Marine Bioinformatics Mid-term exam 2025

Name : 나현성

ID : 202001240

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Q1. (10 points)

1). Change the name “John” to “Clair”.

x <- list(name = "John", age = 19, speaks = c("English", "French"))
x

You can NOT do like this :

x <- list(name = "Clair", age = 19, speaks = c("English", "French"))
x

your answer:

x <- list(name = "John", age = 19, speaks = c("English", "French"))
x$name <- "Clair"
x

2. Change the speaks “English” to “Korean”.

your answer :

x$speaks[x$speaks == "English"] <- "Korean"
x

Q2. Subsetting (10 points)

1. subset linked in data for those linkedin is greater than 10 and facebook is less than 10.

linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)

The output is: [1] 13

your answer:

linkedin[linkedin > 10 & facebook < 10]

2. use which () to subset linkedin for facebook is less than 10.

[1] 9 13 2

linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)

your answer:

linkedin[which(facebook < 10)]

Q3. data drame questions (15 points)

1. Create a data frame as in the output using data.frame().

Output ‘data.frame’: 2 obs. of 3 variables: $ SN :int 1 2 $ Age :num 21 15 $ Name:chr “John” “Dora”

your answer:

df <- data.frame(SN = c(1, 2),
                 Age = c(21, 15),
                 Name = c("John", "Dora"))
str(df)

2. change the name to the following by TWO different ways.

‘data.frame’: 2 obs. of 3 variables: $ SN : num 1 2 $ Age : num 21 15 $ Name: chr “Charlie” “Devy”

your answer:

df$Name <- c("Charlie", "Devy")
str(df)

your answer:

df[["Name"]] <- c("Charlie", "Devy")
 str(df)

3. create a new row and bind it to the data frame by TWO different ways.

str(Output) ‘data.frame’: 4 obs. of 3 variables: $ SN : num 1 2 1 1 $ Age : num 21 15 16 16 $ Name: chr “John” “Dora” “Paul” “Paul”

your answer:

library(dplyr)
new_row <- data.frame(SN = 1, Age = 16, Name = "Paul")
df1 <- bind_rows(df, new_row, new_row)
str(df1)

your answer:

new_row <- data.frame(SN = 1, Age = 16, Name = "Paul")
df2 <- rbind(df, new_row, new_row)
str(df2)

Q4. Function questions. Fill the blank to get the output as : (10 points)

Output

Enter a number: 8 [1] “The factorial of 8 is 40320”

————————————————————————

Example: Find the factorial of a number  

# take input from the user  

num = as.integer(readline(prompt="Enter a number: "))  

factorial = 1  

# check is the number is negative, positive or zero  

if(num < 0) {  
  
     print("Sorry, factorial does not exist for negative numbers")  
  
    } else _____ (num ____ 0) {  
      
         print("The factorial of 0 is 1")  

      } else {  
      
           for(i in _______) {  
      
              factorial = factorial * i  
    
           }  
 
         print(paste("The factorial of", _______ ,"is", ___________))  
      }  

Output  

————————————————————————

your answer:

 num = as.integer(readline(prompt="Enter a number: "))
factorial = 1

if(num < 0) {
  print("Sorry, factorial does not exist for negative numbers")
} else if(num == 0) {
  print("The factorial of 0 is 1")
} else {
  for(i in 1:num) {
    factorial = factorial * i
  }
  print(paste("The factorial of", num, "is", factorial))
}
Output  

Q5. when one is to use pivot_wider format and when one is to use pivot_longer form? (5 points)

your answer: pivot longer : 열 이름이 문자라고 가정하기 때문에 연도 열을 숫자로 변환해야 한다. pivot wider : pivot_longer의 역함수이다. 첫 번째 인수는 데이터에 대한 것이지만 보여주지 않는다. 그리고 names_from 인수는 어떤 변수를 열 이름으로 사용할지 pivot_wider에 알려준준다. 즉 데이터를 분산할 때는 pivot wider, 데이터를 모을 때는 pivot longer를 쓴다.

Q6. Do the following (10 points)

library(tidyverse)
library(dslabs)

data("murders")

murders %>% 
  mutate(rate = total/population*100000) %>%
  group_by(region) %>%
  summarize(avg = mean(rate)) %>%
  mutate(region = factor(region)) %>%

ggplot(aes(region, avg)) +
  geom_bar(stat="identity") +
  ylab("Murder Rate Average")

1. How to reorder from the largest to the smallest? Show the resultant plot.

your answer

library(tidyverse)
library(dslabs)

data("murders")

murders %>%
  mutate(rate = total/population * 100000) %>%
  group_by(region) %>%
  summarize(avg = mean(rate)) %>%
  mutate(region = factor(region, avg, .desc = TRUE)) %>%
  
ggplot(aes(region, avg)) +
  geom_bar(stat = "identity") +
  ylab("Murder Rate Average")

2. How to add error bars on the plot?

your answer

library(tidyverse)
library(dslabs)

data("murders")

murders %>% 
 mutate(rate = total/population * 100000) %>% 
  group_by(region) %>% 
  summarize(avg = mean(rate), sd = sd(rate)) %>% 
  mutate(region = factor(region)) %>%
  
ggplot(aes(region, avg)) + 
  geom_bar(stat = "identity") + 
  geom_errorbar(aes(ymin = avg - sd, ymax = avg + sd)) + 
  ylab("Murder Rate Average")

Q7. Regex questions. 27 points (each 3 points)

1. What should be in x?

⦁ str_subset(x, “.\\(\\^\\\).”)

your answer

 [1] "a$^$z"

⦁ str_subset(x, “[a-z]\\(\\^\\\)[a-z]”)

your answer

[1] "a$^$z"

⦁ str_subset(x, “\[|\]”)

your answer

[1] "[|]"

2. Use str_() to have the same results for grepl()

filter_taxa(grepl(pattern = "^[a-zA-Z]+$", taxon_names)) %>% # remove "odd" taxa
  

your answer

3. What does this mean?

⦁ What does this mean?

"([2-9][0-9]{2})[- .]([0-9]{3})[- .]([0-9]{4})"

: 첫 번째 자리 숫자는 2~9 뒤에 숫자 0~9가 2번 반복되며 총 3자리라는 것을 알 수 있다. 구분은 - . 또는 공백이다. 그리고 두 번째 그룹에서는 숫자가 3자리라는 것을 알 수 있고 세 번째 그룹은 숫자가 4자리라는 것을 알 수 있다.

⦁ What does this mean != “” below?

filter_taxa(taxon_names != "") 

⦁ the meaning ?

paste0("*\\.",ext,"\\.", fext,"$")

⦁ What does this mean?

(\d{1,2}\/\d{1,2}\/\d{4})

4. What is the utility of regex? (i.e., why do we need to learn regex?)

your answer: regex는 텍스트 문자의 특정 패턴을 설명하는 방법이다. 이러한 표현식은 주어진 문자열이 패턴과 일치하는지 여부를 결정하는 데 사용할 수 있다.

Q8. Tidyverse question (10 points)

The American Community Survey (ACS) on languages spoken at home across the U.S. dcldata::languages_states contains data from the 2013 ACS on the number of at-home speakers of non-English speakers by state.

install.packages("remotes")
remotes::install_github("dcl-docs/dcldata")

library(dcldata)
dcldata::languages_states

1. Extract speakers of Korean by state and list the top 10 states of Korean speakers.

your answer :

library(dcldata)
library(dplyr)

dcldata::languages_states %>%
  filter(language == "Korean") %>%
  arrange(desc(speakers)) %>%
   slice(1:10)

2. Can you write without saving as sample_data just using pipe line %>% ? (You DON’T need the actual “sample” data for this question. Just show the command lines.)

ssample_data <- sample_data[sample_data$Experiment == "ecotypes", ]
sample_data <- sample_data <- filter(sample_data,
                              Site %in% c("Mah", "Jam", "Sil"),
                              Age == 3)

your answer :

sample_data %>%
  filter(Experiment == "ecotypes") %>%
  filter(Site %in% c("Mah", "Jam", "Sil"), Age == 3)