[PDF] - Basics of Algorithmics in R a ) T R U E | F A L S E b ) PDF Document

SLIDE 1

Which expression(s) equal to TRUE? (x equals 5) a ) T R U E | F A L S E b ) x > 5 c ) F A L S E & T R U E d ) x < = 1 | x > 5 Answer: a) and d) What is the value of y at the end of the loop if it was 0 and the beginning? How many iterations of the loop occurred? while (y <= 10) { y <- 2*y + 3} Answer: y = 21; the loop ran 3 times.

An introduction to WS 2019/2020

Dr. Noémie Becker
Dr. Eliza Argyridou

Special thanks to:

Dr. Sonja Grath for addition to slides

Basics of Algorithmics in R

3

What you should know after days 7 & 8

Review: Data frames and import your data Conditional execution in R

Logic rules
if(), else(), ifelse()
Example from day 1

Loops Executing a command from a script Writing your own functions How to avoid slow R code 4

Basics

Syntax: m y f u n <

f

u n c t i

n

( a r g 1 , a r g 2 , … ) { c

m

m a n d s } Example: We want to define a function that takes a DNA sequence as input and gives as output the GC content (proportion of G and C in the sequence). How can we name our function? Idea: g c ? g c # T h e r e i s a l r e a d y a f u n c t i

n

g c ( ) Another idea: g c C

n

t e n t ? g c C

n

t e n t N

d
c

u m e n t a t i

n

f

r

‘ g c C

n

t e n t ’ i n s p e c i fj e d p a c k a g e s a n d l i b r a r i e s : y

u

c

u

l d t r y ‘ ? ? g c C

n

t e n t ’ → We can name our function g c C

n

t e n t ( )

5

Our function gcContent() from Day 1

Version 1 g c C

n

t e n t <

f

u n c t i

n

( d n a , c

u

n t e r = ) { d n a <

u

n l i s t ( s t r s p l i t ( d n a , " " ) ) f

r

( i i n 1 : l e n g t h ( d n a ) ) { i f ( d n a [ i ] = = " C " | d n a [ i ] = = " G " ) { c

u

n t e r = c

u

n t e r + 1 } } r e t u r n ( c

u

n t e r / l e n g t h ( d n a ) ) } Does our function works correctly? # T e s t t h e f u n c t i

n

w i t h s

m

e e x a m p l e d a t a g c C

n

t e n t ( " A A C G T G G C T A " ) g c C

n

t e n t ( " A A T A T A T T A T " ) g c C

n

t e n t ( 2 3 ) g c C

n

t e n t ( T R U E ) g c C

n

t e n t ( " n

t

D N A " ) g c C

n

t e n t ( " C

l

" ) YOUR TURN

6

Dealing with problems

Problems:

R gives an error message if the input is not a character value
Our function calculates values if the input is most likely not a DNA

sequence How could we deal with these problems? What do we want our function to output in these cases?

SLIDE 2

7

Error and Warning

There are two types of error messages in R:

Error: Stops execution and returns no value
Warning message: Continues execution

Example: x <

s

u m ( " h e l l

"

) E r r

r

i n s u m ( " h e l l

"

) : i n v a l i d ' t y p e ' ( c h a r a c t e r )

f

a r g u m e n t x <

m

e a n ( " h e l l

"

) W a r n i n g m e s s a g e : I n m e a n . d e f a u l t ( " h e l l

"

) : a r g u m e n t i s n

t

n u m e r i c

r

l

g

i c a l : r e t u r n i n g N A

We can define such messages with the functions s t

p

( ) and w a r n i n g ( ) In our example:

(Specific) Error when argument is not character
Warning if character argument is not DNA

8

Dealing with non-character arguments

Version 2: g c C

n

t e n t <

f

u n c t i

n

( d n a , c

u

n t e r = ) { i f ( ! i s . c h a r a c t e r ( d n a ) ) { s t

p

( " T h e a r g u m e n t m u s t b e

f

t y p e c h a r a c t e r . " ) } d n a <

u

n l i s t ( s t r s p l i t ( d n a , " " ) ) f

r

( i i n 1 : l e n g t h ( d n a ) ) { i f ( d n a [ i ] = = " C " | d n a [ i ] = = " G " ) { c

u

n t e r = c

u

n t e r + 1 } } r e t u r n ( c

u

n t e r / l e n g t h ( d n a ) ) }

Self-defined error message

9

Dealing with input that is not DNA

We define as 'not DNA' any character different from A, C, G or T.
If the input contains any other character, we compute the value but throw

a warning. To solve this task, we can use the function g r e p ( ) as follows: g r e p ( " [ ^ A C G T ] " , " A A T G A C " ) I n t e g e r ( ) # l e n g t h i s g r e p ( " [ ^ A C G T ] " , " N A T G A C " ) [ 1 ] 1 # l e n g t h i s 1

10

Dealing with input that is not DNA

Version 3

g c C

n

t e n t <

f

u n c t i

n

( d n a , c

u

n t e r = ) { i f ( ! i s . c h a r a c t e r ( d n a ) ) { s t

p

( " T h e a r g u m e n t m u s t b e

f

t y p e c h a r a c t e r . " ) } i f ( l e n g t h ( g r e p ( " [ ^ A C G T ] " , d n a ) ) > ) { w a r n i n g ( " T h e i n p u t c

n

t a i n s c h a r a c t e r s

t

h e r t h a n A , C , G

r

T

v

a l u e s h

u

l d n

t

b e t r u s t e d ! " ) } d n a <

u

n l i s t ( s t r s p l i t ( d n a , " " ) ) f

r

( i i n 1 : l e n g t h ( d n a ) ) { i f ( d n a [ i ] = = " C " | d n a [ i ] = = " G " ) { c

u

n t e r = c

u

n t e r + 1 } } r e t u r n ( c

u

n t e r / l e n g t h ( d n a ) ) }

Self-defined warning message

11

Giving several arguments to a function

R functions can have several arguments. You can see them listed in the help page for the function. Example ? m e a n ( ) A frequent argument in R functions is na.rm. This argument (when set to TRUE) removes NA values from vectors. m e a n ( c ( 1 , 2 , N A ) ) [ 1 ] N A m e a n ( c ( 1 , 2 , N A ) , n a . r m = T R U E ) [ 1 ] 1 . 5 We now want to give our function another argument to output the AT content instead of the GC content. YOUR TURN

12

Giving several arguments to a function

Version 4

g c C

n

t e n t <

f

u n c t i

n

( d n a , c

u

n t e r = , A T ) { i f ( ! i s . c h a r a c t e r ( d n a ) ) { s t

p

( " T h e a r g u m e n t m u s t b e

f

t y p e c h a r a c t e r . " ) } i f ( l e n g t h ( g r e p ( " [ ^ A C G T ] " , d n a ) ) > ) { w a r n i n g ( " T h e i n p u t c

n

t a i n s c h a r a c t e r s

t

h e r t h a n A , C , G

r

T

v

a l u e s h

u

l d n

t

b e t r u s t e d ! " ) } d n a <

u

n l i s t ( s t r s p l i t ( d n a , " " ) ) f

r

( i i n 1 : l e n g t h ( d n a ) ) { i f ( d n a [ i ] = = " C " | d n a [ i ] = = " G " ) { c

u

n t e r = c

u

n t e r + 1 } } i f ( A T = = T R U E ) { r e t u r n ( 1

c
u

n t e r / l e n g t h ( d n a ) ) } e l s e { r e t u r n ( c

u

n t e r / l e n g t h ( d n a ) ) } }

SLIDE 3

13

Giving several arguments to a function

Test: g c C

n

t e n t ( " A A C G T G T T T A " , A T = T R U E ) [ 1 ] . 7 g c C

n

t e n t ( " A A C G T G T T T A " ) E r r

r

i n g c C

n

t e n t ( " A A C G T G T T T A " ) : a r g u m e n t " A T " i s m i s s i n g , w i t h n

d

e f a u l t We should give the value F A L S E to the argument AT per default and the argument would only be changed if the user specifies A T = T R U E 14

Calculating GC/AT content – Final version

g c C

n

t e n t <

f

u n c t i

n

( d n a , c

u

n t e r = , A T = F A L S E ) { i f ( ! i s . c h a r a c t e r ( d n a ) ) { s t

p

( " T h e a r g u m e n t m u s t b e

f

t y p e c h a r a c t e r . " ) } i f ( l e n g t h ( g r e p ( " [ ^ A C G T ] " , d n a ) ) > ) { w a r n i n g ( " T h e i n p u t c

n

t a i n s c h a r a c t e r s

t

h e r t h a n A , C , G

r

T

v

a l u e s h

u

l d n

t

b e t r u s t e d ! " ) } d n a <

u

n l i s t ( s t r s p l i t ( d n a , " " ) ) f

r

( i i n 1 : l e n g t h ( d n a ) ) { i f ( d n a [ i ] = = " C " | d n a [ i ] = = " G " ) { c

u

n t e r = c

u

n t e r + 1 } } i f ( A T = = T R U E ) { r e t u r n ( 1

c
u

n t e r / l e n g t h ( d n a ) ) } e l s e { r e t u r n ( c

u

n t e r / l e n g t h ( d n a ) ) } }

15

Returning several values

If we want a function that returns several values (not only one), we can

utput a vector or a list.

Example: We define a function m i n m a x ( ) that determines the minimum and maximum of a given vector and returns these two values. m i n m a x <

f

u n c t i

n

( x ) { r e t u r n ( l i s t ( m i n i m u m = m i n ( x ) , m a x i m u m = m a x ( x ) ) ) } Test: x <

c

( 1 , 2 3 , 2 , 7 , ) m i n m a x ( x ) $ m i n i m u m [ 1 ] $ m a x i m u m [ 1 ] 2 3

16

Returning several values

What type of data does our function return? We can assign the result to a variable to investigate this question further. m y R e s u l t <

m

i n m a x ( x ) t y p e

f

( m y R e s u l t ) [ 1 ] " l i s t " s t r ( m y R e s u l t ) L i s t

f

2 $ m i n : n u m $ m a x : n u m 2 3 How can we access the individual values of a list?

m y R e s u l t [ [ 1 ] ] [ 1 ] m y R e s u l t [ [ 2 ] ] [ 1 ] 2 3 m y R e s u l t $ m i n i m u m [ 1 ] m y R e s u l t $ m a x i m u m [ 1 ] 2 3

17 Almost all functions (e.g., t-test, linear regression, etc.) in R produce output that is stored in a list ...

Remember Lecture 3 – Data types and Structures

➔ We now have everything we need to know to access particular parts

f such an output ☺

18

Example: heartbeat.txt (exercise sheet 7)

# I m p

r

t t h e d a t a h e a r t b e a t s <

r

e a d . t a b l e ( " h e a r t b e a t s . t x t " , h e a d e r = T R U E ) # P e r f

r

m a n u n p a i r e d t

t

e s t t . t e s t ( h e a r t b e a t s $ w g h t i n c r [ h e a r t b e a t s $ t r e a t m e n t = = ] , h e a r t b e a t s $ w g h t i n c r [ h e a r t b e a t s $ t r e a t m e n t = = 1 ] )

W e l c h T w

S

a m p l e t

t

e s t d a t a : h e a r t b e a t s $ w g h t i n c r [ h e a r t b e a t s $ t r e a t m e n t = = ] a n d h e a r t b e a t s $ w g h t i n c r [ h e a r t b e a t s $ t r e a t m e n t = = 1 ] t =

6

. 9 3 7 , d f = 2 6 . 3 , p

v

a l u e = 5 . 2 5 4 e

1

1 a l t e r n a t i v e h y p

t

h e s i s : t r u e d i fg e r e n c e i n m e a n s i s n

t

e q u a l t

9

5 p e r c e n t c

n

fj d e n c e i n t e r v a l :

8

8 . 9 1 9 7 6

4

9 . 5 3 4 7 8 s a m p l e e s t i m a t e s : m e a n

f

x m e a n

f

y

3

1 . 7 2 7 2 7 3 7 . 5

We want to test for a difference in mean of weight increase for the two

groups.

The groups are composed of different individuals, therefore we can apply

an unpaired t-test (more on t-tests tomorrow).

SLIDE 4

19

Example: heartbeat.txt (exercise sheet 7)

# P e r f

r

m a n u n p a i r e d t

t

e s t a n d s a v e t h e r e s u l t i n t

a

# v a r i a b l e t

i

n v e s t i g a t e i t f u r t h e r t _ t e s t _ r e s u l t <

t

. t e s t ( h e a r t b e a t s $ w g h t i n c r [ h e a r t b e a t s $ t r e a t m e n t = = ] , h e a r t b e a t s $ w g h t i n c r [ h e a r t b e a t s $ t r e a t m e n t = = 1 ] ) t y p e

f

( t _ t e s t _ r e s u l t ) [ 1 ] " l i s t " s t r ( t _ t e s t _ r e s u l t )

L i s t

f

9 $ s t a t i s t i c : N a m e d n u m

6

. 9 3 . .

a

t t r ( * , " n a m e s " ) = c h r " t " $ p a r a m e t e r : N a m e d n u m 2 6 . .

a

t t r ( * , " n a m e s " ) = c h r " d f " $ p . v a l u e : n u m 5 . 2 5 e

1

1 $ c

n

f . i n t : a t

m

i c [ 1 : 2 ]

8

8 . 9

4

9 . 5 . .

a

t t r ( * , " c

n

f . l e v e l " ) = n u m . 9 5 $ e s t i m a t e : N a m e d n u m [ 1 : 2 ]

3

1 . 7 3 7 . 5 . .

a

t t r ( * , " n a m e s " ) = c h r [ 1 : 2 ] " m e a n

f

x " " m e a n

f

y " $ n u l l . v a l u e : N a m e d n u m . .

a

t t r ( * , " n a m e s " ) = c h r " d i fg e r e n c e i n m e a n s " $ a l t e r n a t i v e : c h r " t w

.

s i d e d " $ m e t h

d

: c h r " W e l c h T w

S

a m p l e t

t

e s t " $ d a t a . n a m e : c h r " h e a r t b e a t s $ w g h t i n c r [ h e a r t b e a t s $ t r e a t m e n t = = ] a n d h e a r t b e a t s $ w g h t i n c r [ h e a r t b e a t s $ t r e a t m e n t = = 1 ] "

a

t t r ( * , " c l a s s " ) = c h r " h t e s t "

Imagine we are interested in just the p value. How can we extract it? t _ t e s t _ r e s u l t [ [ 3 ] ] t _ t e s t _ r e s u l t $ p . v a l u e

20

What you should know after days 7 & 8

Review: Data frames and import your data Conditional execution in R

Logic rules
if(), else(), ifelse()
Example from day 1

Loops Executing a command from a script Writing your own functions How to avoid slow R code 21

How to avoid slow R code

R has to interpret your commands each time

you run a script and it takes time to determine the type of your variables.

So avoid using loops and calling functions

again and again if possible

When you use loops, avoid increasing the size
f an object (vector ...) at each iteration but

rather define it with full size before.

Think in whole objects such as vectors or lists