Matrices, data frames and lists
A data set that have more than one dimension is conceptually hard to store as a vector. For two-dimensional data set the solution to this is to instead use matrices or data frames. As with vectors all values in a matrix has to be of the same type (eg. you can not mix for example characters and numerics in the same matrix). For data frames this is not a requirement and different columns can have different modes, but all columns in a data frame have the same number of entries. In addition to these R also have objects named lists that can store any type of data set and are not restricted by types or dimensions.
In this exercise you will learn how to:
- Create and work with matrices, data frames and lists
- Perform basic math operator on matrices
- Use functions to summarize information from data frames
- Extract subsets of data from matrices, data frames and lists
- Create S3 object from a list
Matrices in R
The command to create a matrix in R is matrix().
As input it takes a vector of values, the number of
rows and the number of columns.
X <- matrix(1:12, nrow = 4, ncol = 3)
X
[,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12
Note that if one only specify the number of rows or columns the it will infer the size of the matrix automatically using the size of vector and the option given. The default way of filling the matrix is column-wise, so the first values from the vector ends up in column 1 of the matrix. If you instead wants to fill the matrix row by row you can set the byrow flag to TRUE.
X <- matrix(1:12, nrow = 4, ncol = 3, byrow = TRUE)
X
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 6
[3,] 7 8 9
[4,] 10 11 12
Subsetting a matrix is done the same way as for vectors, but you have more than one dimension to work with. So you specify the rows and column needed.
X[1,2]
[1] 2
If one wants all values in a column or a row this can be specified by leaving the other dimension empty, hence this code will print all values in the second column.
X[,2]
[1] 2 5 8 11
Note that if the retrieved part of a matrix can be represented as a vector (eg one of the dimension have the length 1) R will convert it to a vector otherwise it will still be a matrix.
Exercise: Working with matrices in R
Create a matrix containing 1:12 as shown for the matrix X above.
- What is the length and the mode of the matrix?
Click to see howmode(X) length(X) [1] "numeric" [1] 12 - Use similar ideas as when you worked with vectors to extract all
values in the matrix that is larger than 6
Click to see how
X[X>6] [1] 7 10 8 11 9 12 - Shift places of column 1 and 3 in X
Click to see how
X[,c(3,2,1)] [,1] [,2] [,3] [1,] 3 2 1 [2,] 6 5 4 [3,] 9 8 7 [4,] 12 11 10 -
Add a vector with three zeros as a fifth row to the matrix
Click to see howX.2 <- rbind(X, rep(0, 3)) X.2 [,1] [,2] [,3] [1,] 1 2 3 [2,] 4 5 6 [3,] 7 8 9 [4,] 10 11 12 [5,] 0 0 0 - Replace all values the first two columns in your matrix with “NA”.
Click to see how
X[,1:2] <- NA X [,1] [,2] [,3] [1,] NA NA 3 [2,] NA NA 6 [3,] NA NA 9 [4,] NA NA 12 - Replace all values in the matrix with 0 and convert it to a vector
Click to see how
X[] <- 0 as.vector(X) [1] 0 0 0 0 0 0 0 0 0 0 0 0 - In the the exercies earlier you created a vector with the names of
the type Geno_a_1, Geno_a_2, Geno_a_3, Geno_b_1, Geno_b_2…,
Geno_s_3 using vectors. In todays lecture a function named outer
that generate matrixes was mentioned. Try to generate the same
vector, but using the outer function instead. This is a
challenging task
Click to see how
letnum <- outer(paste("Geno",letters[1:19], sep = "_"), 1:3, paste, sep = "_") class(letnum) sort(as.vector(letnum)) [1] "matrix" [1] "Geno_a_1" "Geno_a_2" "Geno_a_3" "Geno_b_1" "Geno_b_2" "Geno_b_3" [7] "Geno_c_1" "Geno_c_2" "Geno_c_3" "Geno_d_1" "Geno_d_2" "Geno_d_3" [13] "Geno_e_1" "Geno_e_2" "Geno_e_3" "Geno_f_1" "Geno_f_2" "Geno_f_3" [19] "Geno_g_1" "Geno_g_2" "Geno_g_3" "Geno_h_1" "Geno_h_2" "Geno_h_3" [25] "Geno_i_1" "Geno_i_2" "Geno_i_3" "Geno_j_1" "Geno_j_2" "Geno_j_3" [31] "Geno_k_1" "Geno_k_2" "Geno_k_3" "Geno_l_1" "Geno_l_2" "Geno_l_3" [37] "Geno_m_1" "Geno_m_2" "Geno_m_3" "Geno_n_1" "Geno_n_2" "Geno_n_3" [43] "Geno_o_1" "Geno_o_2" "Geno_o_3" "Geno_p_1" "Geno_p_2" "Geno_p_3" [49] "Geno_q_1" "Geno_q_2" "Geno_q_3" "Geno_r_1" "Geno_r_2" "Geno_r_3" [55] "Geno_s_1" "Geno_s_2" "Geno_s_3" - Create two different 2 by 2 matrices named A and B. A should
contain the values 1 - 4 and B the values 5 - 8. Try out the
following commands and by looking at the results see if you can
figure out what is going on.
A. A * B
B. A / B
C. A %x% B
D. A + B
E. A - B
F. A == B
Click to see howA <- matrix(1:4, ncol = 2, nrow = 2) B <- matrix(5:8, ncol = 2, nrow = 2) A [,1] [,2] [1,] 1 3 [2,] 2 4 B [,1] [,2] [1,] 5 7 [2,] 6 8 A * B [,1] [,2] [1,] 5 21 [2,] 12 32 A / B [,1] [,2] [1,] 0.2000000 0.4285714 [2,] 0.3333333 0.5000000 A %x% B [,1] [,2] [,3] [,4] [1,] 5 7 15 21 [2,] 6 8 18 24 [3,] 10 14 20 28 [4,] 12 16 24 32 A + B [,1] [,2] [1,] 6 10 [2,] 8 12 A - B [,1] [,2] [1,] -4 -4 [2,] -4 -4 A == B [,1] [,2] [1,] FALSE FALSE [2,] FALSE FALSE - Generate a 10 by 10 matrix with random numbers. Add row and
columnnames and calculate mean and median over rows and save these
in a new matrix.
Click to see how
e <- rnorm(n = 100) E <- matrix(e, nrow = 10, ncol = 10) colnames(E) <- LETTERS[1:10] rownames(E) <- colnames(E) E.means <- rowMeans(E) E.medians <- apply(E, MARGIN = 1, median) E.mm <- rbind(E.means, E.medians) E.mm A B C D E F E.means -0.01902767 0.01075332 -0.4137270 -0.1304978 0.2099126 0.2965743 E.medians 0.53337938 0.18481261 -0.2248858 -0.1139851 0.3269634 0.2601974 G H I J E.means -0.6670421 -0.27378920 -0.1533350 -0.0437610 E.medians -0.5247300 -0.09460231 -0.3547495 -0.2493248
Data frames
Even though vectors are at the very base of R usage, data frames are
central to R as the most common ways to import data into R
(read.table) will create a data frame. Even though a data frame can
itself contain another data frame, the by far, most common data frames
consists of a set of equally long vectors. As data frames can contain
several different data types the command str()
is very useful to run on data frames
vector1 <- 1:10
vector2 <- letters[1:10]
vector3 <- rnorm(10, sd = 10)
df <- data.frame(vector1, vector2, vector3)
str(df)
'data.frame': 10 obs. of 3 variables:
$ vector1: int 1 2 3 4 5 6 7 8 9 10
$ vector2: Factor w/ 10 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10
$ vector3: num 8.463 0.905 -0.255 -6.59 3.369 ...
In the above example we can see that the data frame df contains 10 observations for three variables that all have different modes, column 1 is an integer vector, column 2 a vector with factors! and column 3 a numeric vector. It is noteworthy that the second column is a factor even though we just gave it a character vector. See question 1 below if you wonder why this is the case.
Exercise: Working with data frames
- Use the built-in help in R to figure out what is going on with the
second column in df data frame described above and modify the
creation of the data frame so that the second column is stored as a
character vector.
Click to see how
df <- data.frame(vector1, vector2, vector3, stringsAsFactors = FALSE) str(df) 'data.frame': 10 obs. of 3 variables: $ vector1: int 1 2 3 4 5 6 7 8 9 10 $ vector2: chr "a" "b" "c" "d" ... $ vector3: num 8.463 0.905 -0.255 -6.59 3.369 ... - One can select columns from a data frame using either the name or
the position. Use both methods to print the last two columns from
the df data frame.
Click to see how
df[,2:3] df[,c("vector2", "vector3")] vector2 vector3 1 a 8.4628687 2 b 0.9046253 3 c -0.2549117 4 d -6.5902581 5 e 3.3685362 6 f 16.7773472 7 g 9.3203649 8 h -10.4333097 9 i 2.9716131 10 j 8.1402695 vector2 vector3 1 a 8.4628687 2 b 0.9046253 3 c -0.2549117 4 d -6.5902581 5 e 3.3685362 6 f 16.7773472 7 g 9.3203649 8 h -10.4333097 9 i 2.9716131 10 j 8.1402695 - Print all letters in the vector2 column of the data frame where the
vector3 column has a positive value.
Click to see how
df[df$vector3>0,2] df$vector2[df$vector3>0] [1] "a" "b" "e" "f" "g" "i" "j" [1] "a" "b" "e" "f" "g" "i" "j" - Create a new vector where the number of elements is equal to the number of rows in the original data frame. Each element of this vector combines values from the three columns of df separated by an underscore. For example, the first element of the vector should be: 1_a_8.46286871843976.
Click to see how
paste(df$vector1, df$vector2, df$vector3, sep = "_") [1] "1_a_8.46286871843976" "2_b_0.904625308313597" "3_c_-0.25491171338376" [4] "4_d_-6.59025808447186" "5_e_3.36853617579661" "6_f_16.7773472039123" [7] "7_g_9.32036493453533" "8_h_-10.4333097064694" "9_i_2.97161306345798" [10] "10_j_8.14026953369552" - There is a data frame of car information that comes with the base
installation of R. Have a look at this data by typing
mtcars. Click to see howmtcars mpg cyl disp hp drat wt qsec vs am gear carb Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 AMC Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 Camaro Z28 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 - Re-arrange the row names of this data frame and save as a vector.
Click to see how
car.names <- sample(row.names(mtcars)) - Create a data frame containg the vector from the previous question
and two vectors with random numbers named random1 and random2.
Click to see how
random1 <- rnorm(length(car.names)) random2 <- rnorm(length(car.names)) mtcars2 <- data.frame(car.names, random1, random2) mtcars2 car.names random1 random2 1 Toyota Corona 0.2672093 0.748625274 2 Duster 360 -0.4127061 -0.289656962 3 Hornet Sportabout -0.6291955 1.154517511 4 Volvo 142E 1.4695465 1.822855299 5 Lotus Europa -0.1088715 -0.688590021 6 Hornet 4 Drive -0.4359612 -0.274399856 7 Valiant -0.9114306 -0.552239587 8 Merc 450SLC 0.1083370 0.212631221 9 Fiat X1-9 -0.3422226 -1.991076826 10 Cadillac Fleetwood 0.4657490 0.779438149 11 Toyota Corolla 1.1136944 -0.949605064 12 Mazda RX4 Wag -0.6442193 -0.353000665 13 Ferrari Dino 0.7393240 -0.157842460 14 Mazda RX4 -0.0431834 1.428955430 15 Datsun 710 1.1788716 -0.056881290 16 Merc 280 0.8434795 -1.676932154 17 Fiat 128 0.5203762 -1.540330757 18 Merc 450SE -0.6783654 -1.088913643 19 Honda Civic 0.9413628 -0.689011222 20 Porsche 914-2 -1.7112856 -0.279261819 21 Pontiac Firebird 0.7238131 0.980874293 22 Merc 230 0.4692142 0.417665142 23 Maserati Bora -0.6522722 0.394803085 24 Lincoln Continental 1.3341690 -0.008482409 25 Chrysler Imperial -1.7568138 0.231171108 26 AMC Javelin -0.3436457 -0.801661343 27 Dodge Challenger 0.9847896 0.240541233 28 Ford Pantera L 0.1812936 -2.391389388 29 Camaro Z28 0.2731022 -0.562270119 30 Merc 240D -1.3300011 0.941390495 31 Merc 280C -0.1134380 -1.051899224 32 Merc 450SL 1.0369179 -0.256698993 - Now you have two data frames that both contains information on a
set of cars. A collaborator asks you to create a new data frame
with all this information combined. Create this
merged data frame and make sure that it is combined
in the correct way.
Click to see how
mt.merged <- merge(mtcars, mtcars2, by.x = "row.names", by.y = "car.names") mt.merged Row.names mpg cyl disp hp drat wt qsec vs am gear carb 1 AMC Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 2 Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 3 Camaro Z28 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 4 Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 5 Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 6 Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 7 Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 8 Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 9 Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 10 Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 11 Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 12 Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 13 Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 14 Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 15 Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 16 Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 17 Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 18 Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 19 Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 20 Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 21 Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 22 Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 23 Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 24 Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 25 Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 26 Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 27 Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 28 Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 29 Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 30 Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 31 Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 32 Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 random1 random2 1 -0.3436457 -0.801661343 2 0.4657490 0.779438149 3 0.2731022 -0.562270119 4 -1.7568138 0.231171108 5 1.1788716 -0.056881290 6 0.9847896 0.240541233 7 -0.4127061 -0.289656962 8 0.7393240 -0.157842460 9 0.5203762 -1.540330757 10 -0.3422226 -1.991076826 11 0.1812936 -2.391389388 12 0.9413628 -0.689011222 13 -0.4359612 -0.274399856 14 -0.6291955 1.154517511 15 1.3341690 -0.008482409 16 -0.1088715 -0.688590021 17 -0.6522722 0.394803085 18 -0.0431834 1.428955430 19 -0.6442193 -0.353000665 20 0.4692142 0.417665142 21 -1.3300011 0.941390495 22 0.8434795 -1.676932154 23 -0.1134380 -1.051899224 24 -0.6783654 -1.088913643 25 1.0369179 -0.256698993 26 0.1083370 0.212631221 27 0.7238131 0.980874293 28 -1.7112856 -0.279261819 29 1.1136944 -0.949605064 30 0.2672093 0.748625274 31 -0.9114306 -0.552239587 32 1.4695465 1.822855299 - Calculate the mean value for the two columns that you added to the
mtcars data frame.
Click to see how
colMeans(mtcars2[, c("random1", "random2")]) random1 random2 0.07930118 -0.19708361Try to modify so you get the mean by cylinder number instead.
Click to see howaggregate(mtcars2$random1, by = list(mtcars$cyl), FUN = mean) Group.1 x 1 4 0.02470902 2 6 0.16250439 3 8 0.08059342
Lists
The last data structure that we will explore are lists, which is a very flexible structure. Lists can in R combine different data structures and they do not have to be of equal dimensions or have other restrictions. The drawback with a flexible structure is that it requires a bit more work to interact with.
The syntax to create a list is similar to creation of the other data structures in R.
l <- list(1, 2, 3)
As with the data frames the str() command is very useful for the sometimes fairly complex lists instances.
str(l)
List of 3
$ : num 1
$ : num 2
$ : num 3
This example containing only numeric vector is not very exciting example given the flebility a list structure offers so lets create a more complex example
vec1 <- letters
vec2 <- 1:4
mat1 <- matrix(1:100, nrow = 5)
df1 <- as.data.frame(cbind(10:1, 91:100))
u.2 <- list(vec1, vec2, mat1, df1, l)
As you can see a list can not only contain other data structures, but can also contain other lists.
Looking at the str command reveals much of the details of a list
str(u.2)
List of 5
$ : chr [1:26] "a" "b" "c" "d" ...
$ : int [1:4] 1 2 3 4
$ : int [1:5, 1:20] 1 2 3 4 5 6 7 8 9 10 ...
$ :'data.frame': 10 obs. of 2 variables:
..$ V1: int [1:10] 10 9 8 7 6 5 4 3 2 1
..$ V2: int [1:10] 91 92 93 94 95 96 97 98 99 100
$ :List of 3
..$ : num 1
..$ : num 2
..$ : num 3
With this more complex object subsetting are slighty trickier than with more the more homogenous objects we have looked at so far.
To look at the first entry of a list one can use the same syntax as for the simplier structures, but note that this will give you a list of length 1 irrespective of the actual type of data structure found.
u.2[1]
str(u.2[1])
[[1]]
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s"
[20] "t" "u" "v" "w" "x" "y" "z"
List of 1
$ : chr [1:26] "a" "b" "c" "d" ...
If one instead wants to extract the list entry as the structure that is stored, one needs to “dig” deeper in the object.
u.2[[1]]
str(u.2[[1]])
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s"
[20] "t" "u" "v" "w" "x" "y" "z"
chr [1:26] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" ...
This means that the syntax to extract to exact specific value from a data structure stored in a list can be daunting, examplified by extracting the second column of the data frame stored at position 4 in the list u.2.
u.2[[4]][,2]
[1] 91 92 93 94 95 96 97 98 99 100
Exercise: Working with lists
- Create a list containing 1 character vector, a numeric vector, a
character matrix.
Click to see how
list.2 <- list(vec1 = c("hi", "ho", "merry", "christmas"), vec2 = 4:19, mat1 = matrix(as.character(100:81), nrow = 4)) list.2 $vec1 [1] "hi" "ho" "merry" "christmas" $vec2 [1] 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 $mat1 [,1] [,2] [,3] [,4] [,5] [1,] 100 96 92 88 84 [2,] 99 95 91 87 83 [3,] 98 94 90 86 82 [4,] 97 93 89 85 81 - Create a data fram and add this to the list.
Click to see how
df <- data.frame(letters, LETTERS, letters == LETTERS) list.2[[4]] <- df - Remove the the second entry of your list
Click to see how
list.2[-2] $vec1 [1] "hi" "ho" "merry" "christmas" $mat1 [,1] [,2] [,3] [,4] [,5] [1,] 100 96 92 88 84 [2,] 99 95 91 87 83 [3,] 98 94 90 86 82 [4,] 97 93 89 85 81 [[3]] letters LETTERS letters....LETTERS 1 a A FALSE 2 b B FALSE 3 c C FALSE 4 d D FALSE 5 e E FALSE 6 f F FALSE 7 g G FALSE 8 h H FALSE 9 i I FALSE 10 j J FALSE 11 k K FALSE 12 l L FALSE 13 m M FALSE 14 n N FALSE 15 o O FALSE 16 p P FALSE 17 q Q FALSE 18 r R FALSE 19 s S FALSE 20 t T FALSE 21 u U FALSE 22 v V FALSE 23 w W FALSE 24 x X FALSE 25 y Y FALSE 26 z Z FALSE - Create a new list that contain 20 entries, with each entry holding
a numeric vector.
Click to see how
vec1 <- rnorm(1000) list.a <- split(vec1, 1:20) - How long is your list, and how long are each of the vectors
that are part of the list?
Click to see how
length(list.a) lapply(list.a, FUN = "length") [1] 20 $`1` [1] 50 $`2` [1] 50 $`3` [1] 50 $`4` [1] 50 $`5` [1] 50 $`6` [1] 50 $`7` [1] 50 $`8` [1] 50 $`9` [1] 50 $`10` [1] 50 $`11` [1] 50 $`12` [1] 50 $`13` [1] 50 $`14` [1] 50 $`15` [1] 50 $`16` [1] 50 $`17` [1] 50 $`18` [1] 50 $`19` [1] 50 $`20` [1] 50 -
Figure out what the main differences are between the function lapply and sapply are and use both of them with the function summary on your newly created list. What are the pros and cons of the two approaches to calculate the same summary statistics?
Click to see howlapply(X = list.a, FUN = "summary") sapply(X = list.a, FUN = "summary") $`1` Min. 1st Qu. Median Mean 3rd Qu. Max. -1.91700 -0.98430 -0.10330 -0.09407 0.64310 2.57300 $`2` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.55600 -0.59190 0.06890 0.09146 0.98040 2.40500 $`3` Min. 1st Qu. Median Mean 3rd Qu. Max. -1.7200 -0.2922 0.3422 0.4497 1.0440 3.4580 $`4` Min. 1st Qu. Median Mean 3rd Qu. Max. -1.65400 -0.77660 -0.06379 0.05182 0.68320 2.72800 $`5` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.11200 -0.67370 0.09657 0.08760 0.78310 2.42000 $`6` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.3730 -1.1960 -0.1069 -0.1600 0.7839 2.5650 $`7` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.08900 -0.84710 -0.31490 -0.25480 0.03034 1.86400 $`8` Min. 1st Qu. Median Mean 3rd Qu. Max. -3.13100 -0.74770 0.25510 -0.03403 0.75410 1.98000 $`9` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.58600 -0.36920 0.02267 0.10700 0.48530 2.19900 $`10` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.0500 -1.0210 -0.4427 -0.2017 0.5982 2.4700 $`11` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.00300 -0.65670 -0.02114 0.04536 0.54900 2.47800 $`12` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.08200 -0.76080 -0.17120 -0.09029 0.36670 2.58100 $`13` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.42300 -0.66920 0.02297 -0.01248 0.63560 2.35000 $`14` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.18400 -0.99050 -0.06705 -0.18770 0.43920 2.40500 $`15` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.194000 -0.638600 0.090650 -0.006298 0.599600 2.537000 $`16` Min. 1st Qu. Median Mean 3rd Qu. Max. -1.9650 -0.8252 -0.1867 -0.1255 0.4426 2.4360 $`17` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.5890 -0.8900 -0.3218 -0.3507 0.3900 1.8250 $`18` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.17500 -0.52770 0.05985 -0.07110 0.41190 1.66200 $`19` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.42600 -0.65740 0.06455 0.02680 0.48520 2.75000 $`20` Min. 1st Qu. Median Mean 3rd Qu. Max. -2.5350 -0.4091 0.2411 0.1381 0.6583 2.7100 1 2 3 4 5 6 7 8 Min. -1.91700 -2.55600 -1.7200 -1.65400 -2.11200 -2.3730 -2.08900 -3.13100 1st Qu. -0.98430 -0.59190 -0.2922 -0.77660 -0.67370 -1.1960 -0.84710 -0.74770 Median -0.10330 0.06890 0.3422 -0.06379 0.09657 -0.1069 -0.31490 0.25510 Mean -0.09407 0.09146 0.4497 0.05182 0.08760 -0.1600 -0.25480 -0.03403 3rd Qu. 0.64310 0.98040 1.0440 0.68320 0.78310 0.7839 0.03034 0.75410 Max. 2.57300 2.40500 3.4580 2.72800 2.42000 2.5650 1.86400 1.98000 9 10 11 12 13 14 15 16 Min. -2.58600 -2.0500 -2.00300 -2.08200 -2.42300 -2.18400 -2.194000 -1.9650 1st Qu. -0.36920 -1.0210 -0.65670 -0.76080 -0.66920 -0.99050 -0.638600 -0.8252 Median 0.02267 -0.4427 -0.02114 -0.17120 0.02297 -0.06705 0.090650 -0.1867 Mean 0.10700 -0.2017 0.04536 -0.09029 -0.01248 -0.18770 -0.006298 -0.1255 3rd Qu. 0.48530 0.5982 0.54900 0.36670 0.63560 0.43920 0.599600 0.4426 Max. 2.19900 2.4700 2.47800 2.58100 2.35000 2.40500 2.537000 2.4360 17 18 19 20 Min. -2.5890 -2.17500 -2.42600 -2.5350 1st Qu. -0.8900 -0.52770 -0.65740 -0.4091 Median -0.3218 0.05985 0.06455 0.2411 Mean -0.3507 -0.07110 0.02680 0.1381 3rd Qu. 0.3900 0.41190 0.48520 0.6583 Max. 1.8250 1.66200 2.75000 2.7100
Extra exercises
- Design a S3 class that should hold information on human
proteins. The data needed for each protein is:
- The gene that encodes it
- The molecular weight of the protein
- The length of the protein sequence
- Information on who and when it was discovered
- Protein assay data
Create this hypethetical S3 object in R.
- Among the test data sets that are part of base R, there is one called iris. It contains measurements on set of plants. You can access the data using by typing iris in R. Explore this data set and calculate some useful summary statistics, like SD, mean and median for the parts of the data where this makes sense. Calculate the same statistics for any grouping that you can find in the data.