Assignment 2

Question 1

Solution 1(a)

Separate all the strings and store it in a variable “gene_seqs”

given_combination = "ATCGATCGATCG-ATCGAT-CGATC-GATCGAT-CGATCG-ATCGATCG-CGATCG"
gene_seqs=unlist(strsplit(given_combination, "-"))
cat("(a) gene_seqs=",gene_seqs, '\n')

(a) gene_seqs= ATCGATCGATCG ATCGAT CGATC GATCGAT CGATCG ATCGATCG CGATCG 

Solution 1(b)

Using a loop reverse all the strings in gene_seqs

library(stringi) 
i=1
for(r in gene_seqs){

 cat(i, r, "=", stri_reverse(r),'\n')
 i=1+i
}

1 ATCGATCGATCG = GCTAGCTAGCTA 
2 ATCGAT = TAGCTA 
3 CGATC = CTAGC 
4 GATCGAT = TAGCTAG 
5 CGATCG = GCTAGC 
6 ATCGATCG = GCTAGCTA 
7 CGATCG = GCTAGC 

Solution 1(c)

Using a loop calculate length of each string in gene_seqs

i=1
for(g in gene_seqs) {

  cat(i, g, "=",nchar(g),'\n')
  i=1+i
}

1 ATCGATCGATCG = 12 
2 ATCGAT = 6 
3 CGATC = 5 
4 GATCGAT = 7 
5 CGATCG = 6 
6 ATCGATCG = 8 
7 CGATCG = 6 

Solution 1(d)

In gene_seqs, if a string length is greater than 7, print “Successful” else print “Error”. (use loop and condition)

i=1
for(s in gene_seqs) {
  
  if(nchar(s) > 7) {
    cat(i, s,"=",nchar(s), ",successful \n")
  } 
  else {
    cat(i, s,"=",(nchar(s)),",error \n")
  }
  i=1+i
  
}

1 ATCGATCGATCG = 12 ,successful 
2 ATCGAT = 6 ,error 
3 CGATC = 5 ,error 
4 GATCGAT = 7 ,error 
5 CGATCG = 6 ,error 
6 ATCGATCG = 8 ,successful 
7 CGATCG = 6 ,error 

Solution 1(e)

Find the locations and number of occurrences of pattern “GATC” in each string of the gene_seqs

pattern = ('GATC')
i=1
for (n in gene_seqs){
  locations = gregexpr(pattern, n)
locations = unlist(locations)
cat(i,n,":",locations,'\n')
i=1+i
}

1 ATCGATCGATCG : 4 8 
2 ATCGAT : -1 
3 CGATC : 2 
4 GATCGAT : 1 
5 CGATCG : 2 
6 ATCGATCG : 4 
7 CGATCG : 2 

Bonus (+5): Solve b to e questions using one loop.

given_combination = "ATCGATCGATCG-ATCGAT-CGATC-GATCGAT-CGATCG-ATCGATCG-CGATCG"
gene_seqs=unlist(strsplit(given_combination, "-"))
i=1
for (gene in gene_seqs){
  cat("[",i,"]",'\n')
  i=1+i
  cat("(b)reverse=", stri_reverse(gene),'\n')
  cat("(c) length of ",gene, "=",nchar(gene),'\n')
  if(nchar(gene) > 7) {
    cat("(d)   successful",'\n')
  } 
  else {
    cat("(d)   error",'\n')
  }
pattern = ('GATC')
  locations = gregexpr(pattern, gene)
locations = unlist(locations)
nm= sum(locations != -1)
cat("(e) Locations of GATC in ",gene,":",locations,", number of occurrences =",nm,'\n','\n')
}  

[ 1 ] 
(b)reverse= GCTAGCTAGCTA 
(c) length of  ATCGATCGATCG = 12 
(d)   successful 
(e) Locations of GATC in  ATCGATCGATCG : 4 8 , number of occurrences = 2 
 
[ 2 ] 
(b)reverse= TAGCTA 
(c) length of  ATCGAT = 6 
(d)   error 
(e) Locations of GATC in  ATCGAT : -1 , number of occurrences = 0 
 
[ 3 ] 
(b)reverse= CTAGC 
(c) length of  CGATC = 5 
(d)   error 
(e) Locations of GATC in  CGATC : 2 , number of occurrences = 1 
 
[ 4 ] 
(b)reverse= TAGCTAG 
(c) length of  GATCGAT = 7 
(d)   error 
(e) Locations of GATC in  GATCGAT : 1 , number of occurrences = 1 
 
[ 5 ] 
(b)reverse= GCTAGC 
(c) length of  CGATCG = 6 
(d)   error 
(e) Locations of GATC in  CGATCG : 2 , number of occurrences = 1 
 
[ 6 ] 
(b)reverse= GCTAGCTA 
(c) length of  ATCGATCG = 8 
(d)   successful 
(e) Locations of GATC in  ATCGATCG : 4 , number of occurrences = 1 
 
[ 7 ] 
(b)reverse= GCTAGC 
(c) length of  CGATCG = 6 
(d)   error 
(e) Locations of GATC in  CGATCG : 2 , number of occurrences = 1 
 

Question 2

Solution 2(a)

Create the data frame. Add 2 new rows. Add a new column called “Income”. This column should be numerical.

exam_score = data.frame(
  ID = c(1, 2, 3, 4, 5),
  Name = c("Alice", "Bob", "David", "John", "Jenny"),
  Age = c(20, 25, 30, 22, 18),
  Score = c(100, 78, 90, 55, 81)

)
exam_score

new_row1 = data.frame(
ID = c(6,7),
  Name = c("Aria","Diana"),
  Age = c(22,24),
  Score = c(76, 78)  
)
new_row1

exam_score_2 = rbind(exam_score, new_row1)

exam_score_2

Income = data.frame(
  Income=c(15000, 20000,12000,25000,30000,22000,18000))
exam_score_3 = cbind(exam_score_2,Income)
exam_score_3

NA

Solution 2(b)

Find max, min, median, sum, mean, standard deviation, variance, quantiles of column Age, Score, and Income.

exam_score_3[ , 3:5 ]

cat("max_age=",max(exam_score_3$Age),'\n',"max_score=",max(exam_score_3$Score),'\n',"max_income=",max(exam_score_3$Income),'\n','\n')

max_age= 30 
 max_score= 100 
 max_income= 30000 
 

cat("min_age=", min(exam_score_3$Age),'\n',"min_score=", min(exam_score_3$Score),'\n',"min_income=", min(Income),'\n','\n')

min_age= 18 
 min_score= 55 
 min_income= 12000 
 

cat("median_age=",median(exam_score_3$Age),'\n',"median_score=", median(exam_score_3$Score),'\n',"median_income=", median(exam_score_3$Income),'\n','\n')

median_age= 22 
 median_score= 78 
 median_income= 20000 
 

cat("mean_age=", mean(exam_score_3$Age),'\n',"mean_score=", mean(exam_score_3$Score),'\n',"mean_income=",mean(exam_score_3$Income),'\n','\n')

mean_age= 23 
 mean_score= 79.71429 
 mean_income= 20285.71 
 

cat("sd_age=", sd(exam_score_3$Age),'\n',"sd_score=", sd(exam_score_3$Score),'\n',"sd_income=", sd(exam_score_3$Income),'\n','\n')

sd_age= 3.872983 
 sd_score= 13.84093 
 sd_income= 6074.929 
 

cat("var_age=", var(exam_score_3$Age),'\n',"var_score=", var(exam_score_3$Score),'\n',"var_income=", var(exam_score_3$Income),'\n','\n')

var_age= 15 
 var_score= 191.5714 
 var_income= 36904762 
 

#   quantile(exam_score_3$Age)
#   quantile(exam_score_3$Score)
#   quantile(exam_score_3$Income)

cat("quantile_age=", quantile(exam_score_3$Age),'\n', "quantile_score=", quantile(exam_score_3$Score),'\n',
"quantile_income=", quantile(exam_score_3$Income),'\n','\n')

quantile_age= 18 21 22 24.5 30 
 quantile_score= 55 77 78 85.5 100 
 quantile_income= 12000 16500 20000 23500 30000 
 

Solution 2(c)

Find correlation between i. Age and score, ii. Age and income, iii. Score and income

cat("correlation between Age and score =",cor(exam_score_3$Age, exam_score_3$Score),'\n')

correlation between Age and score = 0.08705541 

cat("correlation between Age and income = ",cor(exam_score_3$Age, exam_score_3$Income),'\n')

correlation between Age and income =  -0.7083729 

cat("correlation between Score and income =",cor(exam_score_3$Score, exam_score_3$Income),'\n')

correlation between Score and income = -0.5915379 

Solution 2(d)

Select rows where the score is greater than or equal to 80

exam_score_3[ exam_score_3$Score >= 80, ]

Solution 2(e)

Select rows with age range 20 to 30

exam_score_3[exam_score_3$Age >= 20 & exam_score_3$Age <= 30, ]