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library(dslabs)

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data(murders)

#q1 which students have cgpa greater equal to 3.10?

BS5th<-data.frame(name=c("ali","ahmad","sania","sara","adil","sharjeel","subhan","arbaz","asad","hassan","waleed"),cgpa=c(2.5,3.2,3.9,2.99,3.10,2.99,2.91,2.82,3.2,3.52,2.50),grade=c("D","B","A","C","B","A","B","B","B","A","D"))
BS5th
##        name cgpa grade
## 1       ali 2.50     D
## 2     ahmad 3.20     B
## 3     sania 3.90     A
## 4      sara 2.99     C
## 5      adil 3.10     B
## 6  sharjeel 2.99     A
## 7    subhan 2.91     B
## 8     arbaz 2.82     B
## 9      asad 3.20     B
## 10   hassan 3.52     A
## 11   waleed 2.50     D
ind<-BS5th$cgpa>=3.10
BS5th$name[ind]
## [1] "ahmad"  "sania"  "adil"   "asad"   "hassan"

#q2 which students have got “B” grade?

ind<-BS5th$grade=="B"
BS5th$name[ind]
## [1] "ahmad"  "adil"   "subhan" "arbaz"  "asad"

#q3………cgpa>3.1 and got “A”?

cgpa_ind<-BS5th$cgpa>3.1
grade_ind<-BS5th$grade=="B"
BS5th$name[cgpa_ind&grade_ind]
## [1] "ahmad" "asad"

#Q1: Compute the per 100,000 murder rate for each state and store it in an object called murder_rate. Then use logical operators to create a logical vector named low that tells us which entries of murder_rate are lower than 1.

murder_rate <- murders$total / murders$population * 100000
murder_rate
##  [1]  2.8244238  2.6751860  3.6295273  3.1893901  3.3741383  1.2924531
##  [7]  2.7139722  4.2319369 16.4527532  3.3980688  3.7903226  0.5145920
## [13]  0.7655102  2.8369608  2.1900730  0.6893484  2.2081106  2.6732010
## [19]  7.7425810  0.8280881  5.0748655  1.8021791  4.1786225  0.9992600
## [25]  4.0440846  5.3598917  1.2128379  1.7521372  3.1104763  0.3798036
## [31]  2.7980319  3.2537239  2.6679599  2.9993237  0.5947151  2.6871225
## [37]  2.9589340  0.9396843  3.5977513  1.5200933  4.4753235  0.9825837
## [43]  3.4509357  3.2013603  0.7959810  0.3196211  3.1246001  1.3829942
## [49]  1.4571013  1.7056487  0.8871131
low <- murder_rate < 1
low
##  [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE
## [13]  TRUE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE
## [25] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE
## [37] FALSE  TRUE FALSE FALSE FALSE  TRUE FALSE FALSE  TRUE  TRUE FALSE FALSE
## [49] FALSE FALSE  TRUE

#Q2: Now use the results from the previous exercise and the function which to determine the indices of murder_rate associated with values lower than 1.

murder_rate <- murders$total/murders$population*100000
low <- murder_rate < 1
which(low)
##  [1] 12 13 16 20 24 30 35 38 42 45 46 51

#Q3: Use the results from the previous exercise to report the names of the states with murder rates lower than 1.

murder_rate <- murders$total/murders$population*100000
low <- murder_rate < 1
murders$state[low]
##  [1] "Hawaii"        "Idaho"         "Iowa"          "Maine"        
##  [5] "Minnesota"     "New Hampshire" "North Dakota"  "Oregon"       
##  [9] "South Dakota"  "Utah"          "Vermont"       "Wyoming"

#Q4: Now extend the code from exercise 2 and 3 to report the states in the Northeast with murder rates lower than 1. Hint: use the previously defned logical vector low and the logical operator &.

murder_rate <- murders$total/murders$population*100000
low <- murder_rate < 1
ind <- low & murders$region=='Northeast'
murders$state[ind] 
## [1] "Maine"         "New Hampshire" "Vermont"

#Q5: In a previous exercise we computed the murder rate for each state and the average of these numbers. How many states are below the average?

murder_rate <- murders$total/murders$population*100000
avg <- mean(murder_rate)
sum(murder_rate<avg)
## [1] 27

#Q6: Use the match function to identify the states with abbreviations AK, MI, and IA. Hint: start by defining an index of the entries of murders$abb that match the three abbreviations, then use the [ operator to extract the states.

abbs <- c('AK','MI','IA')
ind <- match(abbs , murders$abb)
murders$state[ind]
## [1] "Alaska"   "Michigan" "Iowa"

#Q7: Use the %in% operator to create a logical vector that answers the question: which of the following are actual abbreviations: MA, ME, MI, MO, MU ?

abbs <- c('MA', 'ME', 'MI', 'MO', 'MU')
abbs%in%murders$abb
## [1]  TRUE  TRUE  TRUE  TRUE FALSE

#Q8: Extend the code you used in exercise 7 to report the one entry that is not an actual abbreviation. Hint: use the ! operator, which turns FALSE into TRUE and vice versa, then which to obtain an index.

abbs <- c("MA", "ME", "MI", "MO", "MU") 
ind <- which(!abbs%in%murders$abb)
abbs[ind]
## [1] "MU"

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