Feel free to mark using the rpubs document. https://rpubs.com/PontSatyre11119/basic-r_eeb313

To submit this assignment, upload the full document to Quercus, including the original questions, your code, and the output. Submit your assignment as a knitted .pdf (preferred), .docx, or .html file.

  1. Vectors (2 marks)

    1. Create a vector v with all integers 0-20, and a vector w with every third integer in the same range. (0.25 marks)
v <- seq(0,20)
w <- seq(0,20,3)
v
##  [1]  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
w
## [1]  0  3  6  9 12 15 18
b.  How much longer is vector `v` compared with `w`?
    (0.25 marks)
    
length(v) - length(w)
## [1] 14
c.  Create a new vector, `v_square`, with the square of elements at indices
    3, 6, 7, 10, and 15 from the variable `v`. _Hint:
    Use indexing rather than a for loop._ (0.5 marks)
    
v_square <- (v[c(4, 7, 8, 11, 16)])^2
v_square
## [1]   9  36  49 100 225
d.  Calculate the mean and median of the first four values from
    `v_square`. (0.5 marks)
    
mean(v_square[seq(1,4)])
## [1] 48.5
median(v_square[seq(1,4)])
## [1] 42.5
e.  Create a logical (or boolean) vector `v_bool` to indicate which vector `v_square`
    elements are bigger than 30. How many values are over 30? _Hint:
    In R, TRUE = 1, and FALSE = 0, so you can use simple arithmetic
    to find this out._ (0.5 marks)
    
v_bool <- length(v_square > 30)
v_bool
## [1] 5
  1. Functions (2 marks)

Write a function that calculates the mean of the last two elements of any numeric vector. Test this function with the v and v_square vectors.

last_2_ele_func <- function(num_vec){
  ((num_vec[length(num_vec)])+(num_vec[length(num_vec)-1]))/2
} # adds last and second last values, divides by 2

last_2_ele_func(v)
## [1] 19.5
last_2_ele_func(v_square)
## [1] 162.5
  1. Loops (2 marks)

Turn the given data frame into an exponent table, a table that shows you results of raising one number (the base, rows) to the power of another (the exponent, columns). See an example here. Use this data frame to find 5 to the power 9. Hint: use a nested for-loop to fill your data frame

exp_table <- data.frame(matrix(ncol=10, nrow=10))

n <- 10 # how many values we want (n-val)
exp_table <- data.frame() # making new df

for(i in 0:n){ # for every 0 to n-val,
  df <- (0+i)^(0:n) # calculate the exp from 0 to n-val, of each 0 to the i-th value
  exp_table <- rbind(exp_table, df) # makes a table
}

colnames(exp_table) <- c(0:n) # renames columns
rownames(exp_table) <- c(0:n) # renames rows

exp_table
##    0  1   2    3     4      5       6        7         8          9          10
## 0  1  0   0    0     0      0       0        0         0          0           0
## 1  1  1   1    1     1      1       1        1         1          1           1
## 2  1  2   4    8    16     32      64      128       256        512        1024
## 3  1  3   9   27    81    243     729     2187      6561      19683       59049
## 4  1  4  16   64   256   1024    4096    16384     65536     262144     1048576
## 5  1  5  25  125   625   3125   15625    78125    390625    1953125     9765625
## 6  1  6  36  216  1296   7776   46656   279936   1679616   10077696    60466176
## 7  1  7  49  343  2401  16807  117649   823543   5764801   40353607   282475249
## 8  1  8  64  512  4096  32768  262144  2097152  16777216  134217728  1073741824
## 9  1  9  81  729  6561  59049  531441  4782969  43046721  387420489  3486784401
## 10 1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000 10000000000
exp_calculator <- function(b, e){
  exp_table[b+1,e+1]
}

exp_calculator(b = 5, e = 9) # where b is base and e is exponent
## [1] 1953125

  1. Data frames (2 marks)

    1. There are many built-in data frames in R, which you can find more details about these online. What are the column names of the built-in dataframe beaver1? How many observations (rows) and variables (columns) are there? (0.5 marks)
colnames(beaver1)
## [1] "day"   "time"  "temp"  "activ"
dim(beaver1)
## [1] 114   4
b.  Display both the first 6 and last 6 rows of this data frame.
    Show how to do so with both indexing as well as specialized functions. (0.5 marks)
head(beaver1)
##   day time  temp activ
## 1 346  840 36.33     0
## 2 346  850 36.34     0
## 3 346  900 36.35     0
## 4 346  910 36.42     0
## 5 346  920 36.55     0
## 6 346  930 36.69     0
tail(beaver1)
##     day time  temp activ
## 109 347  250 36.84     0
## 110 347  300 36.86     0
## 111 347  310 36.88     0
## 112 347  320 36.93     0
## 113 347  330 36.97     0
## 114 347  340 37.15     1
beaver1[1:6,]
##   day time  temp activ
## 1 346  840 36.33     0
## 2 346  850 36.34     0
## 3 346  900 36.35     0
## 4 346  910 36.42     0
## 5 346  920 36.55     0
## 6 346  930 36.69     0
beaver1[(nrow(beaver1)-6):(nrow(beaver1)),]
##     day time  temp activ
## 108 347  240 36.82     0
## 109 347  250 36.84     0
## 110 347  300 36.86     0
## 111 347  310 36.88     0
## 112 347  320 36.93     0
## 113 347  330 36.97     0
## 114 347  340 37.15     1
c.  What is the difference in mean body temperature of the beaver when it was inside versus outside of the retreat? (0.5 marks)
library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
# assuming variables 0 and 1 for beaver1$activ, where 0 is inside and 1 is outside

(beaver1 %>% 
  filter(activ == c(0,1)) %>% # concatenate 0 and 1
  group_by(activ) %>% # group by 0 and 1
  summarize(mean = mean(temp)) %>% # mean of groups 0 and 1
  mutate(difference = diff(mean)))[2,3] # subset the new column with difference
## # A tibble: 1 x 1
##   difference
##        <dbl>
## 1      0.375
## # A tibble: 2 x 3
##   activ  mean difference
##   <dbl> <dbl>      <dbl>
## 1     0  36.8      0.375
## 2     1  37.2      0.375
d.  How much did the body temperature of the beaver fluctuate (i.e., range) from 9:00-10:00 AM on the first day of the study? (0.5 marks)
beaver1 %>%
  filter(day == 346 & between(time, 900, 1000)) %>% # only first day & between 9am to 10am
  select(temp) %>% # only temp
  range() %>% # max and min in temp
  diff() %>% # difference between max and min
  abs() # makes fluctuation result always positive 
## [1] 0.46