The document demonstrates various operations that can be performed on vectors and data frames in R. It shows how to create, subset, reorder, and modify vectors and data frames. Key operations include subsetting vectors and data frames using indices or logical vectors, applying functions to entire vectors or selected elements, and reordering a data frame based on the values in one of its columns.

6. > v <- c(1, 43, 100, 3, 55)
> is.vector(v)
[1] TRUE
> is(v)
[1] "numeric" "vector"
> length(v)
[1] 5
> v
[1] 1 43 100 3 55

7. > v <- 1
> is.vector(v)
[1] TRUE
> length(v)
[1] 1
> v
[1] 1
Note that a scalar is a vector of length 1.

9. ,
> a = c(1, 3, 5, 7)
> b = c(1, 2, 4, 8)

10. ,
> a = c(1, 3, 5, 7)
> b = c(1, 2, 4, 8)
> 5 * a
[1] 5 15 25 35

11. > a + b
[1] 2 5 9 15

12. > a - b
[1] 0 1 1 -1
> a * b
[1] 1 6 20 56
> a / b
[1] 1.000 1.500 1.250 0.875

13. > a = c(10, 20, 30)
> b = c(1, 2, 3, 4, 5, 6, 7, 8, 9)
> a + b
[1] 11 22 33 14 25 36 17 28 39

14. > x=c(2,4,6,8,12)
> y=c(2,1,4,7,10)
> x%%y
[1] 0 0 2 1 2
> x %/% y
[1] 1 4 1 1 1

15. ?"+"

16. > v <- c(1, 34, 100, 3, 26)
> sum(v)
[1] 164
> prod(v)
[1] 265200
> quantile(v)
0% 25% 50% 75% 100%
1 3 26 34 100

17. …

18. > w <- c(10, 10.2, 34, 7.35, 0)
> is(w)
[1] "numeric" "vector"
> w <- seq(0,10,2)
> w <- 1:10
> w
[1] 1 2 3 4 5 6 7 8 9 10

19. > v <- c(TRUE, FALSE, TRUE, TRUE)
> is(v)
[1] "logical" "vector"

20. > v <- c(TRUE, FALSE, TRUE, TRUE)
> as.numeric(v)
[1] 1 0 1 1
> v <- c(0, 0, 1, 1)
> as.logical(v)
[1] FALSE FALSE TRUE TRUE

21. > v <- c(10, 10.2, 34, 7.35, 0)
> v < 5
[1] FALSE FALSE FALSE FALSE TRUE
> v >= 10
[1] TRUE TRUE TRUE FALSE FALSE
> v == 0
[1] FALSE FALSE FALSE FALSE TRUE
> v!=0
[1] TRUE TRUE TRUE TRUE FALSE

22. > v <- c("a", "b", "c", "d", "e")
> is(v)
[1] "character" "vector"
[3] "data.frameRowLabels" "SuperClassMethod"

23. > v1 <- "ASP Training Workshop"
> v2 <- "24-29 April 2017"
> paste(v1,v2)
[1] "ASP Training Workshop 24-29 April 2017"

24. > as.numeric(v1)
[1] NA
Warning message:
NAs introduced by coercion
> as.logical(v1)
[1] NA

25. > v1 <- c(v, 0, 0, v)
> length(v1)
[1] 8
> v
[1] 1 34 100
> v1
[1] 1 34 100 0 0 1 34 100

27. > v <- c(10, 10.2, 34, 7.35, 0)
> v[1]
[1] 10
> v[c(2, 4)]
[1] 10.20 7.35
> v[c(4, 2)]
[1] 7.35 10.20
> v[-c(2:5)]
[1] 10

28. > v <- c(10, 10.2, 34, 7.35, 0)
> v[c(TRUE, FALSE, FALSE, FALSE, FALSE)]
[1] 10

29. v <- c(10, 10.2, 34, 7.35, 0)
# which elements are larger than 9
v > 9
[1] TRUE TRUE TRUE FALSE FALSE
# select elements larger than 9
v[v > 9]
[1] 10.0 10.2 34.0
# or
idx <- v > 9
v[idx]
[1] 10.0 10.2 34.0

30. v <- c(NA, 10.2, 34, NA, 0)
# select the NA elements
v==NA
[1] NA NA NA NA NA
v=="NA"
[1] NA FALSE FALSE NA FALSE
# neither works because NA is special

31. v
[1] NA 10.2 34.0 NA 0.0
is.na(v)
[1] TRUE FALSE FALSE TRUE FALSE
# now it's possible to select the NA
v[is.na(v)]
[1] NA NA

32. v
[1] NA 10.2 34.0 NA 0.0
v[is.na(v)] <- 200
v
[1] 200.0 10.2 34.0 200.0 0.0

33. v
[1] NA 10.2 34.0 NA 0.0
v[is.na(v)] <- 200
v
[1] 200.0 10.2 34.0 200.0 0.0

34. v <- rnorm(1000)
plot(v, main="My scatter plot")

35. v <- rnorm(1000)
plot(v, main="My scatter plot")

36. hist(v, main="My histogram")

37. > v <- rnorm(1000, mean=40, sd=5)
> hist(v, main="My histogram")

38. > plot(density(v), main="My density plot")

39. v1 <- rnorm(1000)
v2 <- rnorm(1000)
plot(v1, v2, main="Independent variables")

40. v1 <- rnorm(1000)
v2 <- rnorm(1000, v1)
plot(v1, v2, main="Dependent variables")

42. v <- c(10, 3, 0, 54.2, 1)
names(v) <- letters[1:5]
v
a b c d e
10.0 3.0 0.0 54.2 1.0
v["c"]
c
0
v[3]
c
0

56. •
•
•
•

57. > SOC <- read.csv("MASIS_SOC.csv")
> SOC
Id UpperDepth LowerDepth SOC Lambda tsme
1 4 0 30 12.00032455 0.01 0.003985153
2 7 0 30 3.48365276 0.01 0.002502976
3 8 0 30 2.31341405 0.01 0.002504971
4 9 0 30 1.94142743 0.01 0.002508691
5 10 0 30 1.34296903 0.01 0.002509177
6 11 0 30 2.28793284 0.01 0.002509360
7 12 0 30 2.71584298 0.01 0.002518323
8 13 0 30 4.34011158 0.01 0.002515760
...

58. > summary(SOC)
Id UpperDepth LowerDepth SOC Lambda
Min. : 4 Min. :0 Min. :30 Min. : 0.000 Min. :0.01
1st Qu.:1878 1st Qu.:0 1st Qu.:30 1st Qu.: 1.006 1st Qu.:0.01
Median :3214 Median :0 Median :30 Median : 1.495 Median :0.01
Mean :3198 Mean :0 Mean :30 Mean : 1.916 Mean :0.01
3rd Qu.:4502 3rd Qu.:0 3rd Qu.:30 3rd Qu.: 2.268 3rd Qu.:0.01
Max. :6539 Max. :0 Max. :30 Max. :50.205 Max. :0.01
NA's :1
tsme
Min. :0.002472
1st Qu.:0.002502
Median :0.002504
Mean :0.002507
3rd Qu.:0.002507
Max. :0.003985

59. > tail(SOC)
Id UpperDepth LowerDepth SOC Lambda tsme
3257 6531 0 30 0.5698581 0.01 0.002503761
3258 6532 0 30 5.7547935 0.01 0.002505020
3259 6533 0 30 1.6636972 0.01 0.002506451
3260 6535 0 30 1.9226001 0.01 0.002502052
3261 6537 0 30 1.7165334 0.01 0.002502749
3262 6539 0 30 1.3633083 0.01 0.002502855

60. > dim(SOC)
[1] 3262 7

61. > dimnames(SOC)
[[1]]
[1] "1" "2" "3" "4" "5" "6" "7" "8" "9" "10"
[11] "11" "12" "13" "14" "15" "16" "17" "18" "19" "20"
…
[[2]]
[1] "Id" "UpperDepth" "LowerDepth" "SOC" "Lambda"
[6] "tsme"

62. > SOC$SOC
[1] 12.00032455 3.48365276 2.31341405 1.94142743 1.34296903 2.28793284
[7] 2.71584298 4.34011158 5.77118126 4.54692240 4.63597793 2.10768409
[13] 3.96522026 4.80577783 3.08891798 4.59635072 1.51213851 1.31937774
[19] 1.64608828 1.63183332 3.93370051 1.89931487 1.70009503 1.68627536
...

63. > SOC[3, "tsme"]
[1] 0.002504971

64. > SOC[3,6]
[1] 0.002504971

65. SOC[, 6]
[1] 0.003985153 0.002502976 0.002504971 0.002508691 0.002509177 0.002509360
[7] 0.002518323 0.002515760 0.002509165 0.002514908 0.002517316 0.002510426
[13] 0.002506018 0.002503509 0.002505200 0.002503669 0.002505983 0.002504586
[19] 0.002504090 0.002506831 0.002507821 0.002505357 0.002505301 0.002507838
[25] 0.002507646 0.002502572 0.002504063 0.002505777 0.002506869 0.002502164
...

66. You can select multiple rows and columns using
vectors.
> SOC[1:5, c("SOC","tsme")]
SOC tsme
1 12.000325 0.003985153
2 3.483653 0.002502976
3 2.313414 0.002504971
4 1.941427 0.002508691
5 1.342969 0.002509177

67. Of course you can write values into a data.frame.
We make a copy of SOC (so we don't mess up
the original)
> SOCtemp <- SOC
> SOCtemp[3,"tsme"]
[1] 0.002504971
> SOCtemp[3,"tsme"] <- 1
> SOCtemp[3,"tsme"]
[1] 1

68. Recycling rules apply if less values are supplied
than selected
> SOCtemp[1:5,"tsme"] <- 1
> SOCtemp[1:5,]
Id UpperDepth LowerDepth SOC Lambda tsme
1 4 0 30 12.000325 0.01 1
2 7 0 30 3.483653 0.01 1
3 8 0 30 2.313414 0.01 1
4 9 0 30 1.941427 0.01 1
5 10 0 30 1.342969 0.01 1

69. For example, look at the ‘SOC’column. Which of
these is higher than 2?
> SOC$SOC > 2
[1] TRUE TRUE TRUE FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[13] TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE
[25] TRUE FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[37] FALSE FALSE TRUE FALSE TRUE TRUE TRUE FALSE TRUE FALSE TRUE TRUE
[49] FALSE TRUE TRUE FALSE FALSE TRUE TRUE FALSE TRUE FALSE TRUE TRUE
[61] TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE TRUE FALSE
...

70. We can now use this variable to access only the
rows that have SOC > 2
> SOCHigh <- SOC$SOC > 2
> SOC[SOCHigh,]
Id UpperDepth LowerDepth SOC Lambda tsme
1 4 0 30 12.000325 0.01 0.003985153
2 7 0 30 3.483653 0.01 0.002502976
3 8 0 30 2.313414 0.01 0.002504971
6 11 0 30 2.287933 0.01 0.002509360
7 12 0 30 2.715843 0.01 0.002518323
...

71. Or do it all at once (in pure R fashion)
> SOC[SOC$SOC > 2,]
Id UpperDepth LowerDepth SOC Lambda tsme
1 4 0 30 12.000325 0.01 0.003985153
2 7 0 30 3.483653 0.01 0.002502976
3 8 0 30 2.313414 0.01 0.002504971
6 11 0 30 2.287933 0.01 0.002509360
7 12 0 30 2.715843 0.01 0.002518323
8 13 0 30 4.340112 0.01 0.002515760
...

72. Ordering
You can reorder the data.frame by one or more
columns using the order() function
> SOC[order(SOC$tsme),]
Id UpperDepth LowerDepth SOC Lambda tsme
2268 4039 0 30 0.00000000 0.01 0.002472194
145 396 0 30 1.17823531 0.01 0.002479430
1527 2803 0 30 0.45000000 0.01 0.002482055
471 1032 0 30 0.62766244 0.01 0.002482168
1581 3101 0 30 0.92922471 0.01 0.002484241
2237 3996 0 30 1.44240409 0.01 0.002484922
2629 5048 0 30 0.85421359 0.01 0.002485544
1910 3590 0 30 1.45097492 0.01 0.002486062
1650 3234 0 30 0.00000000 0.01 0.002486621
2536 4632 0 30 1.76030792 0.01 0.002486932
...

73. Ordering
You can reorder the data.frame by one or more
columns using the order() function
> SOC[order(SOC$tsme, SOC$SOC),]
Id UpperDepth LowerDepth SOC Lambda tsme
2268 4039 0 30 0.00000000 0.01 0.002472194
145 396 0 30 1.17823531 0.01 0.002479430
1527 2803 0 30 0.45000000 0.01 0.002482055
471 1032 0 30 0.62766244 0.01 0.002482168
1581 3101 0 30 0.92922471 0.01 0.002484241
2237 3996 0 30 1.44240409 0.01 0.002484922
2629 5048 0 30 0.85421359 0.01 0.002485544

74. Making your own data.frame is straightforward
using the data.frame() function. For example:
> year <- 2000:2010
> catch <- c(900, 1230, 1400, 930, 670, 1000, 960, 840, 900, 500,400)
> dat <- data.frame(year=year, catch=catch)
> head(dat)
year catch
1 2000 900
2 2001 1230
3 2002 1400
4 2003 930
5 2004 670
6 2005 1000

75. It's possible to add extra columns of various types
> dat$area <- c("N","S","N","S","N","S","N","S","N","S","N")
> dat$survey <- c(TRUE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE, FALSE, TRUE,
TRUE)
> head(dat)
year catch area survey
1 2000 900 N TRUE
2 2001 1230 S FALSE
3 2002 1400 N FALSE
4 2003 930 S TRUE
5 2004 670 N TRUE
6 2005 1000 S TRUE

76. To add an extra row or rows use rbind and pass in
a data.frame with the exact same column names
and types
> dat2 <- data.frame(year = 1920, catch = 666, area = "N", survey = FALSE)
> dat <- rbind(dat, dat2)
> dat
year catch area survey
1 2000 900 N TRUE
2 2001 1230 S FALSE
3 2002 1400 N FALSE
4 2003 930 S TRUE
5 2004 670 N TRUE
6 2005 1000 S TRUE
7 2006 960 N TRUE
8 2007 840 S TRUE
9 2008 900 N FALSE
10 2009 500 S TRUE
11 2010 400 N TRUE
12 1920 666 N FALSE

77. Ask at least 4 people near you and make a data.frame to
hold the following information about them: Name, hair
colour, height, shoe size, how long they can hold
their breath for.
• Reorder the data.frame by height.
• Subset the data.frame to only include people taller
than 1m 70.
• What is the mean shoe size of the people in the
data.frame?
• Whose shoe size is closest to the mean shoe size?

78. We need to talk about factors.
Macedonian Soil Data data.frame
> head(MSoil)
Id UpperDepth LowerDepth SOC Lambda tsme Region
1 4 0 30 12.000325 0.01 0.003985153 A
2 7 0 30 3.483653 0.01 0.002502976 B
3 8 0 30 2.313414 0.01 0.002504971 B
4 9 0 30 1.941427 0.01 0.002508691 B
5 10 0 30 1.342969 0.01 0.002509177 B
6 11 0 30 2.287933 0.01 0.002509360 B

79. We need to talk about factors. In the Macedonian
Soil Data data.frame Take a look at the ‘Region’
column
> head(MSoil$Region)
[1] A B B B B B
Levels: A B
They look like characters, but no quotes. There are two
“levels”: A and B. What does this mean?

80. They look like characters, but no quotes. There are two
“levels”: A and B. What does this mean?
> class(MSoil$Region)
[1] "factor"

81. Factors
Factors are a way of encoding data that can be used for
grouping variables.
Values can only be one of the defined 'levels'. This
allows you to keep track of what the values could be.
They can be used to ensure that a data set is coherent.

82. For example, if we try to set a value in the
“Region” column to something other than A or B,
we get a warning
> MSoil[1,"Region"] <- "20"
Warning message:
In `[<-.factor`(`*tmp*`, iseq, value = "20") :
invalid factor level, NA generated

83. And a broken data.frame
> MSoil[1,]
Id UpperDepth LowerDepth SOC Lambda tsme Region
1 4 0 30 12.00032 0.01 0.003985153 <NA>

84. Let's fix it :)
> MSoil[1,"Region"] <- "A"
> MSoil[1,]
Id UpperDepth LowerDepth SOC Lambda tsme Region
1 4 0 30 12.00032 0.01 0.003985153 A

85. Factors
If you really wanted to change the value to
something not in the levels you need to change
the levels too (the names of the factors)
> levels(MSoil$Region)
[1] "A" "B"

86. Factors
If you really wanted to change the value to
something not in the levels you need to change
the levels too (the names of the factors)
> MSoil[1,"Region"] <- "C"
>
> head(MSoil)
Id UpperDepth LowerDepth SOC Lambda tsme Region
1 4 0 30 12.000325 0.01 0.003985153 C
2 7 0 30 3.483653 0.01 0.002502976 B
3 8 0 30 2.313414 0.01 0.002504971 B
4 9 0 30 1.941427 0.01 0.002508691 B
5 10 0 30 1.342969 0.01 0.002509177 B
6 11 0 30 2.287933 0.01 0.002509360 B

87. Factors
If you really wanted to change the value to
something not in the levels you need to change
the levels too (the names of the factors)
MSoil[, "Region"]
[1] C B B B B B B B B B B B B B B B B B B B B B A A A A A A A A A A A A A A
[37] A B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B A A
...
[973] B B B B B B B B B B B B B B B B B B B B B B B B B B B B
[ reached getOption("max.print") -- omitted 2262 entries ]
Levels: A B C

88. Factors
Factors are used for many methods and
functions in R, such as linear analysis.

89. let's make another data set that only includes
Region == B
> MSoilB <- subset(MSoil, Region=="B")
>
> MSoilB$Region
[1] B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B
...
[937] B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B
[973] B B B B B B B B B B B B B B B B B B B B B B B B B B B B
[ reached getOption("max.print") -- omitted 1147 entries ]
Levels: A B C

90. You can see that we have no observations for A
but you know that there could be. This might be
important for data management. When you import
data into R, some of your columns may be read in
as factors even if you did not intend them to.

91. The by() function can be used to split the data and apply
a function to each chunk. This can be very useful for
summarising the data. For example, to split the data by
the 'Region' column (into A and B chunks) and take the
mean of the column of each chunk you can do
> by(MSoil$SOC, MSoil$Region, mean)
MSoil$Region: A
[1] NaN
----------------------------------------------------------
MSoil$Region: B
[1] 1.839508
----------------------------------------------------------
MSoil$Region: C
[1] 12.00032

92. aggregate() does something similar but can be
used to operate on multiple columns in a data
frame.Learning how to manipulate data frames is
a very useful skill.The plyr and reshape packages
are worth your time getting to know.

93. What is the mean height by hair
colour of the people in your
data.frame?

94. A list is a very flexible container.
It's like a vector, but the elements can be
objects of any class and size - even lists
(lists f lists of lists of …).
This makes them very handy for moving big
chunks of data around (particularly returning
output from a function).

95. Here we make two objects to put into a list.
> best_food <- c("cake", "banana")
> odd_numbers <- c(1,3,5,7,9)
> notes <- "Something interesting"

96. To make the list, we use the list() function.
When you create a list, you should give the
elements names (they don't have to be the name
of the object).
> my_list <- list(food = best_food, numbers = odd_numbers, note
= notes)
> class(my_list)
[1] "list"

97. Getting the length of the list and the names of the
elements is straightforward
> length(my_list)
[1] 3

98. Getting the length of the list and the names of the
elements is straightforward
> length(my_list)
[1] 3
> names(my_list)
[1] "food" "numbers" "note

99. Elements in a list can be extracted using two
methods. By name, using $ and the element
name.
> my_list$food
[1] "cake" "banana"

100. Accessing data in a list
Using [[ and the element position or name.
> my_list[[1]]
[1] "cake" "banana"
>
> my_list[["food"]]
[1] "cake" "banana"

101. Modifying lists
Lists can be easily extended - just add an extra
element.
> my_list[["new"]] <- c(1,3,5,7)
> summary(my_list)
Length Class Mode
food 2 -none- character
numbers 5 -none- numeric
note 1 -none- character
new 4 -none- numeric

102. Processing lists
lapply - apply the same function to each element
in a list.
> vec1 <- seq(from=1, to = 10, length = 7)
> vec2 <- seq(from=12, to = 20, length = 6)
> lst <- list(vec1 = vec1, vec2 = vec2)
> lapply(lst, sum)
$vec1
[1] 38.5
$vec2
[1] 96

103. Processing lists
This only makes sense if the same function can be
applied to all elements. For example, if we add a
character vector to the list, we can't use sum.But
length makes sense.
> lapply(lst, length)
$vec1
[1] 7
$vec2
[1] 6
$str1
[1] 3

107. # Exercise 4
Newlist <- list(a=1:10, b="Good morning", c="Hi")
Newlist$a <- Newlist$a + 1
Newlist
## $a
## [1] 2 3 4 5 6 7 8 9 10 11
##
## $b
## [1] "Good morning"
##
## $c
## [1] "Hi"