Original Authors: Martin Morgan, Sonali Arora, Lori Shepherd
Presenting Author: Maria Doyle
Date: 23-28 June, 2024
Back: Monday labs
Objective: Gain confidence working with base R commands and data structures.
Lessons learned:
factor()
, NA
?factor
browseVignettes()
Efficient vectorized calculations on ‘atomic’ vectors logical
,
integer
, numeric
, complex
, character
, raw
character_vector <- c("January", "February", "March", "April", "May")
logical_vector <- c(FALSE, FALSE, TRUE, TRUE, TRUE)
integer_vector <- 1:5 # c(1, 2, 3, 4, 5)
Atomic vectors are building blocks for more complicated objects
factor
– enumeration of possible levels
months <- factor(
character_vector, # values realized in 'months'
levels = c( # possible values
"January", "February", "March", "April", "May", "June", "July",
"August", "September", "October", "November", "December"
)
)
matrix
– atomic vector with ‘dim’ attribute
matrix(1:6, nrow = 3) # n.b., 'column-major' order
## [,1] [,2]
## [1,] 1 4
## [2,] 2 5
## [3,] 3 6
data.frame
– list of equal length atomic vectors
data_frame <- data.frame(
month = months,
is_spring = logical_vector,
month_of_year = integer_vector
)
Formal classes represent complicated combinations of vectors,
e.g., the return value of lm()
, below
Functions transform inputs to outputs, perhaps with side effects
rnorm(5)
## [1] -0.8834978 -1.1483506 2.4169771 1.0662281 2.4484377
Argument matching first by name, then by position
Functions may define (some) arguments to have default values
log(1:5) # default base = exp(1)
## [1] 0.0000000 0.6931472 1.0986123 1.3862944 1.6094379
log(1:5, base = 10)
## [1] 0.0000000 0.3010300 0.4771213 0.6020600 0.6989700
log(base = 10, 1:5) # named arguments match before unnamed
## [1] 0.0000000 0.3010300 0.4771213 0.6020600 0.6989700
Generic functions dispatch to specific methods based on class of
argument(s), e.g., print()
.
Methods are functions that implement specific generics, e.g.,
print.factor
; methods are invoked indirectly, via the generic.
?print # what does the generic 'print()' do?
?print.factor # what does the method 'print(x)', when x is a factor, do?
Many but not all functions able to manipulate a particular class are
methods, e.g., abline()
used below is a plain-old-function.
Iteration:
lapply()
args(lapply)
## function (X, FUN, ...)
## NULL
X
(an atomic vector or list()
), apply a
function FUN
to each vector element, returning the result as a
list. ...
are additional arguments to FUN
.FUN
can be built-in, or a user-defined functionlst <- list(a=1:2, b=2:4)
lapply(lst, log) # 'base' argument default; natural log
## $a
## [1] 0.0000000 0.6931472
##
## $b
## [1] 0.6931472 1.0986123 1.3862944
lapply(lst, log, 10) # '10' is second argument to 'log()', i.e., log base 10
## $a
## [1] 0.00000 0.30103
##
## $b
## [1] 0.3010300 0.4771213 0.6020600
sapply()
– like lapply()
, but simplify the result to a
vector, matrix, or array, if possible.vapply()
– like sapply()
, but requires that the return
type of FUN
is specified; this can be safer – an error when
the result is of an unexpected type.mapply()
(also Map()
)
args(mapply)
## function (FUN, ..., MoreArgs = NULL, SIMPLIFY = TRUE, USE.NAMES = TRUE)
## NULL
...
are one or more vectors, recycled to be of the same
length. FUN
is a function that takes as many arguments as
there are components of ...
. mapply
returns the result of
applying FUN
to the elements of the vectors in ...
.
mapply(seq, 1:3, 4:6, SIMPLIFY=FALSE) # seq(1, 4); seq(2, 5); seq(3, 6)
## [[1]]
## [1] 1 2 3 4
##
## [[2]]
## [1] 2 3 4 5
##
## [[3]]
## [1] 3 4 5 6
apply()
args(apply)
## function (X, MARGIN, FUN, ..., simplify = TRUE)
## NULL
For a matrix or array X
, apply FUN
to each MARGIN
(dimension, e.g., MARGIN=1
means apply FUN
to each row,
MARGIN=2
means apply FUN
to each column)
Traditional iteration programming constructs repeat {}
, for () {}
lapply()
!Conditional
if (test) {
## code if TEST == TRUE
} else {
## code if TEST == FALSE
}
Functions (see table below for a few favorites)
fun <- function(x) {
length(unique(x))
}
## list of length 5, each containsing a sample (with replacement) of letters
lets <- replicate(5, sample(letters, 50, TRUE), simplify=FALSE)
sapply(lets, fun)
## [1] 21 24 22 22 23
Introspection
class()
, str()
dim()
Help
?"print"
: help on the generic print?"print.data.frame"
: help on print method for objects of class
data.frame.help(package="GenomeInfoDb")
browseVignettes("GenomicRanges")
methods("plot")
methods(class="lm")
The following code chunk illustrates R vectors, vectorized
operations, objects (e.g., data.frame()
), formulas, functions,
generics (plot
) and methods (plot.formula
), class and method
discovery (introspection).
x <- rnorm(1000) # atomic vectors
y <- x + rnorm(1000, sd=.5) # vectorized computation
df <- data.frame(x=x, y=y) # object of class 'data.frame'
plot(y ~ x, df) # generic plot, method plot.formula
fit <- lm(y ~x, df) # object of class 'lm'
anova(fit) # see help with ?anova.lm
## Analysis of Variance Table
##
## Response: y
## Df Sum Sq Mean Sq F value Pr(>F)
## x 1 987.59 987.59 4103.4 < 2.2e-16 ***
## Residuals 998 240.19 0.24
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
plot(y ~ x, df) # methods(plot); ?plot.formula
abline(fit, col="red", lwd=3, lty=2) # a function, not generic.method
Use methods()
for introspection 9calss and method discovery), e.g.,
methods(class=class(fit)) # introspection
## [1] add1 alias anova case.names coerce
## [6] confint cooks.distance deviance dfbeta dfbetas
## [11] drop1 dummy.coef effects extractAIC family
## [16] formula hatvalues influence initialize kappa
## [21] labels logLik model.frame model.matrix nobs
## [26] plot predict print proj qr
## [31] residuals rstandard rstudent show simulate
## [36] slotsFromS3 summary variable.names vcov
## see '?methods' for accessing help and source code
Programming example – group 1000 gene SYMBOLs into GO identifiers
The file ‘symgo.csv’ is from an Excel spreadsheet (exported as ‘csv’
– comma-separated value – format) with four columns – the gene
‘SYMBOL’ (e.g., SOX17
), the gene ontology (GO) term(s) that the
symbol has been associated with, and additional gene ontology information.
## example data
fl <- file.choose() ## symgo.csv
symgo <- read.csv(fl, row.names=1, stringsAsFactors=FALSE)
head(symgo)
## SYMBOL GO EVIDENCE ONTOLOGY
## 1 PPIAP28 <NA> <NA> <NA>
## 2 PTLAH <NA> <NA> <NA>
## 3 HIST1H2BC GO:0000786 NAS CC
## 4 HIST1H2BC GO:0000788 IBA CC
## 5 HIST1H2BC GO:0002227 IDA BP
## 6 HIST1H2BC GO:0003677 IBA MF
dim(symgo)
## [1] 5041 4
length(unique(symgo$SYMBOL))
## [1] 1000
head(symgo[symgo$SYMBOL == "SOX17",])
## SYMBOL GO EVIDENCE ONTOLOGY
## 4576 SOX17 GO:0000122 IEA BP
## 4577 SOX17 GO:0001525 ISS BP
## 4578 SOX17 GO:0001570 ISS BP
## 4579 SOX17 GO:0001706 IDA BP
## 4580 SOX17 GO:0001828 IEA BP
## 4581 SOX17 GO:0001947 ISS BP
How many gene SYMBOLs are associated with each GO term? There are several ways to calculate this…
## split + length
go2sym <- split(symgo$SYMBOL, symgo$GO)
len1 <- lengths(go2sym)
head(len1)
## GO:0000049 GO:0000050 GO:0000060 GO:0000077 GO:0000086 GO:0000118
## 3 2 1 1 3 1
## smarter built-in functions, e.g., omiting NAs
len2 <- aggregate(SYMBOL ~ GO, symgo, length)
head(len1)
## GO:0000049 GO:0000050 GO:0000060 GO:0000077 GO:0000086 GO:0000118
## 3 2 1 1 3 1
In aggregate()
, the third argument is FUN
. The value of FUN
is
the function that is applied to each group defined by the formula of
the first argument. Provide a ‘custom’ function that uses the unique
lower-case values
## your own function -- unique, lower-case identifiers
uidfun <- function(x)
unique(tolower(x))
This illustrates how one is not restricted to ‘built-in’ solutions for solving biological problems.
head(aggregate(SYMBOL ~ GO , symgo, uidfun))
## GO SYMBOL
## 1 GO:0000049 yars2, eef1a1
## 2 GO:0000050 asl
## 3 GO:0000060 oprd1
## 4 GO:0000077 pea15
## 5 GO:0000086 tubb4a, cenpf, clasp1
## 6 GO:0000118 cir1
These case studies serve as refreshers on R input and manipulation of data.
Input a file that contains ALL (acute lymphoblastic leukemia) patient information
fname <- file.choose() ## "ALLphenoData.tsv"
stopifnot(file.exists(fname))
pdata <- read.delim(fname)
Check out the help page ?read.delim
for input options, and explore
basic properties of the object you’ve created, for instance…
class(pdata)
## [1] "data.frame"
colnames(pdata)
## [1] "id" "diagnosis" "sex" "age"
## [5] "BT" "remission" "CR" "date.cr"
## [9] "t.4.11." "t.9.22." "cyto.normal" "citog"
## [13] "mol.biol" "fusion.protein" "mdr" "kinet"
## [17] "ccr" "relapse" "transplant" "f.u"
## [21] "date.last.seen"
dim(pdata)
## [1] 127 21
head(pdata)
## id diagnosis sex age BT remission CR date.cr t.4.11. t.9.22. cyto.normal
## 1 1005 5/21/1997 M 53 B2 CR CR 8/6/1997 FALSE TRUE FALSE
## 2 1010 3/29/2000 M 19 B2 CR CR 6/27/2000 FALSE FALSE FALSE
## 3 3002 6/24/1998 F 52 B4 CR CR 8/17/1998 NA NA NA
## 4 4006 7/17/1997 M 38 B1 CR CR 9/8/1997 TRUE FALSE FALSE
## 5 4007 7/22/1997 M 57 B2 CR CR 9/17/1997 FALSE FALSE FALSE
## 6 4008 7/30/1997 M 17 B1 CR CR 9/27/1997 FALSE FALSE FALSE
## citog mol.biol fusion.protein mdr kinet ccr relapse transplant
## 1 t(9;22) BCR/ABL p210 NEG dyploid FALSE FALSE TRUE
## 2 simple alt. NEG <NA> POS dyploid FALSE TRUE FALSE
## 3 <NA> BCR/ABL p190 NEG dyploid FALSE TRUE FALSE
## 4 t(4;11) ALL1/AF4 <NA> NEG dyploid FALSE TRUE FALSE
## 5 del(6q) NEG <NA> NEG dyploid FALSE TRUE FALSE
## 6 complex alt. NEG <NA> NEG hyperd. FALSE TRUE FALSE
## f.u date.last.seen
## 1 BMT / DEATH IN CR <NA>
## 2 REL 8/28/2000
## 3 REL 10/15/1999
## 4 REL 1/23/1998
## 5 REL 11/4/1997
## 6 REL 12/15/1997
summary(pdata$sex)
## Length Class Mode
## 127 character character
summary(pdata$cyto.normal)
## Mode FALSE TRUE NA's
## logical 69 24 34
Remind yourselves about various ways to subset and access columns of a data.frame
pdata[1:5, 3:4]
## sex age
## 1 M 53
## 2 M 19
## 3 F 52
## 4 M 38
## 5 M 57
pdata[1:5, ]
## id diagnosis sex age BT remission CR date.cr t.4.11. t.9.22. cyto.normal
## 1 1005 5/21/1997 M 53 B2 CR CR 8/6/1997 FALSE TRUE FALSE
## 2 1010 3/29/2000 M 19 B2 CR CR 6/27/2000 FALSE FALSE FALSE
## 3 3002 6/24/1998 F 52 B4 CR CR 8/17/1998 NA NA NA
## 4 4006 7/17/1997 M 38 B1 CR CR 9/8/1997 TRUE FALSE FALSE
## 5 4007 7/22/1997 M 57 B2 CR CR 9/17/1997 FALSE FALSE FALSE
## citog mol.biol fusion.protein mdr kinet ccr relapse transplant
## 1 t(9;22) BCR/ABL p210 NEG dyploid FALSE FALSE TRUE
## 2 simple alt. NEG <NA> POS dyploid FALSE TRUE FALSE
## 3 <NA> BCR/ABL p190 NEG dyploid FALSE TRUE FALSE
## 4 t(4;11) ALL1/AF4 <NA> NEG dyploid FALSE TRUE FALSE
## 5 del(6q) NEG <NA> NEG dyploid FALSE TRUE FALSE
## f.u date.last.seen
## 1 BMT / DEATH IN CR <NA>
## 2 REL 8/28/2000
## 3 REL 10/15/1999
## 4 REL 1/23/1998
## 5 REL 11/4/1997
head(pdata[, 3:5])
## sex age BT
## 1 M 53 B2
## 2 M 19 B2
## 3 F 52 B4
## 4 M 38 B1
## 5 M 57 B2
## 6 M 17 B1
tail(pdata[, 3:5], 3)
## sex age BT
## 125 M 19 T2
## 126 M 30 T3
## 127 M 29 T2
head(pdata$age)
## [1] 53 19 52 38 57 17
head(pdata$sex)
## [1] "M" "M" "F" "M" "M" "M"
head(pdata[pdata$age > 21,])
## id diagnosis sex age BT remission CR date.cr t.4.11. t.9.22. cyto.normal
## 1 1005 5/21/1997 M 53 B2 CR CR 8/6/1997 FALSE TRUE FALSE
## 3 3002 6/24/1998 F 52 B4 CR CR 8/17/1998 NA NA NA
## 4 4006 7/17/1997 M 38 B1 CR CR 9/8/1997 TRUE FALSE FALSE
## 5 4007 7/22/1997 M 57 B2 CR CR 9/17/1997 FALSE FALSE FALSE
## 10 8001 1/15/1997 M 40 B2 CR CR 3/26/1997 FALSE FALSE FALSE
## 11 8011 8/21/1998 M 33 B3 CR CR 10/8/1998 FALSE FALSE FALSE
## citog mol.biol fusion.protein mdr kinet ccr relapse transplant
## 1 t(9;22) BCR/ABL p210 NEG dyploid FALSE FALSE TRUE
## 3 <NA> BCR/ABL p190 NEG dyploid FALSE TRUE FALSE
## 4 t(4;11) ALL1/AF4 <NA> NEG dyploid FALSE TRUE FALSE
## 5 del(6q) NEG <NA> NEG dyploid FALSE TRUE FALSE
## 10 del(p15) BCR/ABL p190 NEG <NA> FALSE TRUE FALSE
## 11 del(p15/p16) BCR/ABL p190/p210 NEG dyploid FALSE FALSE TRUE
## f.u date.last.seen
## 1 BMT / DEATH IN CR <NA>
## 3 REL 10/15/1999
## 4 REL 1/23/1998
## 5 REL 11/4/1997
## 10 REL 7/11/1997
## 11 BMT / DEATH IN CR <NA>
It seems from below that there are 17 females over 40 in the data set,
but when sub-setting pdata
to contain just those individuals 19 rows
are selected. Why? What can we do to correct this?
idx <- pdata$sex == "F" & pdata$age > 40
table(idx)
## idx
## FALSE TRUE
## 108 17
dim(pdata[idx,])
## [1] 19 21
Use the mol.biol
column to subset the data to contain just
individuals with ‘BCR/ABL’ or ‘NEG’, e.g.,
bcrabl <- pdata[pdata$mol.biol %in% c("BCR/ABL", "NEG"),]
The mol.biol
column is a factor, and retains all levels even after
subsetting. How might you drop the unused factor levels?
bcrabl$mol.biol <- factor(bcrabl$mol.biol)
The BT
column is a factor describing B- and T-cell subtypes
levels(bcrabl$BT)
## NULL
How might one collapse B1, B2, … to a single type B, and likewise for T1, T2, …, so there are only two subtypes, B and T
table(bcrabl$BT)
##
## B B1 B2 B3 B4 T T1 T2 T3 T4
## 4 9 35 22 9 4 1 15 9 2
levels(bcrabl$BT) <- substring(levels(bcrabl$BT), 1, 1)
table(bcrabl$BT)
##
## B B1 B2 B3 B4 T T1 T2 T3 T4
## 4 9 35 22 9 4 1 15 9 2
Use xtabs()
(cross-tabulation) to count the number of samples with
B- and T-cell types in each of the BCR/ABL and NEG groups
xtabs(~ BT + mol.biol, bcrabl)
## mol.biol
## BT BCR/ABL NEG
## B 2 2
## B1 1 8
## B2 19 16
## B3 8 14
## B4 7 2
## T 0 4
## T1 0 1
## T2 0 15
## T3 0 9
## T4 0 2
Use aggregate()
to calculate the average age of males and females in
the BCR/ABL and NEG treatment groups.
aggregate(age ~ mol.biol + sex, bcrabl, mean)
## mol.biol sex age
## 1 BCR/ABL F 39.93750
## 2 NEG F 30.42105
## 3 BCR/ABL M 40.50000
## 4 NEG M 27.21154
Use t.test()
to compare the age of individuals in the BCR/ABL versus
NEG groups; visualize the results using boxplot()
. In both cases,
use the formula
interface. Consult the help page ?t.test
and re-do
the test assuming that variance of ages in the two groups is
identical. What parts of the test output change?
t.test(age ~ mol.biol, bcrabl)
##
## Welch Two Sample t-test
##
## data: age by mol.biol
## t = 4.8172, df = 68.529, p-value = 8.401e-06
## alternative hypothesis: true difference in means between group BCR/ABL and group NEG is not equal to 0
## 95 percent confidence interval:
## 7.13507 17.22408
## sample estimates:
## mean in group BCR/ABL mean in group NEG
## 40.25000 28.07042
boxplot(age ~ mol.biol, bcrabl)
This case study is a second walk through basic data manipulation and visualization skills. We use data from the US Center for Disease Control’s Behavioral Risk Factor Surveillance System (BRFSS) annual survey. Check out the web page for a little more information. We are using a small subset of this data, including a random sample of 10000 observations from each of 1990 and 2010.
Input the data using read.csv()
, creating a variable brfss
to hold
it. Use file.choose()
to locate the data file BRFSS-subset.csv
fname <- file.choose() ## BRFSS-subset.csv
stopifnot(file.exists(fname))
brfss <- read.csv(fname)
Base plotting functions
Explore the data using class()
, dim()
, head()
, summary()
,
etc. Use xtabs()
to summarize the number of males and females in
the study, in each of the two years.
Use aggregate()
to summarize the average weight in each sex and
year.
Create a scatterplot showing the relationship between the square
root of weight and height, using the plot()
function and the
main
argument to annotate the plot. Note the transformed
Y-axis. Experiment with different plotting symbols (try the command
example(points)
to view different points).
plot(sqrt(Weight) ~ Height, brfss, main="All Years, Both Sexes")
Color the female and male points differently. To do this, use the
col
argument to plot()
. Provide as a value to that argument a
vector of colors, subset by brfss$Sex
.
Create a subset of the data containing only observations from
brfss2010 <- brfss[brfss$Year == "2010", ]
Create the figure below (two panels in a single figure). Do this by
using the par()
function with the mfcol
argument before calling
plot()
. You’ll need to create two more subsets of data, perhaps
when you are providing the data to the function plot
.
opar <- par(mfcol=c(1, 2))
plot(sqrt(Weight) ~ Height, brfss2010[brfss2010$Sex == "Female", ],
main="2010, Female")
plot(sqrt(Weight) ~ Height, brfss2010[brfss2010$Sex == "Male", ],
main="2010, Male")
par(opar) # reset 'par' to original value
Plotting large numbers of points means that they are often
over-plotted, potentially obscuring important patterns. Experiment
with arguments to plot()
to address over-plotting, e.g.,
pch='.'
or alpha=.4
. Try using the smoothScatter()
function
(the data have to be presented as x
and y
, rather than as a
formula). Try adding the hexbin library to your R session
(using library()
) and creating a hexbinplot()
.
ggplot2 graphics
Create a scatterplot showing the relationship between the square root of weight and height, using the ggplot2 library, and the annotate the plot. Two equivalent ways to create the plot are show in the solution.
library(ggplot2)
## 'quick' plot
qplot(Height, sqrt(Weight), data=brfss)
## Warning: `qplot()` was deprecated in ggplot2 3.4.0.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was
## generated.
## Warning: Removed 735 rows containing missing values or values outside the scale range
## (`geom_point()`).
## specify the data set and 'aesthetics', then how to plot
ggplot(brfss, aes(x=Height, y=sqrt(Weight))) +
geom_point()
## Warning: Removed 735 rows containing missing values or values outside the scale range
## (`geom_point()`).
qplot()
gives us a warning which states that it has removed rows
containing missing values. This is actually very helpful because we
find out that our dataset contains NA
’s and we can take a design
decision here about what we’d like to do these NA
’s. We can find
the indicies of the rows containing NA
using is.na()
, and count
the number of rows with NA
values using sum()
:
sum(is.na(brfss$Height))
## [1] 184
sum(is.na(brfss$Weight))
## [1] 649
drop <- is.na(brfss$Height) | is.na(brfss$Weight)
sum(drop)
## [1] 735
Remove the rows which contain NA
’s in Height and Weight.
brfss <- brfss[!drop,]
Plot is annotated with
qplot(Height, sqrt(Weight), data=brfss) +
ylab("Square root of Weight") +
ggtitle("All Years, Both Sexes")
Color the female and male points differently.
ggplot(brfss, aes(x=Height, y=sqrt(Weight), color=Sex)) +
geom_point()
One can also change the shape of the points for the female and male groups
ggplot(brfss, aes(x=Height, y = sqrt(Weight), color=Sex, shape=Sex)) +
geom_point()
or plot Male and Female in different panels using facet_grid()
ggplot(brfss, aes(x=Height, y = sqrt(Weight), color=Sex)) +
geom_point() +
facet_grid(Sex ~ .)
Create a subset of the data containing only observations from 2010
and make density curves for male and female groups. Use the fill
aesthetic to indicate that each sex is to be calculated separately,
and geom_density()
for the density plot.
brfss2010 <- brfss[brfss$Year == "2010", ]
ggplot(brfss2010, aes(x=sqrt(Weight), fill=Sex)) +
geom_density(alpha=.25)
Plotting large numbers of points means that they are often over-plotted, potentially obscuring important patterns. Make the points semi-transparent using alpha. Here we make them 60% transparent. The solution illustrates a nice feature of ggplot2 – a partially specified plot can be assigned to a variable, and the variable modified at a later point.
sp <- ggplot(brfss, aes(x=Height, y=sqrt(Weight)))
sp + geom_point(alpha=.4)
Add a fitted regression model to the scatter plot.
sp + geom_point() + stat_smooth(method=lm)
## `geom_smooth()` using formula = 'y ~ x'
By default, stat_smooth()
also adds a 95% confidence region for
the regression fit. The confidence interval can be changed by
setting level, or it can be disabled with se=FALSE
.
sp + geom_point() + stat_smooth(method=lm + level=0.95)
sp + geom_point() + stat_smooth(method=lm, se=FALSE)
How do you fit a linear regression line for each group? First we’ll make the base plot object sps, then we’ll add the linear regression lines to it.
sps <- ggplot(brfss, aes(x=Height, y=sqrt(Weight), colour=Sex)) +
geom_point() +
scale_colour_brewer(palette="Set1")
sps + geom_smooth(method="lm")
## `geom_smooth()` using formula = 'y ~ x'
sessionInfo()
## R version 4.4.0 (2024-04-24)
## Platform: x86_64-apple-darwin20
## Running under: macOS Sonoma 14.5
##
## Matrix products: default
## BLAS: /Library/Frameworks/R.framework/Versions/4.4-x86_64/Resources/lib/libRblas.0.dylib
## LAPACK: /Library/Frameworks/R.framework/Versions/4.4-x86_64/Resources/lib/libRlapack.dylib; LAPACK version 3.12.0
##
## locale:
## [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
##
## time zone: America/New_York
## tzcode source: internal
##
## attached base packages:
## [1] stats graphics grDevices utils datasets methods base
##
## other attached packages:
## [1] ggplot2_3.5.1 BiocStyle_2.32.0
##
## loaded via a namespace (and not attached):
## [1] Matrix_1.7-0 gtable_0.3.5 jsonlite_1.8.8
## [4] crayon_1.5.2 dplyr_1.1.4 compiler_4.4.0
## [7] BiocManager_1.30.23 highr_0.11 tidyselect_1.2.1
## [10] tinytex_0.51 jquerylib_0.1.4 splines_4.4.0
## [13] scales_1.3.0 yaml_2.3.8 fastmap_1.2.0
## [16] lattice_0.22-6 R6_2.5.1 labeling_0.4.3
## [19] generics_0.1.3 knitr_1.47 tibble_3.2.1
## [22] bookdown_0.39 munsell_0.5.1 RColorBrewer_1.1-3
## [25] bslib_0.7.0 pillar_1.9.0 rlang_1.1.4
## [28] utf8_1.2.4 cachem_1.1.0 xfun_0.44
## [31] sass_0.4.9 cli_3.6.2 mgcv_1.9-1
## [34] withr_3.0.0 magrittr_2.0.3 digest_0.6.35
## [37] grid_4.4.0 rstudioapi_0.16.0 nlme_3.1-165
## [40] lifecycle_1.0.4 vctrs_0.6.5 evaluate_0.23
## [43] glue_1.7.0 farver_2.1.2 codetools_0.2-20
## [46] fansi_1.0.6 colorspace_2.1-0 rmarkdown_2.27
## [49] tools_4.4.0 pkgconfig_2.0.3 htmltools_0.5.8.1
Research reported in this tutorial was supported by the National Human Genome Research Institute and the National Cancer Institute of the National Institutes of Health under award numbers U24HG004059 (Bioconductor), U24HG010263 (AnVIL) and U24CA180996 (ITCR).