class: title-slide

<br>
<br>
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# The Pipe Operator
## Dr. Mine Dogucu
]

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.font50[Three solutions to a single problem]

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What is the average of 4, 8, 16 approximately?

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1.What is the average of **4, 8, 16** approximately?

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2.What is the **average** of 4, 8, 16 approximately?

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3.What is the average of 4, 8, 16 **approximately**?

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.font50[Solution 1: Functions within Functions]

---

```r
c(4, 8, 16)
```

```
## [1]  4  8 16
```

--

<hr>

```r
mean(c(4, 8, 16))
```

```
## [1] 9.333333
```

--

<hr>

```r
round(mean(c(4, 8, 16)))
```

```
## [1] 9
```

---

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**Problem with writing functions within functions**

Things will get messy and more difficult to read and debug as we deal with more complex operations on data.

---

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.font50[Solution 2: Creating Objects]

---

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```r
numbers <- c(4, 8, 16)
numbers
```

```
## [1]  4  8 16
```

--

<hr>

```r
avg_number <- mean(numbers)
avg_number
```

```
## [1] 9.333333
```

--

<hr>

```r
round(avg_number)
```

```
## [1] 9
```

---

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**Problem with creating many objects**

We will end up with too many objects in `Environment`.

---

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.font50[Solution 3: The (forward) Pipe Operator %>% ]

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.font75[Shortcut: <br>Ctrl (Command) + Shift + M]

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.pull-left[

```r
c(4, 8, 16) %>% 
  mean() %>% 
  round()
```

```
## [1] 9
```
]

.pull-right[

Combine 4, 8, and 16 `and then`  
Take the mean   `and then`  
Round the output

]

---

.pull-left[

```r
c(4, 8, 16) %>% 
  mean() %>% 
  round()
```
]

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The output of the first function is the first argument of the second function.

---

Do you recall composite functions such as `\(f \circ g(x)\)`?

--

Now we have `\(f \circ g \circ h (x)\)` or `round(mean(c(4, 8, 16)))`

--

.pull-left[

```r
h(x) %>% 
  g() %>% 
  f()
```

]

.pull-right[

```r
c(4, 8, 16) %>% 
  mean() %>% 
  round()
```

]

Notes for current slide

Notes for next slide

The Pipe Operator

Dr. Mine Dogucu

1 / 24

Three solutions to a single problem

2 / 24

What is the average of 4, 8, 16 approximately?

3 / 24

1.What is the average of 4, 8, 16 approximately?

4 / 24

2.What is the average of 4, 8, 16 approximately?

5 / 24

3.What is the average of 4, 8, 16 approximately?

6 / 24

Solution 1: Functions within Functions

7 / 24

c(4, 8, 16)

## [1]  4  8 16

8 / 24

c(4, 8, 16)

## [1]  4  8 16

mean(c(4, 8, 16))

## [1] 9.333333

9 / 24

c(4, 8, 16)

## [1]  4  8 16

mean(c(4, 8, 16))

## [1] 9.333333

round(mean(c(4, 8, 16)))

## [1] 9

10 / 24

Problem with writing functions within functions

Things will get messy and more difficult to read and debug as we deal with more complex operations on data.

11 / 24

Solution 2: Creating Objects

12 / 24

numbers <- c(4, 8, 16)
numbers

## [1]  4  8 16

13 / 24

numbers <- c(4, 8, 16)
numbers

## [1]  4  8 16

avg_number <- mean(numbers)
avg_number

## [1] 9.333333

14 / 24

numbers <- c(4, 8, 16)
numbers

## [1]  4  8 16

avg_number <- mean(numbers)
avg_number

## [1] 9.333333

round(avg_number)

## [1] 9

15 / 24

Problem with creating many objects

We will end up with too many objects in Environment.

16 / 24

Solution 3: The (forward) Pipe Operator %>%

17 / 24

Shortcut:
Ctrl (Command) + Shift + M

18 / 24

c(4, 8, 16) %>% 
  mean() %>% 
  round()

## [1] 9

Combine 4, 8, and 16 and then
Take the mean and then
Round the output

19 / 24

c(4, 8, 16) %>% 
  mean() %>% 
  round()

20 / 24

The output of the first function is the first argument of the second function.

21 / 24

Do you recall composite functions such as $f \circ g (x)$ ?

22 / 24

Do you recall composite functions such as $f \circ g (x)$ ?

Now we have $f \circ g \circ h (x)$ or round(mean(c(4, 8, 16)))

23 / 24

Do you recall composite functions such as $f \circ g (x)$ ?

Now we have $f \circ g \circ h (x)$ or round(mean(c(4, 8, 16)))

h(x) %>% 
  g() %>% 
  f()

c(4, 8, 16) %>% 
  mean() %>% 
  round()

24 / 24

Three solutions to a single problem

2 / 24

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