5.1.2.1 Consider 3 vectors, day, month and year:

year=c(2005:2016)
month=c(1:12)
day=c(1:31)

Define a list Date such as:

Date=

$year
[1] 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016

$month
[1] 1 2 3 4 5 6 7 8 9 10 11 12

$day
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

year=c(2005:2016)
month=c(1:12)
day=c(1:31)

Date <- list(year = year,month = month, day = day)

Date
## $year
##  [1] 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016
## 
## $month
##  [1]  1  2  3  4  5  6  7  8  9 10 11 12
## 
## $day
##  [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
## [24] 24 25 26 27 28 29 30 31

5.1.2.2 write an R statement that will replace the values of year element in Date list for c(2000:2010).

Date$year <- c(2000:2010)

5.1.2.3 write an R statement that will delete the value 4 of the month component of the list Date.

Date$month <- Date$month[-4]

5.1.2.4 Consider a vector x such that: x=c(1,3,4,7,11,18,29). Write an R statement that will return a list X2 with components of value: x*2,x/2,sqrt(x) and names "x*2","x/2","sqrt(x)"

x=c(1,3,4,7,11,18,29)

X2 <- list("x*2" = x*2, "x/2" = x/2, "sqrt(x)" = sqrt(x))

X2
## $`x*2`
## [1]  2  6  8 14 22 36 58
## 
## $`x/2`
## [1]  0.5  1.5  2.0  3.5  5.5  9.0 14.5
## 
## $`sqrt(x)`
## [1] 1.000000 1.732051 2.000000 2.645751 3.316625 4.242641 5.385165

5.1.2.5 Consider the X2 list. Write an R statement that will return a vector: 2.000000 2.645751 3.316625

X2 # As you can see, these are the 3rd,4th and 5th elements of X2$sqrt(x)
## $`x*2`
## [1]  2  6  8 14 22 36 58
## 
## $`x/2`
## [1]  0.5  1.5  2.0  3.5  5.5  9.0 14.5
## 
## $`sqrt(x)`
## [1] 1.000000 1.732051 2.000000 2.645751 3.316625 4.242641 5.385165
X2$`sqrt(x)`[3:5]
## [1] 2.000000 2.645751 3.316625

5.1.2.6 Write an R statement that will return a concatenation M, of Date and X2 lists.

M <- c(Date,X2) #Yes, you can use c for concatenation of lists. You can also try list(Date,X2) to see the difference.

5.1.2.7 Write an R statement that will return a sublist N of M, with components year,x*2 and day.

N <- M[c("year","x*2","day")]
N
## $year
##  [1] 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010
## 
## $`x*2`
## [1]  2  6  8 14 22 36 58
## 
## $day
##  [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
## [24] 24 25 26 27 28 29 30 31

5.1.2.8 Consider the N list. Write an R statement that will return:
-the length of x*2 vector in N
-the value of the second element of vector year in N

length(N$`x*2`) # or N[[2]]
## [1] 7
N$year[2]
## [1] 2001

5.1.2.9 Consider 3 letters vectors, and 2 numeric vectors:
A=letters[1:4],B=letters[5:10],C=letters[11:15] D=c(1:10),E=c(20:5)

Define a list z, with elementes x and y,such that x is a list with elements A,B,and C; and y is a list with elements D and E.

A <- letters[1:4]
B <- letters[5:10]
C <- letters[11:15]
D <- c(1:10)
E <- c(20:5)

x=list(A,B,C)

y=list(D,E)

z=list(x,y)

z
## [[1]]
## [[1]][[1]]
## [1] "a" "b" "c" "d"
## 
## [[1]][[2]]
## [1] "e" "f" "g" "h" "i" "j"
## 
## [[1]][[3]]
## [1] "k" "l" "m" "n" "o"
## 
## 
## [[2]]
## [[2]][[1]]
##  [1]  1  2  3  4  5  6  7  8  9 10
## 
## [[2]][[2]]
##  [1] 20 19 18 17 16 15 14 13 12 11 10  9  8  7  6  5

5.1.2.10 Write an R statement that will return:
-the number on third position on the second vector of the first list of z
-the letter on fifth position on the third vector of the second list of z

z[[1]][[2]][3]
## [1] "g"
# z[[2]][[3]][5] The second list is y which has only 2 vectors. So, this statement will give an error instead of an output. If it was the first list of z, the below statement could be used.

z[[1]][[3]][5]
## [1] "o"