Scott W. Hegerty ECON 346
Here,we run through some basic R operations to get a sense of what the syntax looks like. We use more complicated tools in some of the other examples.
Here are three ways to create an object in R. These are simply single values. These objects are stired for future use.
v1<-2
v2=9
4->v3Generally the top method is preferred. You can think of it like an “arrow” that “shoots” the 2 into the new object, v1.
Next, we can try some orders of operation:
v2-v1## [1] 7
v3/v2## [1] 0.4444444
v3/v2-v1## [1] -1.555556
One exercise is to see what formula you could use to turn v1, v2, and v3 into the value 14.
While there are ways to not have to do this (such as dplyr), I focus on the “telescoping” nature of R functions. You start from the inside (within the most parentheses) and work outward. The parentheses are very important, so make sure they match (in number, as well as including what you want to do). For example, there is a difference in order if you add 6 and 4 and multiply by 2.
(6+4)*2## [1] 20
6+4*2## [1] 14
Here divide, add, and multiply – in that order. Then we round the result to two decimal places. The first two steps are “included” in the final result (the third line).
v3/v2-v1## [1] -1.555556
v1*(v3/v2-v1)## [1] -3.111111
round(v1*(v3/v2-v1),2)## [1] -3.11
We can also try it a different way:
v4<-v1*(v3/v2-v1)
round(v4,2)## [1] -3.11
A simple exercise is to find out how could you round v4 to three decimal places. In general, thinking through what you want to do (and modifying your code to do it) is more important than pure memorization.
Now we make a matrix; it has 12 elements. We start with a list of numbers, 1-12. Note how the series has a start and an end value. This just fills in the numbers in order.
m1<-c(1:12)
m1## [1] 1 2 3 4 5 6 7 8 9 10 11 12
This isn’t a matrix yet; it’s a 12x1 vector. Remember that matrices are represented as (#rows, #columns)
dim(m1)## NULL
Because it’s not a matrix,R says it doesn’t have dimensions: it has a length.
length(m1)## [1] 12
(Here’s a way to make numbers “skip” by twos. There are a lot of ways to make different sequences of numbers.) This one starts at 0, ends at 100, and has a “step” of 2. R functions have parameters that generaly make sense. Remember that R cannot read your mind. It won’t assume anything, and might not run if you miss something. You need to tell it exactly what you want it to do.
seq(0,100,2)## [1] 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
## [20] 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74
## [39] 76 78 80 82 84 86 88 90 92 94 96 98 100
One exercise is, how could you get the exact sequence 4,7,10,13,16,19,22?
Going back to the object we saved, m1, we can convert it into a matrix using the as.matrix() function. There are others, especially as.dataframe and as.character. Now it has dimensions.
m2<-as.matrix(x=m1)
m2## [,1]
## [1,] 1
## [2,] 2
## [3,] 3
## [4,] 4
## [5,] 5
## [6,] 6
## [7,] 7
## [8,] 8
## [9,] 9
## [10,] 10
## [11,] 11
## [12,] 12
dim(m2)## [1] 12 1
We can change these dimensions by assigning new numbers for the rows and columns:
dim(m2)<-c(4,3)
m2## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
How would you make a 2 x 6 matrix using these commands?
One important set of procedures involves selecting part of a dataset. This can be a single row or column, or only some of the columns, for example. Or, you can choose only rows with values above a certain value.
Here, we extract a vector ftom our new matrix– the middle column. Notice the brackets in the command. This again is written as (rows, columns). The “blank” row number means “all.” So this gives all rows for the second column.
m2[,2]## [1] 5 6 7 8
If we want all rows but the second one, we can use the - sign for “not.”
m2[,-2]## [,1] [,2]
## [1,] 1 9
## [2,] 2 10
## [3,] 3 11
## [4,] 4 12
We can also extract the bottom left corner of our original matrix: This is the fourth row, third column:
m2## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
m2[4,3]## [1] 12
Transposing the matrix switches the rows and columns, so (4,3) becomes (3,4):
t(m2)## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 5 6 7 8
## [3,] 9 10 11 12
We can also apply a function to every element in a matrix. Here, we double every value in our original m2:
2*m2## [,1] [,2] [,3]
## [1,] 2 10 18
## [2,] 4 12 20
## [3,] 6 14 22
## [4,] 8 16 24
There are a lot of other operations that we will see elsewhere.
One important set of tools involves selecting elements that meet a certain condition. You could “eyeball” a dataset and choose the rows and/or columns you need, but that gets really hard, really quick. We want R to choose the rows and columns for us.
In our original matrix m2, we can turn it into a dataframe amd note how it looks:
m2<-as.data.frame(m2)
m2## V1 V2 V3
## 1 1 5 9
## 2 2 6 10
## 3 3 7 11
## 4 4 8 12
The numbers on the left are row names, not numbers. It is not an additional variable. This is a coincidence that they match the numbers in the first actual column of data. In time series, these could be dates.
In this format, we can also select columns using the $ sign. (This doesn’t work in time series, though, so I use the brackets a lot more.). In this example, the first column is assigned the label “V1.”
m2$V1## [1] 1 2 3 4
We can choose only the rows where V1 is greater than 2. Eyeballing it tells us that it’s Rows 3 and 4, but we can only do that because the dataset is so small. So while we could just write m2[c(3,4),], that only works in one particular case.
m2[c(3,4),]## V1 V2 V3
## 3 3 7 11
## 4 4 8 12
We can break the procedure down into steps, by first stating the condition:
m2$V1>2## [1] FALSE FALSE TRUE TRUE
This gives us a TRUE/FALSE (Boolean) list of rows that show where the condition is/is not met.
To get the corresponding row numbers, we use the which() command:
which(m2$V1>2)## [1] 3 4
The next step replaces the numbers c(3,4) (the one-time case) with what R calculated. This would work on any matrix, no matter what the numbers or the order.
m2[m2$V1>2,]## V1 V2 V3
## 3 3 7 11
## 4 4 8 12
We can do multiple conditions as well. If we want to see which rows have the third column less than 12, we simply can enter:
m2[which(m2$V3<12),]## V1 V2 V3
## 1 1 5 9
## 2 2 6 10
## 3 3 7 11
For both conditions to be true, we want the overlap. First we had Rows 3 and 4; here we have 1, 2, and 3. This command gives us the one row that matches both:
m2[which(m2$V3<12 & m2$V1>2),]## V1 V2 V3
## 3 3 7 11
You can also do the Union (where either condition is met) with the | sign, but here that would give us all the rows!
Onother good way to find the largest or smallest numbers in a dataset are which.max() and which.min():
which.max(m2$V3)## [1] 4
That gives the position, not the value. So it has to be placed in the matrix:
m2[which.max(m2$V3),]## V1 V2 V3
## 4 4 8 12
If you just want the second column from that, you can put it in the column position of (rows, columns)
m2[which.max(m2$V3),2]## [1] 8
The next step is do open and work with “real” data; the other examples do a lot of this.