In-class activity #1

#Let us complete some basic opereation Using R
1+2

[1] 3

1-4

[1] -3

2*5

[1] 10

sqrt(9)

[1] 3

log(10)

[1] 2.302585

log(2.72)#Natural Log (Ln)

[1] 1.000632

log10(5)

[1] 0.69897

log10(10)

[1] 1

#Batting Average=(No. of Hits)/(No. of At Bats)
#What is the batting average of a player that bats 29 hits in 112 at bats?
BA=(29)/(112)
BA

[1] 0.2589286

Batting_Average=round(BA,digits = 3)
Batting_Average

[1] 0.259

#Question_2:What is the batting average of a player that bats 42 hits in 212 at bats?
BA=(42)/(212)
BA

[1] 0.1981132

Batting_Average=round(BA,digits = 3)
Batting_Average

[1] 0.198

OBP

#On Base Percentage
#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+172+84+5+6)
OBP

[1] 0.3337596

On_Base_Percentage=round(OBP,digits = 3)
On_Base_Percentage

[1] 0.334

OBP

#Compute the OBP for a player with the following general stats:
#AB=565,H=156,BB=65,HBP=3,SF=7
OBP=(152+65+3)/(565+152+65+3+6)
OBP

[1] 0.278129

On_Base_Percentage=round(OBP,digits = 3)
On_Base_Percentage

[1] 0.278

3 == 8# Does 3 equals 8?

[1] FALSE

3 !=8# Is 3 different From 8?

[1] TRUE

3>4

[1] FALSE

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

[1] FALSE

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

[1] FALSE

# Negation
! FALSE # Not False

[1] TRUE

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

[1] TRUE

Total_Bases <- 6 + 5
Total_Bases*3

[1] 33

ls()

[1] "BA"                 "Batting_Average"   
[3] "OBP"                "On_Base_Percentage"
[5] "pitches_by_innings" "strikes_by_innings"

rm("Total_Bases")

Warning in rm("Total_Bases") : object 'Total_Bases' not found

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_9innings and hits_per_9innings, each with five elements.
runs_per_9innings <- c(11, 13, 6, 9, 11)
runs_per_9innings

[1] 11 13  6  9 11

hits_per_9innings <- c(13, 5, 7, 10, 20)
hits_per_9innings

[1] 13  5  7 10 20

# replicate function
rep(2, 5)

[1] 2 2 2 2 2

# consecutive numbers
0:5

[1] 0 1 2 3 4 5

# sequence from 1 to 20 with a step of 5
seq(1, 20, by=5)

[1]  1  6 11 16

# add vectors
pitches_by_innings+ runs_per_9innings

[1] 23 28 16 29 21

# compare vectors
hits_per_9innings == strikes_by_innings

[1] FALSE FALSE FALSE FALSE FALSE

# find length of vector
length(hits_per_9innings)

[1] 5

# find minimum value in vector
min(strikes_by_innings)

[1] 6

# find average value in vector
mean(pitches_by_innings)

[1] 13.4

pitches_by_innings

[1] 12 15 10 20 10

#Question_5: Get the first element of hits_per_9innings.
hits_per_9innings[1]

[1] 13

#Question_6: Get the last element of hits_per_9innings.
hits_per_9innings[length(hits_per_9innings)]

[1] 20

data.frame(bonus = c(2, 3, 1),#in millions 
           active_roster = c("yes", "no", "yes"), 
           salary = c(1.5, 2.5, 1))#in millions 

sample(1:16, size=5)

[1] 14  5  4  7  3

bar <- data.frame(var1 = LETTERS[1:10], var2 = 1:10)
# Check data frame
bar

samplerows <- sample(1:nrow(bar), size=n) 
# print sample rows
samplerows

[1]  5  9 10  1  7

# extract rows
barsample <- bar[samplerows, ]
# print sample
print(barsample)

bar[sample(1:nrow(bar), n), ]

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

x
 No Yes 
  2   3 

sals <- c(12, .4, 5, 2, 50, 8, 3, 1, 4, 0.25)
# the average
mean(sals) 

[1] 8.565

# the variance
var(sals)

[1] 225.5145

# the standard deviation
sd(sals)

[1] 15.01714

# the median
median(sals)

[1] 3.5

# Tukey's five number summary, usefull for boxplots
# five numbers: min, lower hinge, median, upper hinge, max
fivenum(sals)

[1]  0.25  1.00  3.50  8.00 50.00

# summary statistics
summary(sals)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.250   1.250   3.500   8.565   7.250  50.000 

# Most frequent value inhits_per_9innings
getMode(hits_per_9innings)

[1] 13

#Question_8: Summarize the following survey with the `table()` command:
table (game_day <-c("Saturday", "Saturday", "Sunday", "Monday", "Saturday","Tuesday", "Sunday", "Friday", "Friday", "Monday"))

  Friday   Monday Saturday   Sunday  Tuesday 
       2        2        3        2        1 

#Question_9: What is the most frequent answer recorded in the survey? Use the getMode function to compute results. 
getMode(game_day)

[1] "Saturday"