Getting Started with R

#Addition
2+2

[1] 4

#Subtraction
4-2

[1] 2

#Multiplication
9*6

[1] 54

#Division
25/5

[1] 5

25/2

[1] 12.5

#Exponentiation
2^2

[1] 4

5^3

[1] 125

#Square root
sqrt(25)

[1] 5

# Logarithms
log(2)

[1] 0.6931472

Question_1: Compute the log base 5 of 10 and the log of 10

log10(5)

[1] 0.69897

log10(10)

[1] 1

log(10,5) #log of 10, base 5

[1] 1.430677

log(10,10) #log of 10, base 10

[1] 1

log(100,4) #log of 100, base 4

[1] 3.321928

#Batting Average=(No. of Hits)/(No. of At Bats)
#What is the batting average of a player that bats 129 hits in 412 at bats?
BA=129/412
BA

[1] 0.3131068

#Alternative Solution
N_Hits=129
At_Bats=412
BA<-N_Hits/At_Bats
BA

[1] 0.3131068

Batting_Average=round(BA,digits = 3)
Batting_Average

[1] 0.313

Question_2:What is the batting average of a player that bats 42 hits in 212 at bats?

#Answers
N_Hits=42
At_Bats1=212
Batting_Average<-N_Hits/At_Bats1
BattingAverage=round(Batting_Average,digits = 3)
BattingAverage

[1] 0.198

#On Base Percentage
#OBP=(H+BB+HBP)/(At Bats+BB+HBP+SF)
#Let us compute the OBP for a player with the following general stats
#AB=515,H=172,BB=84,HBP=5,SF=6
OBP=(172+84+5)/(515+84+5+6)
OBP

[1] 0.4278689

OBP_Adj=round(OBP,digits = 3)
OBP_Adj

[1] 0.428

Question_3:Compute the OBP for a player with the following general stats:

#AB=565,H=156,BB=65,HBP=3,SF=7
OBP=(156+65+3)/(565+65+3+156+7)
OBP_ad=round(OBP,digits = 3)
OBP_ad

[1] 0.281

Often you will want to test whether something is less than, greater than or equal to something.

3==8

[1] FALSE

2==3

[1] FALSE

1==1

[1] TRUE

3>=1

[1] TRUE

3>=9

[1] FALSE

7<=10

[1] TRUE

7<=6

[1] FALSE

3!=4

[1] TRUE

The logical operators are & for logical AND, | for logical OR, and ! for NOT. These are some examples:

# Logical Disjunction (or)
FALSE | FALSE # False OR False

[1] FALSE

FALSE | TRUE

[1] TRUE

# Logical Conjunction (and)
TRUE & FALSE #True AND False

[1] FALSE

# Negation
! FALSE # Not False

[1] TRUE

! TRUE # Not True

[1] FALSE

# Combination of statements
2 < 3 | 1 == 5 # 2<3 is True, 1==5 is False, True OR False is True

[1] TRUE

2<1|2==3

[1] FALSE

total_bases <- 7 + 4
total_bases*4

[1] 44

ls()

 [1] "At_Bats"            "At_Bats1"           "BA"                
 [4] "Batting_Average"    "BattingAverage"     "ceo_salaries"      
 [7] "hits_per_innings"   "N_Hits"             "OBP"               
[10] "OBP_ad"             "OBP_Adj"            "pitches_by_innings"
[13] "player_positions"   "runs_per_innings"   "strikes_by_innings"
[16] "total_bases"        "x"                 

rm(total_bases)
ls()

 [1] "At_Bats"            "At_Bats1"           "BA"                
 [4] "Batting_Average"    "BattingAverage"     "ceo_salaries"      
 [7] "hits_per_innings"   "N_Hits"             "OBP"               
[10] "OBP_ad"             "OBP_Adj"            "pitches_by_innings"
[13] "player_positions"   "runs_per_innings"   "strikes_by_innings"
[16] "x"                 

Vectors

pitches_by_innings<- c(12, 15, 10, 20, 10)
pitches_by_innings

[1] 12 15 10 20 10

strikes_by_innings <- c(9, 12, 6, 14, 9)
strikes_by_innings

[1]  9 12  6 14  9

Question_4: Define two vectors,runs_per_innings and hits_per_innings, each with five elements.

hits_per_innings <- c(3, 3, 1, 4, 1)
hits_per_innings

[1] 3 3 1 4 1

runs_per_innings <- c(1, 1, 0, 2, 0)
runs_per_innings

[1] 1 1 0 2 0

# replicate function
rep(2,5)

[1] 2 2 2 2 2

rep(3,3)

[1] 3 3 3

#consecutive numbers
1:6 

[1] 1 2 3 4 5 6

2:7

[1] 2 3 4 5 6 7

# sequence from 1 to 10 with a step of 3
seq(1,10,by=3)

[1]  1  4  7 10

# sequence from 2 to 13 with a step of 3
seq(2,13,by=3)

[1]  2  5  8 11

#adding vectors
pitches_by_innings+strikes_by_innings # + operator

[1] 21 27 16 34 19

#compare two vectors
pitches_by_innings

[1] 12 15 10 20 10

strikes_by_innings

[1]  9 12  6 14  9

pitches_by_innings==strikes_by_innings

[1] FALSE FALSE FALSE FALSE FALSE

length(pitches_by_innings)

[1] 5

min(pitches_by_innings)

[1] 10

mean(pitches_by_innings)

[1] 13.4

pitches_by_innings[1]

[1] 12

hits_per_innings[1]

[1] 3

pitches_by_innings[length(pitches_by_innings)]

[1] 10

hits_per_innings

[1] 3 3 1 4 1

hits_per_innings[length(hits_per_innings)]

[1] 1

pitches_by_innings

[1] 12 15 10 20 10

pitches_by_innings[c(1:3)]

[1] 12 15 10

hits_per_innings

[1] 3 3 1 4 1

hits_per_innings[c(1:4)]

[1] 3 3 1 4

player_positions<-c("catcher", "pitcher", "infielders", "outfielders")
player_positions

[1] "catcher"     "pitcher"     "infielders"  "outfielders"

Data Frames

data.frame(bonus=c(2,3,1), active_roster=c("yes","No","Yes"),salary=c(1.5, 2.5, 1))

Using Tables

x<-c("Yes", "No", "No", "Yes", "Yes")
table(x)

x
 No Yes 
  2   3 

Numerical measures and center of a spread

ceo_salaries<-c(12, .4, 2, 15, 8, 3, 1, 4, .25)
mean(ceo_salaries)

[1] 5.072222

var(ceo_salaries)

[1] 28.95944

sd(ceo_salaries)

[1] 5.381398

median(ceo_salaries)

[1] 3

fivenum(ceo_salaries)

[1]  0.25  1.00  3.00  8.00 15.00

# 12, .4, 2, 15, 8, 3, 1, 4, .25
# .25, .4, 1, 2, 3, 4, 8, 12, 15

getMode<-function(x) {
  ux<-unique(x)
  ux[which.max(tabulate(match(x, ux)))]
}

pitches_by_innings

[1] 12 15 10 20 10

getMode(pitches_by_innings)

[1] 10

#Question_7: Find the most frequent value of hits_per_innings & Find the most frequent value of strikes_per_innings..

hits_per_innings

[1] 3 3 1 4 1

getMode(hits_per_innings)

[1] 3

strikes_by_innings

[1]  9 12  6 14  9

getMode(strikes_by_innings)

[1] 9

#Question_8: Summarize the following survey with the table() command:

game_day<-c("Saturday", "Saturday", "Sunday", "Monday", "Saturday", "Tuesday", "Sunday", "Friday", "Monday")
table(game_day)

game_day
  Friday   Monday Saturday   Sunday  Tuesday 
       1        2        3        2        1 

getMode(game_day)

[1] "Saturday"