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Day 14 Swirl

install.packages("swirl")

https://github.com/swirldev/swirl_courses#swirl-courses

https://en.wikipedia.org/wiki/YAML
http://yaml.org/

| You can exit swirl and return to the R prompt (>) at any time by pressing the Esc key. If you are
| already at the prompt, type bye() to exit and save your progress. When you exit properly, you'll see a
| short message letting you know you've done so.

| When you are at the R prompt (>):
| -- Typing skip() allows you to skip the current question.
| -- Typing play() lets you experiment with R on your own; swirl will ignore what you do...
| -- UNTIL you type nxt() which will regain swirl's attention.
| -- Typing bye() causes swirl to exit. Your progress will be saved.
| -- Typing main() returns you to swirl's main menu.
| -- Typing info() displays these options again.

| Let's get started!

sqrt() function and to take the absolute value, use the abs() function

Vector of unequal length Artihmetic Operations
If the vectors are of different lengths,
| R 'recycles' the shorter vector until it is the same length as the longer vector.
| When we did z * 2 + 100 in our earlier example, z was a vector of length 3, but technically 2
| and 100 are each vectors of length 1.

...

  |==========================================================================                       |  76%

| Behind the scenes, R is 'recycling' the 2 to make a vector of 2s and the 100 to make a vector
| of 100s. In other words, when you ask R to compute z * 2 + 100, what it really computes is
| this: z * c(2, 2, 2) + c(100, 100, 100).


Shortcuts
 In many programming environments, the up arrow will cycle through previous commands. Try
| hitting the up arrow on your keyboard until you get to this command (z * 2 + 100), then
| change 100 to 1000 and hit Enter. If the up arrow doesn't work for you, just type the
| corrected command.

| You can type the first two letters of the variable name, then hit the Tab key (possibly more
| than once). Most programming environments will provide a list of variables that you've
| created that begin with 'my'. This is called auto-completion and can be quite handy when you
| have many variables in your workspace. Give it a try. (If auto-completion doesn't work for
| you, just type my_div and press Enter.)

List all the files in your working directory using list.files() or dir().

Use dir.create() to create a directory in the current working directory called "testdir".

> dir.create("testdir")
Create a file in your working directory called "mytest.R" using the file.create() function.

>
> file.create("mytest.R")
list.files() shows that the directory only contains mytest.R.

> list.files()

| Create a directory in the current working directory called "testdir2" and a subdirectory for
| it called "testdir3", all in one command by using dir.create() and file.path().

> dir.create(file.path('testdir2','testdir3'))
Warning message:
In dir.create(file.path("testdir2", "testdir3")) :
  cannot create dir 'testdir2\testdir3', reason 'No such file or directory'

| You're close...I can feel it! Try it again. Or, type info() for more options.

| dir.create(file.path('testdir2', 'testdir3'), recursive = TRUE) will do the trick. If you
| forgot the recursive argument, the command may have appeared to work, but it didn't create
| the nested directory.

> dir.create(file.path('testdir2','testdir3'),recursive = TRUE)

 That gave us every integer between (and including) 1 and 20. We could also use it to create a
| sequence of real numbers. For example, try pi:10.

> pi:10

| Remember that if you have questions about a particular R function, you can access its
| documentation with a question mark followed by the function name: ?function_name_here.
| However, in the case of an operator like the colon used above, you must enclose the symbol in
| backticks like this: ?`:`. (NOTE: The backtick (`) key is generally located in the top left
| corner of a keyboard, above the Tab key. If you don't have a backtick key, you can use
| regular quotes.)

This gives us the same output as 1:20. However, let's say that instead we want a vector of
| numbers ranging from 0 to 10, incremented by 0.5. seq(0, 10, by=0.5) does just that. Try it
| out.

| Another option is to use seq(along.with = my_seq).

If we're interested in creating a vector that contains 40 zeros, we can use rep(0, times =
| 40). Try it out.

> rep(0,time = 40)
 [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

| Try again. Getting it right on the first try is boring anyway! Or, type info() for more
| options.

| Type rep(0, times = 40) to make a vector containing 40 zeros.

> rep(0,times = 40)

Right now, my_char is a character vector of length 3. Let's say we want to join the elements
| of my_char together into one continuous character string (i.e. a character vector of length
| 1). We can do this using the paste() function.

...

  |======================================================                                |  63%

| Type paste(my_char, collapse = " ") now. Make sure there's a space between the double quotes
| in the `collapse` argument. You'll see why in a second.

> paste(my_char,collapse = " ")

 For a slightly more complicated example, we can join two vectors, each of length 3. Use
| paste() to join the integer vector 1:3 with the character vector c("X", "Y", "Z"). This time,
| use sep = "" to leave no space between the joined elements.

> paste(c(1:3),c("X","Y","Z"),sep = "")
[1] "1X" "2Y" "3Z"

| That's a job well done!

LETTERS is a predefined variable
| in R containing a character vector of all 26 letters in the English alphabet.


To make things a little more interesting, lets create a vector containing 1000 draws from a
| standard normal distribution with y <- div="" rnorm="">

> y <- div="" rnorm="">

| That's a job well done!

  |==========================                                                            |  30%

| Next, let's create a vector containing 1000 NAs with z <- 1000="" div="" rep="">

> z <- 1000="" div="" rep="">

| That's correct!

  |==============================                                                        |  35%

| Finally, let's select 100 elements at random from these 2000 values (combining y and z) such
| that we don't know how many NAs we'll wind up with or what positions they'll occupy in our
| final vector -- my_data <- 100="" c="" div="" sample="" y="" z="">

> my_data <- 100="" c="" div="" sample="" y="" z="">

In our previous discussion of logical operators, we introduced the `==` operator as a method
| of testing for equality between two objects. So, you might think the expression my_data == NA
| yields the same results as is.na(). Give it a try.

> my_data==NA
  [1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
 [31] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
 [61] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
 [91] NA NA NA NA NA NA NA NA NA NA

| Keep up the great work!

  |====================================================                                  |  60%

| The reason you got a vector of all NAs is that NA is not really a value, but just a
| placeholder for a quantity that is not available. Therefore the logical expression is
| incomplete and R has no choice but to return a vector of the same length as my_data that
| contains all NAs.

x[!is.na(x) & x > 0]

Now, let's check that vect and vect2 are the same by passing them as arguments to the
| identical() function.

> identical(vect,vect2)

Now we'll use the cbind() function to 'combine columns'. Don't worry about storing the result
| in a new variable. Just call cbind() with two arguments -- the patients vector and my_matrix.

> cbind(patients,my_matrix)
     patients                       
[1,] "Bill"   "1" "5" "9"  "13" "17"
[2,] "Gina"   "2" "6" "10" "14" "18"
[3,] "Kelly"  "3" "7" "11" "15" "19"
[4,] "Sean"   "4" "8" "12" "16" "20"

| You are quite good my friend!

Subsetting Function
http://www.statmethods.net/management/subset.html


http://stackoverflow.com/questions/3445590/how-to-extract-a-subset-of-a-data-frame-based-on-a-condition-involving-a-field

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