# plots with a blue color
plot(days, area, type="b", col="blue")
# plots with filled circles
plot(days, area, type="b", col="blue", pch=19)
# plots with labels
plot(days, area, type="b",
xlab="Day", ylab="Area", main="Area of water lilies")
# limits on the y axis and x axis
plot(days, area, type="b",
xlim=c(0, 8), ylim=c(0, 1))
Formula for area when the growth is by 75% per day:
\[a(n) = 0.01 \times 1.75^{n-1}\]
area_slow <- 0.01 * 1.75^(days - 1)
plot(days, area, type="b", col="blue")
# add another data series to the plot
lines(days, area_slow, type="b", col="red")
# generate random readouts
readouts <- runif(48)
# create the matrix
res_mtx <- matrix(readouts, nrow=6)
# always check your results!
dim(res_mtx)[1] 6 8To change the “borders” to NA:
res_mtx[1,] <- NA
res_mtx[,1] <- NA
res_mtx[nrow(res_mtx), ] <- NA
res_mtx[, ncol(res_mtx)] <- NAAbove, we could have used res_mtx[6, ] <- NA instead of res_mtx[nrow(res_mtx), ] <- NA, but the latter is more general and will keep working if you decide to use a matrix with a different layout.
Row and column names:
row_names <- c("control", "low", "medium", "high")
row_names <- paste0("Inh1_", row_names)
row_names <- c(NA, row_names, NA)
rownames(res_mtx) <- row_names
col_names <- c("control", "low", "high")
col_names <- rep(col_names, 2)
inh <- rep(c("Inh2_", "Inh3_"), each=3)
col_names <- paste0(inh, col_names)
col_names <- c(NA, col_names, NA)
colnames(res_mtx) <- col_namesAbove, we use NA for row names and column names of the border row / column. We could have used another value as well, it doesn’t matter.
Selecting from the matrix:
# wells with inhibitor 3
res_mtx[ 2:5, 5:6 ] Inh3_control Inh3_low
Inh1_control 0.71422992 0.6657274
Inh1_low 0.52711822 0.6521810
Inh1_medium 0.40945933 0.8049797
Inh1_high 0.07877701 0.4460096# inhibitor 1 as well as 2
inh1 <- c("Inh1_low", "Inh1_medium", "Inh1_high")
inh2 <- c("Inh2_low", "Inh2_high")
res_mtx[ inh1, inh2 ] Inh2_low Inh2_high
Inh1_low 0.9786829 0.6754574
Inh1_medium 0.7269673 0.4473528
Inh1_high 0.1366912 0.1345950mtx <- matrix(rnorm(15), nrow=5)
df <- as.data.frame(mtx)
colnames(df) <- c("X", "Y", "Z")
rownames(df) <- c("a", "b", "c", "d", "e")
df$A <- rep("A", 5)
df$sequence <- seq(0, 1, length.out=5)
df X Y Z A sequence
a -1.90635742 -0.596422448 0.7761556 A 0.00
b 0.75975917 0.161859283 -0.9482953 A 0.25
c 0.06380527 0.000362048 -1.1097728 A 0.50
d -1.67180221 -1.010560011 -2.3550369 A 0.75
e 0.53106772 0.684821178 -0.5697360 A 1.00If you look at the file, you will find that there are 4 lines with gibberish (which you can skip with skip=4), then the header row, then 10 lines with actual data, and finally a few lines with gibberish again. Thus, we need to read 1 + 10 = 11 lines, starting with line 5, and finishing with line 16.
fn <- readxl_example("deaths.xls")
# one way: use the n_max argument
deaths <- read_excel(fn, skip=4, n_max=11)
# another way: use the excel range specification
deaths <- read_excel(fn, range="A5:F16")meta_data_botched.xlsx (Exercise 3.7)botched <- read_excel("Datasets/meta_data_botched.xlsx")
colnames(botched)[1] "SUBJ" "time point" "AGE" "SEX" "PLACEBO"
[6] "ARM" First thing that we see is that one of the columns is called time point with a space in between. This is neither conforming to the other column names (which are ALL CAPS) nor it is a good idea to have spaces in column names.
summary(botched) SUBJ time point AGE SEX
Min. :4023 Length:122 Length:122 Length:122
1st Qu.:4232 Class :character Class :character Class :character
Median :4429 Mode :character Mode :character Mode :character
Mean :4473
3rd Qu.:4701
Max. :4973
PLACEBO ARM
Length:122 Length:122
Class :character Class :character
Mode :character Mode :character
library(skimr)
skim(botched)| Name | botched |
| Number of rows | 122 |
| Number of columns | 6 |
| _______________________ | |
| Column type frequency: | |
| character | 5 |
| numeric | 1 |
| ________________________ | |
| Group variables | None |
Variable type: character
| skim_variable | n_missing | complete_rate | min | max | empty | n_unique | whitespace |
|---|---|---|---|---|---|---|---|
| time point | 0 | 1 | 2 | 5 | 0 | 8 | 0 |
| AGE | 0 | 1 | 2 | 6 | 0 | 34 | 0 |
| SEX | 0 | 1 | 1 | 8 | 0 | 12 | 0 |
| PLACEBO | 0 | 1 | 1 | 5 | 0 | 10 | 0 |
| ARM | 0 | 1 | 1 | 16 | 0 | 20 | 0 |
Variable type: numeric
| skim_variable | n_missing | complete_rate | mean | sd | p0 | p25 | p50 | p75 | p100 | hist |
|---|---|---|---|---|---|---|---|---|---|---|
| SUBJ | 0 | 1 | 4472.51 | 287.12 | 4023 | 4232 | 4429 | 4701 | 4973 | ▇▇▇▅▆ |
OK, so we see several problems. First, the column “Age” is not numeric, but we would expect it to be. Let’s take a closer look:
age_n <- as.numeric(botched$AGE)Warning: NAs introduced by coercionbotched$AGE[ is.na(age_n) ] [1] "32 ," "23 yrs" "28 ," "30 ," "3 7" "23 yrs" "32 yrs" "40 yrs"
[9] "35 yrs" "22 ," "30 ," Right, so someone added extra text or spaces to the the actual values. And what does “3 0” mean? It could be 30, but it also could mean “3 years 0 months” – who knows?
Next, the SEX column. skim() shows that it has 12 unique values, but this is not the number we expect. What are these unique values?
unique(botched$SEX) [1] "Mann" "M" "männlich" "F" "female" "Female"
[7] "Herr" "Male" "male" "Frau" "Femal" "Weiblich"Maybe the data was entered by hand by different people, and they did not agree beforehand on how to enter the data. We see a similar problem also in the columns PLACEBO, ARM and time point:
unique(botched$PLACEBO) [1] "TRUE" "yes" "T" "1" "true" "FALSE" "false" "0" "no"
[10] "F" unique(botched$ARM) [1] "Placebo" "placebo" "P" "Plazebo"
[5] "control" "Placibo" "F" "fluad vaccine"
[9] "A" "Fl." "agrippal vaccine" "agrippal"
[13] "Agrippal vaccine" "grippal" "AGRP." "Fuad"
[17] "Fluad" "Agrippal" "FLUAD" "Fluad vaccine" unique(botched[["time point"]])[1] "D0" "D1" "Day1" "D 0" "day 1" "day 0" "D 1" "Day0" meta_data_botched.xlsx (Exercise 3.13)Let’s start with the easiest one, the time point. First, we should fix the name of the column:
# either works
botched <- rename(botched, TIMEPOINT=`time point`)
# or
colnames(botched)[2] <- "TIMEPOINT"
colnames(botched)[1] "SUBJ" "TIMEPOINT" "AGE" "SEX" "PLACEBO" "ARM" Now, fix the values in the TIMEPOINT column. We can use the toupper().
# change to upper case
tp <- toupper(botched$TIMEPOINT)
# replace day with "D"
tp <- str_replace_all(tp, "DAY", "D")
# remove all spaces
tp <- str_replace_all(tp, " *", "")
table(tp, botched$TIMEPOINT)
tp D 0 D 1 D0 D1 day 0 day 1 Day0 Day1
D0 5 0 52 0 3 0 1 0
D1 0 5 0 50 0 2 0 4# looks good!
botched$TIMEPOINT <- tpThe ARM column should be easy, too. We only have to pay attention to the first letter. There are two exceptions, though – sometimes control is used instead of placebo, and in some cases grippal has been used instead of agrippal.
arm <- toupper(botched$ARM)
arm <- str_replace_all(arm, "^P.*", "PLACEBO")
arm <- str_replace_all(arm, "^F.*", "FLUAD")
arm <- str_replace_all(arm, "^A.*", "AGRIPPAL")
arm <- str_replace_all(arm, "^C.*", "PLACEBO")
arm <- str_replace_all(arm, "^G.*", "AGRIPPAL")
unique(arm)[1] "PLACEBO" "FLUAD" "AGRIPPAL"botched$ARM <- armProceed with the columns SEX and PLACEBO in the same way.
Finally, AGE. The simplest way would be to replace everything that is not a digit:
age <- botched$AGE
age <- str_replace_all(age, "[^0-9]", "")
age <- as.numeric(age)
# check for NAs
any(is.na(age))[1] FALSE# replace the original column
botched$AGE <- ageSolution to Exercise 4.1.
# Load the data
library(tidyverse)
results <- read_csv("Datasets/transcriptomics_results.csv")
# what columns are there?
colnames(results) [1] "...1" "ProbeName" "GeneName" "SystematicName"
[5] "Description" "logFC.F.D0" "qval.F.D0" "logFC.F.D1"
[9] "qval.F.D1" "logFC.F.D2" "qval.F.D2" "logFC.F.D3"
[13] "qval.F.D3" # Select the columns that we need
results <- select(results, GeneName, Description, logFC.F.D1, qval.F.D1)
colnames(results) <- c("Gene", "Description", "logFC", "qval")Alternatively, we could have specified the new column names directly in the select() function. This is useful if you want to rename only some of the columns:
results <- select(results, Gene=GeneName,
Description,
LFC=logFC.F.D1, FDR=qval.F.D1)
colnames(results)[1] "Gene" "Description" "LFC" "FDR" persons <- c("Henry Fonda", "Bob Marley", "Robert F. Kennedy",
"Bob Dylan", "Alan Rickman")
# first, create a vector with last names only.
# basically, remove everything before the last space
# and the space itself
lastnames <- str_replace_all(persons, ".* ", "")
lastnames[1] "Fonda" "Marley" "Kennedy" "Dylan" "Rickman"# now get the order for the last names
ord <- order(lastnames)
ord[1] 4 1 3 2 5# and use it to sort the original vector
persons[ord][1] "Bob Dylan" "Henry Fonda" "Robert F. Kennedy"
[4] "Bob Marley" "Alan Rickman" significant <- tr_res$FDR < 0.05
interferons <- str_detect(tr_res$Description, "interferon")
both <- significant & interferons
# how many are there?
sum(both)[1] 23# how many are significant, but not interferons?
sum(significant & !interferons)[1] 1455gbps <- str_detect(tr_res$Gene, "^GBP")
sum(gbps & interferons)[1] 4Solution to Exercise 4.9.
First, we load the data. Nothing fancy here. You should examine the resulting data frames with View (or by clicking on the data frame in the Environment pane in RStudio), but in the code below we simply list the available columns.
# The libraries needed
library(tidyverse)
library(readxl)
# Load the data
labresults <- read_csv("Datasets/labresults_full.csv")
targets <- read_excel("Datasets/expression_data_vaccination_example.xlsx",
sheet="targets")
# what columns are there?
colnames(labresults) [1] "USUBJID" "SUBJ" "LBTEST" "LBTESTCD" "LBCAT" "LBORRES"
[7] "LBORRESU" "LBSPEC" "VISIT" "Timepoint"colnames(targets) [1] "USUBJID" "SUBJ" "Batch" "ID" "Timepoint" "ARMCD"
[7] "ARM" "AGE" "SEX" "PLACEBO" intersect(colnames(labresults), colnames(targets))[1] "USUBJID" "SUBJ" "Timepoint"intersectThe intersect() function returns the common elements between two vectors. In this case, it returns the common column names between the two data frames. As you can see, there are USUBJID and SUBJ columns in both data frames, chances are that there are common elements between the data frame here. We will focus at the SUBJ column.
OK, do we see any common subjects between the two data frames?
# how many unique subjects are there in each data frame?
length(unique(labresults$SUBJ))[1] 123length(unique(targets$SUBJ))[1] 61# how many are common?
length(intersect(labresults$SUBJ, targets$SUBJ))[1] 61Right, so there are more unique subjects in the labresults data frame than in the targets data frame.
The other column that is common between the two data frames is Timepoint. That already indicates that each row – in both data frames – corresponds to a sample rather than subject. That is, we need to identify the rows not only by subject, but also by timepoint.
If you use the unique() function as above, you will find that there are only 2 unique timepoints in the targets data frame, but 23 unique timepoints in the labresults data frame.
Now first let us select only the column that we need, as mentioned in the exercise. For the targets data frame, we need the columns SUBJ and Timepoint, as well as ARM, Timepoint, AGE and SEX.
targets <- select(targets, SUBJ, Timepoint, ARM, AGE, SEX)If you inspect the labresults data frame, you will see that it has only one column that looks like a measurement – LBORRES. Also, the column LBTEST ostensibly contains the name of the test that was performed, and LBTESTCD contains the code for the test. However, we also find the Timepoint column here as well.
labresults <- select(labresults, SUBJ, Timepoint, LBTEST, LBTESTCD, LBORRES)Depending on what one wants to do, we can try to get one of the four types of join (inner, left, right, full) between the two data frames. However, assuming that the goal is correlate expression data with lab data, we will need an inner join, so either the tidyverse function inner_join() or the base R function merge() with default parameter values will do the job.
joined <- merge(targets, labresults, by=c("SUBJ", "Timepoint"))
dim(joined)[1] 3782 8colnames(joined)[1] "SUBJ" "Timepoint" "ARM" "AGE" "SEX" "LBTEST"
[7] "LBTESTCD" "LBORRES" Yep, seems to have worked well. The resulting data frame has 3782 rows – since we have many different measurements for each person.