factors in data frames.
dat3 that you have created in Question 4. dat3[3, ] #3rd row## V1 V2
## 3 3 Cdat3[, 2] #2nd column## [1] A B C D E F
## Levels: A B C D E Fdat3$V2 #also 2nd column## [1] A B C D E F
## Levels: A B C D E Fdat3[3,2] #intersection## [1] C
## Levels: A B C D E FThe [3,2] index is very useful in ‘R’. The first number (before the comma) represents the row and the second number (after the comma) represents the column. For a vector there are no two dimensions and only one dimension can be called. For example, vec1[3] would yield 3. Try it.
Columns can also be called by the $ sign, but only if a name has been assigned. With dataframes assigning names happens automatically.
Note that R automatically reports the values the character column can take. This means that the column is indeed a factor (a categorical variable - as it is supposed to be). A useful function to inspect the structure of a dataframe is str(). Try running it.
str(dat3)## 'data.frame': 6 obs. of 2 variables:
## $ V1: num 1 2 3 4 5 6
## $ V2: Factor w/ 6 levels "A","B","C","D",..: 1 2 3 4 5 6Inspecting the structure of your data is vital, as you probably have imported your data from some other source. If we, at a later stage, start analyzing our data without the correct measurement level, we may run into problems. One problem that often occurs is that categorical variables (factors in R) are not coded as such.
V1 in our dataframe dat3 is not coded correctly, but actually represents grouping information about cities. Convert the variable to a factor and add the labels Utrecht, New York, London, Singapore, Rome and Cape Town.dat3$V1 <- factor(dat3$V1, labels = c("Utrecht", "New York", "London", "Singapore", "Rome", "Capetown"))
dat3## V1 V2
## 1 Utrecht A
## 2 New York B
## 3 London C
## 4 Singapore D
## 5 Rome E
## 6 Capetown Fboys.RData. You need to download this file and put it in the project folder.Either run the below code
load("boys.RData")or double-click the boys.RData file on your machine (right-click and open with RStudio if it does not open by default in RStudio, but in R).
You can also import it directly from the internet by running and loading the below code connection
con <- url("https://gerkovink.github.io/Statistical-Programming-with-R/Contents/Material/Part%20B%20-%20How%20is%20R%20organized/boys.RData")
load(con)The boys object will be added to your Global Environment. You can now use the boys data by running
boysboys dataset (it is from package mice, by the way) by typing boys in the console, and subsequently by using the function View(). The output is not displayed here as it is simply too large.
Using View() is preferred for inspecting datasets that are large. View() opens the dataset in a spreadsheet-like window (conform MS Excel, or SPSS). If you View() your own datasets, you can not edit the datasets’ contents.
boys data set and inspect the first and final 6 cases in the data set. To do it numerically, find out what the dimensions of the boys dataset are.
dim(boys)## [1] 748 9There are 748 cases on 9 variables. To select the first and last six cases, use
boys[1:6, ]## age hgt wgt bmi hc gen phb tv reg
## 3 0.035 50.1 3.650 14.54 33.7 <NA> <NA> NA south
## 4 0.038 53.5 3.370 11.77 35.0 <NA> <NA> NA south
## 18 0.057 50.0 3.140 12.56 35.2 <NA> <NA> NA south
## 23 0.060 54.5 4.270 14.37 36.7 <NA> <NA> NA south
## 28 0.062 57.5 5.030 15.21 37.3 <NA> <NA> NA south
## 36 0.068 55.5 4.655 15.11 37.0 <NA> <NA> NA southboys[743:748, ]## age hgt wgt bmi hc gen phb tv reg
## 7410 20.372 188.7 59.8 16.79 55.2 <NA> <NA> NA west
## 7418 20.429 181.1 67.2 20.48 56.6 <NA> <NA> NA north
## 7444 20.761 189.1 88.0 24.60 NA <NA> <NA> NA west
## 7447 20.780 193.5 75.4 20.13 NA <NA> <NA> NA west
## 7451 20.813 189.0 78.0 21.83 59.9 <NA> <NA> NA north
## 7475 21.177 181.8 76.5 23.14 NA <NA> <NA> NA eastor, more efficiently:
head(boys)## age hgt wgt bmi hc gen phb tv reg
## 3 0.035 50.1 3.650 14.54 33.7 <NA> <NA> NA south
## 4 0.038 53.5 3.370 11.77 35.0 <NA> <NA> NA south
## 18 0.057 50.0 3.140 12.56 35.2 <NA> <NA> NA south
## 23 0.060 54.5 4.270 14.37 36.7 <NA> <NA> NA south
## 28 0.062 57.5 5.030 15.21 37.3 <NA> <NA> NA south
## 36 0.068 55.5 4.655 15.11 37.0 <NA> <NA> NA southtail(boys)## age hgt wgt bmi hc gen phb tv reg
## 7410 20.372 188.7 59.8 16.79 55.2 <NA> <NA> NA west
## 7418 20.429 181.1 67.2 20.48 56.6 <NA> <NA> NA north
## 7444 20.761 189.1 88.0 24.60 NA <NA> <NA> NA west
## 7447 20.780 193.5 75.4 20.13 NA <NA> <NA> NA west
## 7451 20.813 189.0 78.0 21.83 59.9 <NA> <NA> NA north
## 7475 21.177 181.8 76.5 23.14 NA <NA> <NA> NA eastThe functions head() and tail() are very useful functions. For example, from looking at both functions we can observe that the data are very likely sorted based on age.
boys data are sorted based on age. Verify this.To verify if the data are indeed sorted, we can run the following command to test the complement of that statement. Remember that we can always search the help for functions: e.g. we could have searched here for ?sort and we would quickly have ended up at function is.unsorted() as it tests whether an object is not sorted.
is.unsorted(boys$age)## [1] FALSEwhich returns FALSE, indicating that boys’ age is indeed sorted (we asked if it was unsorted!). To directly test if it is sorted, we could have used
!is.unsorted(boys$age)## [1] TRUEwhich tests if data data are not unsorted. In other words the values TRUE and FALSE under is.unsorted() turn into FALSE and TRUE under !is.unsorted(), respectively.
boys dataset with str(). Use one or more functions to find distributional summary information (at least information about the minimum, the maximum, the mean and the median) for all of the variables. Give the standard deviation for age and bmi. Tip: make use of the help (?) and help search (??) functionality in R.str(boys)## 'data.frame': 748 obs. of 9 variables:
## $ age: num 0.035 0.038 0.057 0.06 0.062 0.068 0.068 0.071 0.071 0.073 ...
## $ hgt: num 50.1 53.5 50 54.5 57.5 55.5 52.5 53 55.1 54.5 ...
## $ wgt: num 3.65 3.37 3.14 4.27 5.03 ...
## $ bmi: num 14.5 11.8 12.6 14.4 15.2 ...
## $ hc : num 33.7 35 35.2 36.7 37.3 37 34.9 35.8 36.8 38 ...
## $ gen: Ord.factor w/ 5 levels "G1"<"G2"<"G3"<..: NA NA NA NA NA NA NA NA NA NA ...
## $ phb: Ord.factor w/ 6 levels "P1"<"P2"<"P3"<..: NA NA NA NA NA NA NA NA NA NA ...
## $ tv : int NA NA NA NA NA NA NA NA NA NA ...
## $ reg: Factor w/ 5 levels "north","east",..: 4 4 4 4 4 4 4 3 3 2 ...summary(boys) #summary info## age hgt wgt bmi
## Min. : 0.035 Min. : 50.00 Min. : 3.14 Min. :11.77
## 1st Qu.: 1.581 1st Qu.: 84.88 1st Qu.: 11.70 1st Qu.:15.90
## Median :10.505 Median :147.30 Median : 34.65 Median :17.45
## Mean : 9.159 Mean :132.15 Mean : 37.15 Mean :18.07
## 3rd Qu.:15.267 3rd Qu.:175.22 3rd Qu.: 59.58 3rd Qu.:19.53
## Max. :21.177 Max. :198.00 Max. :117.40 Max. :31.74
## NA's :20 NA's :4 NA's :21
## hc gen phb tv reg
## Min. :33.70 G1 : 56 P1 : 63 Min. : 1.00 north: 81
## 1st Qu.:48.12 G2 : 50 P2 : 40 1st Qu.: 4.00 east :161
## Median :53.00 G3 : 22 P3 : 19 Median :12.00 west :239
## Mean :51.51 G4 : 42 P4 : 32 Mean :11.89 south:191
## 3rd Qu.:56.00 G5 : 75 P5 : 50 3rd Qu.:20.00 city : 73
## Max. :65.00 NA's:503 P6 : 41 Max. :25.00 NA's : 3
## NA's :46 NA's:503 NA's :522sd(boys$age) #standard deviation for age## [1] 6.894052sd(boys$bmi, na.rm = TRUE) #standard deviation for bmi## [1] 3.053421Note that bmi contains 21 missing values, e.g. by looking at the summary information. Therefor we need to use na.rm = T to calculate the standard deviation on the observed cases only.
The logical operators (TRUE vs FALSE) are a very powerful tool in R. For example, we can just select the rows (respondents) in the data that are older than 20 by putting the logical operater within the row index of the dataset:
boys2 <- boys[boys$age >= 20, ]
nrow(boys2)## [1] 12or, alternatively,
boys2.1 <- subset(boys, age >= 20)
nrow(boys2.1)## [1] 12boys3 <- boys[boys$age > 19 & boys$age < 19.5, ]
nrow(boys3)## [1] 18or, alternatively,
boys3.2 <- subset(boys, age > 19 & age < 19.5)
nrow(boys3.2)## [1] 18north?mean(boys$age[boys$age < 15 & boys$reg != "north" ], na.rm = TRUE)## [1] 6.044461or, alternatively,
mean(subset(boys, age < 15 & reg != "north")$age, na.rm=TRUE)## [1] 6.044461The mean age is 6.0444609 years
Today we have learned to use R at its basics. This offers tremendous flexibility, but may also be inefficient when our aim is some complex analysis, data operation of data manipulation. Doing advanced operations in basic R may require lots and lots of code. Tomorrow we will start using packages that allow us to do complicated operations with just a few lines of code.
As you start using R in your own research, you will find yourself in need of packages that are not part of the default R installation. The beauty of R is that its functionality is community-driven. People can add packages to CRAN that other people can use and improve. Chances are that a function and/or package has been already developed for the analysis or operation you plan to carry out. If not, you are of course welcome to fill the gap by submitting your own package.
End of practical