1 What are apply functions?

Apply functions are a family of functions in base R which allow you to repetitively perform an action on multiple chunks of data. An apply function is essentially a loop, but run faster than loops and often require less code.

The apply functions that this chapter will address are apply, lapply, sapply, vapply, tapply, and mapply. There are so many different apply functions because they are meant to operate on different types of data.

2 The apply function

First, let’s go over the basic apply function. You can use the help section to get a description of this function.

?apply

the apply function looks like this: apply(X, MARGIN, FUN).

  • X is an array or matrix (this is the data that you will be performing the function on)
  • Margin specifies whether you want to apply the function across rows (1) or columns (2)
  • FUN is the function you want to use

2.1 apply examples

my.matrx is a matrix with 1-10 in column 1, 11-20 in column 2, and 21-30 in column 3. my.matrx will be used to show some of the basic uses for the apply function.

my.matrx <- matrix(c(1:10, 11:20, 21:30), nrow = 10, ncol = 3)
my.matrx
##       [,1] [,2] [,3]
##  [1,]    1   11   21
##  [2,]    2   12   22
##  [3,]    3   13   23
##  [4,]    4   14   24
##  [5,]    5   15   25
##  [6,]    6   16   26
##  [7,]    7   17   27
##  [8,]    8   18   28
##  [9,]    9   19   29
## [10,]   10   20   30

2.1.1 Example 1: Using apply to find row sums

What if I wanted to summarize the data in matrix m by finding the sum of each row? The arguments are X = m, MARGIN = 1 (for row), and FUN = sum

apply(my.matrx, 1, sum)
##  [1] 33 36 39 42 45 48 51 54 57 60

The apply function returned a vector containing the sums for each row.

2.1.2 Example 2: Creating a function in the arguments

What if I wanted to be able to find how many datapoints (n) are in each column of m? I can use the length function to do this. Because we are using columns, MARGIN = 2.

apply(my.matrx, 2, length)
## [1] 10 10 10

What if instead, I wanted to find n-1 for each column? There isn’t a function in R to do this automatically, so I can create my own function. If the function is simple, you can create it right inside the arguments for apply. In the arguments I created a function that returns length - 1.

apply(my.matrx, 2, function (x) length(x)-1)
## [1] 9 9 9

As you can see, the function correctly returned a vector of n-1 for each column.

2.1.3 Example 3: Using a function defined outside of apply

If you don’t want to write a function inside of the arguments, you can define the function outside of apply, and then use that function in apply later. This may be useful if you want to have the function available to use later. In this example, a function to find standard error was created, then passed into an apply function.

st.err <- function(x){
  sd(x)/sqrt(length(x))
}
apply(my.matrx,2, st.err)
## [1] 0.9574271 0.9574271 0.9574271

2.1.4 Example 4: Transforming data

Now for something a little different. In the previous examples, apply was used to summarize over a row or column. It can also be used to repeat a function on cells within a matrix. In this example, the apply function is used to transform the values in each cell. Pay attention to the MARGIN argument. If you set the MARGIN to 1:2 it will have the function operate on each cell.

my.matrx2 <- apply(my.matrx,1:2, function(x) x+3)
my.matrx2
##       [,1] [,2] [,3]
##  [1,]    4   14   24
##  [2,]    5   15   25
##  [3,]    6   16   26
##  [4,]    7   17   27
##  [5,]    8   18   28
##  [6,]    9   19   29
##  [7,]   10   20   30
##  [8,]   11   21   31
##  [9,]   12   22   32
## [10,]   13   23   33

2.1.5 Example 5: Vectors?

The previous examples showed several ways to use the apply function on a matrix. But what if I wanted to loop through a vector instead? Will the apply function work?

vec <- c(1:10)
vec
##  [1]  1  2  3  4  5  6  7  8  9 10
apply(vec, 1, sum)

If you run this function it will return the error: Error in apply(v, 1, sum) : dim(X) must have a positive length. As you can see, this didn’t work because apply was expecting the data to have at least two dimensions. If your data is a vector you need to use lapply, sapply, or vapply instead.

3 lapply, sapply, and vapply

lapply, sapply, and vapply are all functions that will loop a function through data in a list or vector. First, try looking up lapply in the help section to see a description of all three function.

?lapply

Here are the agruments for the three functions:

  • lapply(X, FUN, …)
  • sapply(X, FUN, …, simplify = TRUE, USE.NAMES = TRUE)
  • vapply(X, FUN, FUN.VALUE, …, USE.NAMES = TRUE)

In this case, X is a vector or list, and FUN is the function you want to use. sapply and vapply have extra arguments, but most of them have default values, so you don’t need to worry about them. However, vapply requires another agrument called FUN.VALUE, which we will look at later.

3.0.1 Example 1: Getting started with lapply

Earlier, we created the vector v. Let’s use that vector to test out the lapply function.

lapply(vec, sum)
## [[1]]
## [1] 1
## 
## [[2]]
## [1] 2
## 
## [[3]]
## [1] 3
## 
## [[4]]
## [1] 4
## 
## [[5]]
## [1] 5
## 
## [[6]]
## [1] 6
## 
## [[7]]
## [1] 7
## 
## [[8]]
## [1] 8
## 
## [[9]]
## [1] 9
## 
## [[10]]
## [1] 10

This function didn’t add up the values like we may have expected it to. This is because lapply applies treats the vector like a list, and applies the function to each point in the vector.

Let’s try using a list instead

A<-c(1:9)
B<-c(1:12)
C<-c(1:15)
my.lst<-list(A,B,C)
lapply(my.lst, sum)
## [[1]]
## [1] 45
## 
## [[2]]
## [1] 78
## 
## [[3]]
## [1] 120

This time, the lapply function seemed to work better. The function summed each vector in the list and returned a list of the 3 sums.

3.0.2 Example 2: sapply

sapply works just like lapply, but will simplify the output if possible. This means that instead of returning a list like lapply, it will return a vector instead if the data is simplifiable.

sapply(vec, sum)
##  [1]  1  2  3  4  5  6  7  8  9 10
sapply(my.lst, sum)
## [1]  45  78 120

See how these two examples gave the same answers, but returned a vector instead?

3.0.3 Example 3: vapply

vapply is similar to sapply, but it requires you to specify what type of data you are expecting the arguments for vapply are vapply(X, FUN, FUN.VALUE). FUN.VALUE is where you specify the type of data you are expecting. I am expecting each item in the list to return a single numeric value, so FUN.VALUE = numeric(1).

vapply(vec, sum, numeric(1))
##  [1]  1  2  3  4  5  6  7  8  9 10
vapply(my.lst, sum, numeric(1))
## [1]  45  78 120

If your function were to return more than one numeric value, FUN.VALUE = numeric(1) will cause the function to return an error. This could be useful if you are expecting only one result per subject.

#vapply(my.lst, function(x) x+2, numeric(1))

3.0.4 Example 4: Transforming data with sapply

Like apply, these functions can also be used for transforming data inside the list

my.lst2 <- sapply(my.lst, function(x) x*2)
my.lst2
## [[1]]
## [1]  2  4  6  8 10 12 14 16 18
## 
## [[2]]
##  [1]  2  4  6  8 10 12 14 16 18 20 22 24
## 
## [[3]]
##  [1]  2  4  6  8 10 12 14 16 18 20 22 24 26 28 30

3.0.5 Which function should I use, lapply, sapply, or vapply?

If you are trying to decide which of these three functions to use, because it is the simplest, I would suggest to use sapply if possible. If you do not want your results to be simplified to a vector, lapply should be used. If you want to specify the type of result you are expecting, use vapply.

4 tapply

Sometimes you may want to perform the apply function on some data, but have it separated by factor. In that case, you should use tapply. Let’s take a look at the information for tapply.

?tapply

The arguments for tapply are tapply(X, INDEX, FUN). The only new argument is INDEX, which is the factor you want to use to separate the data.

4.0.1 Example 1: Means split by condition

First, let’s create data with an factor for indexing. Dataset t will be created by adding a factor to matrix m and converting it to a dataframe.

tdata <- as.data.frame(cbind(c(1,1,1,1,1,2,2,2,2,2), my.matrx))
colnames(tdata)
## [1] "V1" "V2" "V3" "V4"

Now let’s use column 1 as the index and find the mean of column 2

tapply(tdata$V2, tdata$V1, mean)
## 1 2 
## 3 8

4.0.2 Example 2: Combining functions

You can use tapply to do some quick summary statistics on a variable split by condition. In this example, I created a function that returns a vector ofboth the mean and standard deviation. You can create a function like this for any apply function, not just tapply.

summary <- tapply(tdata$V2, tdata$V1, function(x) c(mean(x), sd(x)))
summary
## $`1`
## [1] 3.000000 1.581139
## 
## $`2`
## [1] 8.000000 1.581139

5 mapply

the last apply function I will cover is mapply.

?mapply

the arguments for mapply are mapply(FUN, …, MoreArgs = NULL, SIMPLIFY = TRUE, USE.NAMES = TRUE). First you list the function, followed by the vectors you are using the rest of the arguments have default values so they don’t need to be changed for now. When you have a function that takes 2 arguments, the first vector goes into the first argument and the second vector goes into the second argument.

5.0.1 Example 1: Understanding mapply

In this example, 1:9 is specifying the value to repeat, and 9:1 is specifying how many times to repeat. This order is based on the order of arguments in the rep function itself.

mapply(rep, 1:9, 9:1)
## [[1]]
## [1] 1 1 1 1 1 1 1 1 1
## 
## [[2]]
## [1] 2 2 2 2 2 2 2 2
## 
## [[3]]
## [1] 3 3 3 3 3 3 3
## 
## [[4]]
## [1] 4 4 4 4 4 4
## 
## [[5]]
## [1] 5 5 5 5 5
## 
## [[6]]
## [1] 6 6 6 6
## 
## [[7]]
## [1] 7 7 7
## 
## [[8]]
## [1] 8 8
## 
## [[9]]
## [1] 9

5.0.2 Example 2: Creating a new variable

Another use for mapply would be to create a new variable. For example, using dataset t, I could divide one column by another column to create a new value. This would be useful for creating a ratio of two variables as shown in the example below.

tdata$V5 <- mapply(function(x, y) x/y, tdata$V2, tdata$V4)
tdata$V5
##  [1] 0.04761905 0.09090909 0.13043478 0.16666667 0.20000000 0.23076923
##  [7] 0.25925926 0.28571429 0.31034483 0.33333333

5.0.3 Example 3: Saving data into a premade vector

When using an apply family function to create a new variable, one option is to create a new vector ahead of time with the size of the vector pre-allocated. I created a numeric vector of length 10 using the vector function. The arguments for the vector function are vector(mode, length). Inside mapply I created a function to multiple two variables together. The results of the mapply function are then saved into the vector.

new.vec <- vector(mode = "numeric", length = 10)
new.vec <- mapply(function(x, y) x*y, tdata$V3, tdata$V4)
new.vec
##  [1] 231 264 299 336 375 416 459 504 551 600

6 Using apply functions on real datasets

This last section will be a few examples of using apply functions on real data.This section will make use of the MASS package, which is a collection of publicly available datasets. Please install MASS if you do not already have it. If you do not have MASS installed, you can uncomment the code below.

#install.packages("MASS")
library(MASS)

load the state dataset. It contains information about all 50 states

data(state)

Let’s look at the data we will be using. We will be using the state.x77 dataset

head(state.x77)
##            Population Income Illiteracy Life Exp Murder HS Grad Frost
## Alabama          3615   3624        2.1    69.05   15.1    41.3    20
## Alaska            365   6315        1.5    69.31   11.3    66.7   152
## Arizona          2212   4530        1.8    70.55    7.8    58.1    15
## Arkansas         2110   3378        1.9    70.66   10.1    39.9    65
## California      21198   5114        1.1    71.71   10.3    62.6    20
## Colorado         2541   4884        0.7    72.06    6.8    63.9   166
##              Area
## Alabama     50708
## Alaska     566432
## Arizona    113417
## Arkansas    51945
## California 156361
## Colorado   103766
str(state.x77)
##  num [1:50, 1:8] 3615 365 2212 2110 21198 ...
##  - attr(*, "dimnames")=List of 2
##   ..$ : chr [1:50] "Alabama" "Alaska" "Arizona" "Arkansas" ...
##   ..$ : chr [1:8] "Population" "Income" "Illiteracy" "Life Exp" ...

All the data in the dataset happens to be numeric, which is necessary when the function inside the apply function requires numeric data.

6.0.1 Example 1: using apply to get summary data

You can use apply to find measures of central tendency and dispersion

apply(state.x77, 2, mean)
## Population     Income Illiteracy   Life Exp     Murder    HS Grad 
##  4246.4200  4435.8000     1.1700    70.8786     7.3780    53.1080 
##      Frost       Area 
##   104.4600 70735.8800
apply(state.x77, 2, median)
## Population     Income Illiteracy   Life Exp     Murder    HS Grad 
##   2838.500   4519.000      0.950     70.675      6.850     53.250 
##      Frost       Area 
##    114.500  54277.000
apply(state.x77, 2, sd)
##   Population       Income   Illiteracy     Life Exp       Murder 
## 4.464491e+03 6.144699e+02 6.095331e-01 1.342394e+00 3.691540e+00 
##      HS Grad        Frost         Area 
## 8.076998e+00 5.198085e+01 8.532730e+04

6.0.2 Example 2: Saving the results of apply

In this, I created one function that gives the mean and SD, and another that give min, median, and max. Then I saved them as objects that could be used later.

state.summary<- apply(state.x77, 2, function(x) c(mean(x), sd(x))) 
state.summary
##      Population    Income Illiteracy  Life Exp  Murder   HS Grad     Frost
## [1,]   4246.420 4435.8000  1.1700000 70.878600 7.37800 53.108000 104.46000
## [2,]   4464.491  614.4699  0.6095331  1.342394 3.69154  8.076998  51.98085
##          Area
## [1,] 70735.88
## [2,] 85327.30
state.range <- apply(state.x77, 2, function(x) c(min(x), median(x), max(x)))
state.range
##      Population Income Illiteracy Life Exp Murder HS Grad Frost   Area
## [1,]      365.0   3098       0.50   67.960   1.40   37.80   0.0   1049
## [2,]     2838.5   4519       0.95   70.675   6.85   53.25 114.5  54277
## [3,]    21198.0   6315       2.80   73.600  15.10   67.30 188.0 566432

6.0.3 Example 3: Using mapply to compute a new variable

In this example, I want to find the population density for each state. In order to do this, I want to divide population by area. state.area and state.x77 are not from the same dataset, but that is fine as long as the vectors are the same length and the data is in the same order. Both vectors are alphabetically by state, so mapply can be used.

population <- state.x77[1:50]
area <- state.area
pop.dens <- mapply(function(x, y) x/y, population, area)
pop.dens
##  [1] 0.070045922 0.000618899 0.019419010 0.039733353 0.133578671
##  [6] 0.024374802 0.618886005 0.281477880 0.141342213 0.083752293
## [11] 0.134573643 0.009729885 0.198528369 0.146399934 0.050826079
## [16] 0.027715647 0.083847011 0.078437030 0.031853078 0.389713529
## [21] 0.704129829 0.156503367 0.046640815 0.049061112 0.068406854
## [26] 0.005070070 0.019993008 0.005337434 0.087274291 0.935809086
## [31] 0.009402791 0.364611909 0.103468604 0.009014364 0.260419194
## [36] 0.038830647 0.023551005 0.261619571 0.766886326 0.090677830
## [41] 0.008838761 0.098783259 0.045773344 0.014166941 0.049120616
## [46] 0.122038466 0.052190873 0.074397254 0.081721694 0.003840105

6.0.4 Example 4: Using tapply to explore population by region

In this example, I want to find out some information about the population of states split by region. state.region is a factor with four levels: Northeast, South, North Central, and West. For each region, I want the minimum, median, and maximum populations.

region.info <- tapply(population, state.region, function(x) c(min(x), median(x), max(x)))
region.info
## $Northeast
## [1]   472  3100 18076
## 
## $South
## [1]   579.0  3710.5 12237.0
## 
## $`North Central`
## [1]   637  4255 11197
## 
## $West
## [1]   365  1144 21198

7 References

Here are some sources I used to help me create this chapter:

Datacamp tutorial on apply functions: https://www.datacamp.com/community/tutorials/r-tutorial-apply-family

r-bloggers: Using apply, sapply, and lapply in R: https://www.r-bloggers.com/using-apply-sapply-lapply-in-r/

stackoverflow: Why is vapply safer than sapply?: http://stackoverflow.com/questions/12339650/why-is-vapply-safer-than-sapply

---
title: 'Chapter 4: apply Functions'
author: "Erin Sovansky Winter"
output:
  html_document:
    theme: cerulean
    highlight: textmate
    fontsize: 8pt
    toc: true
    number_sections: true
    code_download: true
    toc_float:
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---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```

#  What are apply functions?
Apply functions are a family of functions in base R which allow you to repetitively perform an
action on multiple chunks of data. An apply function is essentially a loop, but run faster than 
loops and often require less code. 

The apply functions that this chapter will address are apply, lapply, sapply, vapply, tapply, and
mapply. There are so many different apply functions because they are meant to operate on different
types of data. 

#  The apply function
First, let's go over the basic apply function. You can use the help section to get a description
of this function.
```{r, eval=FALSE}
?apply
```
the apply function looks like this: apply(X, MARGIN, FUN). 

* X is an array or matrix (this is the data that you will be performing the function on)
* Margin specifies whether you want to apply the function across rows (1) or columns (2)
* FUN is the function you want to use

## apply examples
my.matrx is a matrix with 1-10 in column 1, 11-20 in column 2, and 21-30 in column 3. 
my.matrx will be used to show some of the basic uses for the apply function.
```{r}
my.matrx <- matrix(c(1:10, 11:20, 21:30), nrow = 10, ncol = 3)
my.matrx
```

### Example 1: Using apply to find row sums
What if I wanted to summarize the data in matrix m by finding the sum of each row? The arguments 
are X = m, MARGIN = 1 (for row), and FUN = sum

```{r}
apply(my.matrx, 1, sum)
```
The apply function returned a vector containing the sums for each row.

### Example 2: Creating a function in the arguments
What if I wanted to be able to find how many datapoints (n) are in each column of m? I can use 
the length function to do this. Because we are using columns, MARGIN = 2.
```{r}
apply(my.matrx, 2, length)
```
What if instead, I wanted to find n-1 for each column? There isn't a function in R to do this
automatically, so I can create my own function. If the function is simple, you can create it
right inside the arguments for apply. In the arguments I created a function that returns
length - 1.
```{r}
apply(my.matrx, 2, function (x) length(x)-1)
```
As you can see, the function correctly returned a vector of n-1 for each column.
 
### Example 3: Using a function defined outside of apply
If you don't want to write a function inside of the arguments, you can define the function 
outside of apply, and then use that function in apply later. This may be useful if you want to 
have the function available to use later. In this example, a function to find standard error was
created, then passed into an apply function.
```{r}
st.err <- function(x){
  sd(x)/sqrt(length(x))
}
apply(my.matrx,2, st.err)
```

### Example 4: Transforming data
Now for something a little different. In the previous examples, apply was used to summarize
over a row or column. It can also be used to repeat a function on cells within a matrix. In this
example, the apply function is used to transform the values in each cell. Pay attention to the
MARGIN argument. If you set the MARGIN to 1:2 it will have the function operate on each cell.
```{r}
my.matrx2 <- apply(my.matrx,1:2, function(x) x+3)
my.matrx2
```

### Example 5: Vectors?
The previous examples showed several ways to use the apply function on a matrix. But what if I 
wanted to loop through a vector instead? Will the apply function work?

```{r, }
vec <- c(1:10)
vec
```
```{r, eval=FALSE}
apply(vec, 1, sum)
```
If you run this function it will return the error: Error in apply(v, 1, sum) : dim(X) must have a positive length. 
As you can see, this didn't work because apply was expecting the data to have at least two dimensions. If your data is a vector you need to use lapply, sapply, or vapply instead.

# lapply, sapply, and vapply
lapply, sapply, and vapply are all functions that will loop a function through data in a list or
vector. First, try looking up lapply in the help section to see a description of all three 
function.

```{r, eval=FALSE}
?lapply
```

Here are the agruments for the three functions:

* lapply(X, FUN, ...)
* sapply(X, FUN, ..., simplify = TRUE, USE.NAMES = TRUE)
* vapply(X, FUN, FUN.VALUE, ..., USE.NAMES = TRUE)

In this case, X is a vector or list, and FUN is the function you want to use. sapply and vapply have extra arguments, but most of them have default values, so you don't need to worry about
them. However, vapply requires another agrument called FUN.VALUE, which we will look at later.

### Example 1: Getting started with lapply
Earlier, we created the vector v. Let's use that vector to test out the lapply function.
```{r}
lapply(vec, sum)
```
This function didn't add up the values like we may have expected it to. This is because lapply
applies treats the vector like a list, and applies the function to each point in the vector.

Let's try using a list instead
```{r}
A<-c(1:9)
B<-c(1:12)
C<-c(1:15)
my.lst<-list(A,B,C)
lapply(my.lst, sum)
```
This time, the lapply function seemed to work better. The function summed each vector in the list
and returned a list of the 3 sums. 

### Example 2: sapply
sapply works just like lapply, but will simplify the output if possible. This means that instead
of returning a list like lapply, it will return a vector instead if the data is simplifiable.

```{r}
sapply(vec, sum)
```

```{r}
sapply(my.lst, sum)
```
See how these two examples gave the same answers, but returned a vector instead?

### Example 3: vapply
vapply is similar to sapply, but it requires you to specify what type of data you are expecting
the arguments for vapply are vapply(X, FUN, FUN.VALUE).
FUN.VALUE is where you specify the type of data you are expecting.
I am expecting each item in the list to return a single numeric value, so FUN.VALUE = numeric(1).

```{r}
vapply(vec, sum, numeric(1))
```

```{r}
vapply(my.lst, sum, numeric(1))
```

If your function were to return more than one numeric value, FUN.VALUE = numeric(1) will cause the function to return an error. This could be useful if you are expecting only one result per subject. 
```{r}
#vapply(my.lst, function(x) x+2, numeric(1))
```

### Example 4: Transforming data with sapply
Like apply, these functions can also be used for transforming data inside the list
```{r}
my.lst2 <- sapply(my.lst, function(x) x*2)
my.lst2
```

### Which function should I use, lapply, sapply, or vapply?

If you are trying to decide which of these three functions to use, because it is the simplest, I would suggest to use sapply if possible. If you do not want your results to be simplified to a vector, lapply should be used. If you want to specify the type of result you are expecting, use vapply.


# tapply

Sometimes you may want to perform the apply function on some data, but have it separated by 
factor. In that case, you should use tapply. Let's take a look at the information for tapply.

```{r, eval=FALSE}
?tapply
```
The arguments for tapply are tapply(X, INDEX, FUN). The only new argument is INDEX, which is the 
factor you want to use to separate the data.

### Example 1: Means split by condition
First, let's create data with an factor for indexing. Dataset t will be created by adding a factor to matrix m and converting it to a dataframe. 

```{r}
tdata <- as.data.frame(cbind(c(1,1,1,1,1,2,2,2,2,2), my.matrx))
colnames(tdata)
```
Now let's use column 1 as the index and find the mean of column 2

```{r}
tapply(tdata$V2, tdata$V1, mean)
```

### Example 2: Combining functions
You can use tapply to do some quick summary statistics on a variable split by condition. In this 
example, I created a function that returns a vector ofboth the mean and standard deviation. You 
can create a function like this for any apply function, not just tapply.
```{r}
summary <- tapply(tdata$V2, tdata$V1, function(x) c(mean(x), sd(x)))
summary
```

# mapply
the last apply function I will cover is mapply.
```{r, eval=FALSE}
?mapply
```
the arguments for mapply are mapply(FUN, ..., MoreArgs = NULL, SIMPLIFY = TRUE, USE.NAMES = TRUE).
First you list the function, followed by the vectors you are using
the rest of the arguments have default values so they don't need to be changed for now. 
When you have a function that takes 2 arguments, the first vector goes into the first argument
and the second vector goes into the second argument.

### Example 1: Understanding mapply
In this example, 1:9 is specifying the value to repeat, and 9:1 is specifying how many times
to repeat. This order is based on the order of arguments in the rep function itself.
```{r}
mapply(rep, 1:9, 9:1)
```

### Example 2: Creating a new variable
Another use for mapply would be to create a new variable. For example, using dataset t, I could
divide one column by another column to create a new value. This would be useful for creating a 
ratio of two variables as shown in the example below. 

```{r}
tdata$V5 <- mapply(function(x, y) x/y, tdata$V2, tdata$V4)
tdata$V5
```

### Example 3: Saving data into a premade vector
When using an apply family function to create a new variable, one option is to create a new vector ahead of time with the size of the vector pre-allocated. I created a numeric vector of length 10 using the vector function. The arguments for the vector function are vector(mode, length). Inside mapply I created a function to multiple two variables together. The results of the mapply function are then saved into the vector.

```{r}
new.vec <- vector(mode = "numeric", length = 10)
new.vec <- mapply(function(x, y) x*y, tdata$V3, tdata$V4)
new.vec
```

# Using apply functions on real datasets
This last section will be a few examples of using apply functions on real data.This section will
make use of the MASS package, which is a collection of publicly available datasets. Please
install MASS if you do not already have it. If you do not have MASS installed, you can uncomment
the code below.

```{r}
#install.packages("MASS")
library(MASS)
```

load the state dataset. It contains information about all 50 states
```{r}
data(state)
```
Let's look at the data we will be using. We will be using the state.x77 dataset
```{r}
head(state.x77)
str(state.x77)
```
All the data in the dataset happens to be numeric, which is necessary when the function inside the apply function requires numeric data.

### Example 1: using apply to get summary data
You can use apply to find measures of central tendency and dispersion
```{r}
apply(state.x77, 2, mean)
apply(state.x77, 2, median)
apply(state.x77, 2, sd)
```

### Example 2: Saving the results of apply

In this, I created one function that gives the mean and SD, and another that give min, median, and max. Then I saved them as objects that could be used later.
```{r}
state.summary<- apply(state.x77, 2, function(x) c(mean(x), sd(x))) 
state.summary
state.range <- apply(state.x77, 2, function(x) c(min(x), median(x), max(x)))
state.range
```

### Example 3: Using mapply to compute a new variable
In this example, I want to find the population density for each state. In order to do this, I 
want to divide population by area. state.area and state.x77 are not from the same dataset, but 
that is fine as long as the vectors are the same length and the data is in the same order. Both
vectors are alphabetically by state, so mapply can be used.
```{r}
population <- state.x77[1:50]
area <- state.area
pop.dens <- mapply(function(x, y) x/y, population, area)
pop.dens
```

### Example 4: Using tapply  to explore population by region
In this example, I want to find out some information about the population of states split by
region. state.region is a factor with four levels: Northeast, South, North Central, and West.
For each region, I want the minimum, median, and maximum populations.

```{r}
region.info <- tapply(population, state.region, function(x) c(min(x), median(x), max(x)))
region.info
```

# References
Here are some sources I used to help me create this chapter:

Datacamp tutorial on apply functions: https://www.datacamp.com/community/tutorials/r-tutorial-apply-family

r-bloggers: Using apply, sapply, and lapply in R: https://www.r-bloggers.com/using-apply-sapply-lapply-in-r/

stackoverflow: Why is vapply safer than sapply?: http://stackoverflow.com/questions/12339650/why-is-vapply-safer-than-sapply


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