How to very efficiently extract specific pattern from characters? - regex

I have big data like this :
> Data[1:7,1]
[1] mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5
[2] mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9
[3] mature=hsa-miR-448|mir_Family=mir-448|Gene=OR4F5
[4] mature=hsa-miR-659-3p|mir_Family=-|Gene=OR4F5
[5] mature=hsa-miR-5197-3p|mir_Family=-|Gene=OR4F5
[6] mature=hsa-miR-5093|mir_Family=-|Gene=OR4F5
[7] mature=hsa-miR-650|mir_Family=mir-650|Gene=OR4F5
what I want to do is that, in every row, I want to select the name after word mature= and also the word after Gene= and then pater them together with
paste(a,b, sep="-")
for example, the expected output from first two rows would be like :
hsa-miR-5087-OR4F5
hsa-miR-26a-1-3p-OR4F9
so, the final implementation is like this:
for(i in 1:nrow(Data)){
Data[i,3] <- sub("mature=([^|]*).*Gene=(.*)", "\\1-\\2", Data[i,1])
Name <- strsplit(as.vector(Data[i,2]),"\\|")[[1]][2]
Data[i,4] <- as.numeric(sub("pvalue=","",Name))
print(i)
}
which work well, but it's very slow. the size of Data is very big and it has 200,000,000 rows. this implementation is very slow for that. how can I speed it up ?

If you can guarantee that the format is exactly as you specified, then a regular expression can capture (denoted by the brackets below) everything from the equals sign upto the pipe symbol, and from the Gene= to the end, and paste them together with a minus sign:
sub("mature=([^|]*).*Gene=(.*)", "\\1-\\2", Data[,1])

Another option is to use read.table with = as a separator then pasting the 2 columns:
res = read.table(text=txt,sep='=')
paste(sub('[|].*','',res$V2), ## get rid from last part here
sub('^ +| +$','',res$V4),sep='-') ## remove extra spaces
[1] "hsa-miR-5087-OR4F5" "hsa-miR-26a-1-3p-OR4F9" "hsa-miR-448-OR4F5" "hsa-miR-659-3p-OR4F5"
[5] "hsa-miR-5197-3p-OR4F5" "hsa-miR-5093-OR4F5" "hsa-miR-650-OR4F5"

The simple sub solution already given looks quite nice but just in case here are some other approaches:
1) read.pattern Using read.pattern in the gsubfn package we can parse the data into a data.frame. This intermediate form, DF, can then be manipulated in many ways. In this case we use paste in essentially the same way as in the question:
library(gsubfn)
DF <- read.pattern(text = Data[, 1], pattern = "(\\w+)=([^|]*)")
paste(DF$V2, DF$V6, sep = "-")
giving:
[1] "hsa-miR-5087-OR4F5" "hsa-miR-26a-1-3p-OR4F9" "hsa-miR-448-OR4F5"
[4] "hsa-miR-659-3p-OR4F5" "hsa-miR-5197-3p-OR4F5" "hsa-miR-5093-OR4F5"
[7] "hsa-miR-650-OR4F5"
The intermediate data frame, DF, that was produced looks like this:
> DF
V1 V2 V3 V4 V5 V6
1 mature hsa-miR-5087 mir_Family - Gene OR4F5
2 mature hsa-miR-26a-1-3p mir_Family mir-26 Gene OR4F9
3 mature hsa-miR-448 mir_Family mir-448 Gene OR4F5
4 mature hsa-miR-659-3p mir_Family - Gene OR4F5
5 mature hsa-miR-5197-3p mir_Family - Gene OR4F5
6 mature hsa-miR-5093 mir_Family - Gene OR4F5
7 mature hsa-miR-650 mir_Family mir-650 Gene OR4F5
Here is a visualization of the regular expression we used:
(\w+)=([^|]*)
Debuggex Demo
1a) names We could make DF look nicer by reading the three columns of data and the three names separately. This also improves the paste statement:
DF <- read.pattern(text = Data[, 1], pattern = "=([^|]*)")
names(DF) <- unlist(read.pattern(text = Data[1,1], pattern = "(\\w+)=", as.is = TRUE))
paste(DF$mature, DF$Gene, sep = "-") # same answer as above
The DF in this section that was produced looks like this. It has 3 instead of 6 columns and remaining columns were used to determine appropriate column names:
> DF
mature mir_Family Gene
1 hsa-miR-5087 - OR4F5
2 hsa-miR-26a-1-3p mir-26 OR4F9
3 hsa-miR-448 mir-448 OR4F5
4 hsa-miR-659-3p - OR4F5
5 hsa-miR-5197-3p - OR4F5
6 hsa-miR-5093 - OR4F5
7 hsa-miR-650 mir-650 OR4F5
2) strapplyc
Another approach using the same package. This extracts the fields coming after a = and not containing a | producing a list. We then sapply over that list pasting the first and third fields together:
sapply(strapplyc(Data[, 1], "=([^|]*)"), function(x) paste(x[1], x[3], sep = "-"))
giving the same result.
Here is a visualization of the regular expression used:
=([^|]*)
Debuggex Demo

Here is one approach:
Data <- readLines(n = 7)
mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5
mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9
mature=hsa-miR-448|mir_Family=mir-448|Gene=OR4F5
mature=hsa-miR-659-3p|mir_Family=-|Gene=OR4F5
mature=hsa-miR-5197-3p|mir_Family=-|Gene=OR4F5
mature=hsa-miR-5093|mir_Family=-|Gene=OR4F5
mature=hsa-miR-650|mir_Family=mir-650|Gene=OR4F5
df <- read.table(sep = "|", text = Data, stringsAsFactors = FALSE)
l <- lapply(df, strsplit, "=")
trim <- function(x) gsub("^\\s*|\\s*$", "", x)
paste(trim(sapply(l[[1]], "[", 2)), trim(sapply(l[[3]], "[", 2)), sep = "-")
# [1] "hsa-miR-5087-OR4F5" "hsa-miR-26a-1-3p-OR4F9" "hsa-miR-448-OR4F5" "hsa-miR-659-3p-OR4F5" "hsa-miR-5197-3p-OR4F5" "hsa-miR-5093-OR4F5"
# [7] "hsa-miR-650-OR4F5"

Maybe not the more elegant but you can try :
sapply(Data[,1],function(x){
parts<-strsplit(x,"\\|")[[1]]
y<-paste(gsub("(mature=)|(Gene=)","",parts[grepl("mature|Gene",parts)]),collapse="-")
return(y)
})
Example
Data<-data.frame(col1=c("mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5","mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9"),col2=1:2,stringsAsFactors=F)
> Data[,1]
[1] "mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5" "mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9"
> sapply(Data[,1],function(x){
+ parts<-strsplit(x,"\\|")[[1]]
+ y<-paste(gsub("(mature=)|(Gene=)","",parts[grepl("mature|Gene",parts)]),collapse="-")
+ return(y)
+ })
mature=hsa-miR-5087|mir_Family=-|Gene=OR4F5 mature=hsa-miR-26a-1-3p|mir_Family=mir-26|Gene=OR4F9
"hsa-miR-5087-OR4F5" "hsa-miR-26a-1-3p-OR4F9"

Related

Aggregate modis list files by month

I am looking for a more efficient way of separating each year from the time series (2002-2016) by month. I've done it by hand, but it takes a lot.
mypath<-"D:/SNOWL"
myras<-list.files(path=mypath,pattern = glob2rx("*.tif$"),
full.names = TRUE, recursive = TRUE)
> myras
[1] "D:/SNOWL/MOYDSL10A1.A2002001.tif" "D:/SNOWL/MOYDSL10A1.A2002002.tif"
[3] "D:/SNOWL/MOYDSL10A1.A2002003.tif" "D:/SNOWL/MOYDSL10A1.A2002004.tif"
[5] "D:/SNOWL/MOYDSL10A1.A2002005.tif" "D:/SNOWL/MOYDSL10A1.A2002006.tif"
[7] "D:/SNOWL/MOYDSL10A1.A2002007.tif" "D:/SNOWL/MOYDSL10A1.A2002008.tif"
[9] "D:/SNOWL/MOYDSL10A1.A2002009.tif" "D:/SNOWL/MOYDSL10A1.A2002010.tif"
[11] "D:/SNOWL/MOYDSL10A1.A2002011.tif" "D:/SNOWL/MOYDSL10A1.A2002012.tif"
serie<-orgTime(myras, nDays = "asIn", begin ="2002-01-01",end = "2016-12-31", pillow = 75, pos1 = 13, pos2 = 19)
filter<-serie$inputLayerDates
> filter
[1] "2002-01-01" "2002-01-02" "2002-01-03" "2002-01-04" "2002-01-05"
[6] "2002-01-06" "2002-01-07" "2002-01-08" "2002-01-09" "2002-01-10"
[11] "2002-01-11" "2002-01-12" "2002-01-13" "2002-01-14" "2002-01-15"
[16] "2002-01-16" "2002-01-17" "2002-01-18" "2002-01-19" "2002-01-20"
[21] "2002-01-21" "2002-01-22" "2002-01-23" "2002-01-24" "2002-01-25"
[26] "2002-01-26" "2002-01-27" "2002-01-28" "2002-01-29" "2002-01-30"
[31] "2002-01-31" "2002-02-01" "2002-02-02" "2002-02-03" "2002-02-04"
[36] "2002-02-05" "2002-02-07" "2002-02-08" "2002-02-09" "2002-02-10"
[41] "2002-02-11" "2002-02-12" "2002-02-13" "2002-02-14" "2002-02-15"
EDIT:
Ok, let's try a full size example and see if it's working for you:
# Here we generate filenames as returned from `list.files`:
rm(list = ls())
myras <- sapply(1:5465, function(i) paste0('D:/SNOWL/MOYDSL10A1.A',sample(2000:2016,1),sample(c(paste0('00',1:9),paste0('0',10:99),100:365),1),'.tif'))
head(myras)
# Let's extract the timestamps
tstmps <- regmatches(myras,regexpr('[[:digit:]]{7}',myras))
head(tstmps,50)
# And now convert the timestamps to dates
dates <- as.Date(as.numeric(substr(tstmps,5,7)) - 1, origin = paste0(substr(tstmps,1,4),"-01-01"))
head(dates,10)
# Last step is to sort the files by month
#check months
print(month.name)
myras_byM = sapply(month.name,function(x) myras[months(dates) == x])
head(myras_byM$January)
head(myras_byM$February)
head(myras_byM$March)
head(myras_byM$April)
head(myras_byM$May)
head(myras_byM$June)
head(myras_byM$July)
head(myras_byM$August)
head(myras_byM$September)
head(myras_byM$October)
head(myras_byM$November)
head(myras_byM$December)
You can easily get the date from your filename, if you have a consistent naming convention.
In your case, I see the files are ordered by year and day of the year. So just strip the date from the filename, and then you can filter it by whatever you need. To do this I'm using regular expressions. In this case, I'm interested in the date and DOY string, which should always be 7 numbers. The corresponding RE is therefore [[:digit:]]{7}, which means 7 consecutive digits. regexpr finds the matches and regmatches returns them.
dts <- regmatches(myras,regexpr('[[:digit:]]{7}',myras))
Then you just use substring to extract the digits you need (this method assumes it's always 4 digits for year followed by 3 for DOY) and convert it to a date:
dts <-as.Date(as.numeric(substr(dts,5,7)) - 1, origin = paste0(substr(dts,1,4),"-01-01"))
That would give you the variable of filter you have in your example.
If you then want to sort the entire time series by month, you could use sapply or lapply with the built-in names month.name. The base function months will return you the name of the month for a given date:
myras_byMonth <- sapply(month.name,function(x) myras[months(dts) == x])
Hope I understood your question correctly and this was what you were looking for.
Best,
Val

R - extract all strings matching pattern and create relational table

I am looking for a shorter and more pretty solution (possibly in tidyverse) to the following problem. I have a data.frame "data":
id string
1 A 1.001 xxx 123.123
2 B 23,45 lorem ipsum
3 C donald trump
4 D ssss 134, 1,45
What I wanted to do is to extract all numbers (no matter if the delimiter is "." or "," -> in this case I assume that string "134, 1,45" can be extracted into two numbers: 134 and 1.45) and create a data.frame "output" looking similar to this:
id string
1 A 1.001
2 A 123.123
3 B 23.45
4 C <NA>
5 D 134
6 D 1.45
I managed to do this (code below) but the solution is pretty ugly for me also not so efficient (two for-loops). Could someone suggest a better way to do do this (preferably using dplyr)
# data
data <- data.frame(id = c("A", "B", "C", "D"),
string = c("1.001 xxx 123.123",
"23,45 lorem ipsum",
"donald trump",
"ssss 134, 1,45"),
stringsAsFactors = FALSE)
# creating empty data.frame
len <- length(unlist(sapply(data$string, function(x) gregexpr("[0-9]+[,|.]?[0-9]*", x))))
output <- data.frame(id = rep(NA, len), string = rep(NA, len))
# main solution
start = 0
for(i in 1:dim(data)[1]){
tmp_len <- length(unlist(gregexpr("[0-9]+[,|.]?[0-9]*", data$string[i])))
for(j in (start+1):(start+tmp_len)){
output[j,1] <- data$id[i]
output[j,2] <- regmatches(data$string[i], gregexpr("[0-9]+[,|.]?[0-9]*", data$string[i]))[[1]][j-start]
}
start = start + tmp_len
}
# further modifications
output$string <- gsub(",", ".", output$string)
output$string <- as.numeric(ifelse(substring(output$string, nchar(output$string), nchar(output$string)) == ".",
substring(output$string, 1, nchar(output$string) - 1),
output$string))
output
1) Base R This uses relatively simple regular expressions and no packages.
In the first 2 lines of code replace any comma followed by a space with a
space and then replace all remaining commas with a dot. After these two lines s will be: c("1.001 xxx 123.123", "23.45 lorem ipsum", "donald trump", "ssss 134 1.45")
In the next 4 lines of code trim whitespace from beginning and end of each string field and split the string field on whitespace producing a
list. grep out those elements consisting only of digits and dots. (The regular expression ^[0-9.]*$ matches the start of a word followed by zero or more digits or dots followed by the end of the word so only words containing only those characters are matched.) Replace any zero length components with NA. Finally add data$id as the names. After these 4 lines are run the list L will be list(A = c("1.001", "123.123"), B = "23.45", C = NA, D = c("134", "1.45")) .
In the last line of code convert the list L to a data frame with the appropriate names.
s <- gsub(", ", " ", data$string)
s <- gsub(",", ".", s)
L <- strsplit(trimws(s), "\\s+")
L <- lapply(L, grep, pattern = "^[0-9.]*$", value = TRUE)
L <- ifelse(lengths(L), L, NA)
names(L) <- data$id
with(stack(L), data.frame(id = ind, string = values))
giving:
id string
1 A 1.001
2 A 123.123
3 B 23.45
4 C <NA>
5 D 134
6 D 1.45
2) magrittr This variation of (1) writes it as a magrittr pipeline.
library(magrittr)
data %>%
transform(string = gsub(", ", " ", string)) %>%
transform(string = gsub(",", ".", string)) %>%
transform(string = trimws(string)) %>%
with(setNames(strsplit(string, "\\s+"), id)) %>%
lapply(grep, pattern = "^[0-9.]*$", value = TRUE) %>%
replace(lengths(.) == 0, NA) %>%
stack() %>%
with(data.frame(id = ind, string = values))
3) dplyr/tidyr This is an alternate pipeline solution using dplyr and tidyr. unnest converts to long form, id is made factor so that we can later use complete to recover id's that are removed by subsequent filtering, the filter removes junk rows and complete inserts NA rows for each id that would otherwise not appear.
library(dplyr)
library(tidyr)
data %>%
mutate(string = gsub(", ", " ", string)) %>%
mutate(string = gsub(",", ".", string)) %>%
mutate(string = trimws(string)) %>%
mutate(string = strsplit(string, "\\s+")) %>%
unnest() %>%
mutate(id = factor(id))
filter(grepl("^[0-9.]*$", string)) %>%
complete(id)
4) data.table
library(data.table)
DT <- as.data.table(data)
DT[, string := gsub(", ", " ", string)][,
string := gsub(",", ".", string)][,
string := trimws(string)][,
string := setNames(strsplit(string, "\\s+"), id)][,
list(string = list(grep("^[0-9.]*$", unlist(string), value = TRUE))), by = id][,
list(string = if (length(unlist(string))) unlist(string) else NA_character_), by = id]
DT
Update Removed assumption that junk words do not have digit or dot. Also added (2), (3) and (4) and some improvements.
We can replace the , in between the numbers with . (using gsub), extract the numbers with str_extract_all (from stringr into a list), replace the list elements that have length equal to 0 with NA, set the names of the list with 'id' column, stack to convert the list to data.frame and rename the columns.
library(stringr)
setNames(stack(setNames(lapply(str_extract_all(gsub("(?<=[0-9]),(?=[0-9])", ".",
data$string, perl = TRUE), "[0-9.]+"), function(x)
if(length(x)==0) NA else as.numeric(x)), data$id))[2:1], c("id", "string"))
# id string
#1 A 1.001
#2 A 123.123
#3 B 23.45
#4 C NA
#5 D 134
#6 D 1.45
Same idea as Gabor's. I had hoped to use R's built-in parsing of strings (type.convert, used in read.table) rather than writing custom regex substitutions:
sp = setNames(strsplit(data$string, " "), data$id)
spc = lapply(sp, function(x) {
x = x[grep("[^0-9.,]$", x, invert=TRUE)]
if (!length(x))
NA_real_
else
mapply(type.convert, x, dec=gsub("[^.,]", "", x), USE.NAMES=FALSE)
})
setNames(rev(stack(spc)), names(data))
id string
1 A 1.001
2 A 123.123
3 B 23.45
4 C <NA>
5 D 134
6 D 1.45
Unfortunately, type.convert is not robust enough to consider both decimal delimiters at once, so we need this mapply malarkey instead of type.convert(x, dec = "[.,]").

Use lapply to plot data in a list and use names of list elements as plot titles [duplicate]

This question already has an answer here:
Adding lists names as plot titles in lapply call in R
(1 answer)
Closed 7 years ago.
If I have the following list:
comp.surv <- list(a = 1:4, b = c(1, 2, 4, 8), c = c(1, 3, 8, 27))
comp.surv
# $a
# [1] 1 2 3 4
#
# $b
# [1] 1 2 4 8
#
# $c
# [1] 1 3 8 27
I can use lapply to plot each list element:
lapply(comp.surv, function(x) plot(x))
However, I want to include the name of each list element as plot title (main). For my example data, the title of each graph would be a,b and c respectively. First thing, is that I have a gsub rule that given comp.surv$a, I return a :
gsub(comp.surv\\$([a-z]+), "\\1", deparse(sustitute((comp.surv$a)))
# "a"
Which is good. However I cannot embed this result into my lapply statement above. Any ideas?
In the mean time I have tried getting round this by creating a function this to include the main parameter:
splot <- function(x){
plot(x, main = gsub(comp.surv\\$([a-z]+), "\\1" deparse(sustitute((x))))
}
lapply(comp.surv, function(x) splot(x))
This will plot each sub-variable of comp.surv, but all the titles are blank.
Can anyone recommend if I am going down the right track?
One possibility would be to loop over the names of the list:
lapply(names(comp.surv), function(x) plot(comp.surv[[x]], main = x))
Or slightly more verbose, loop over the list indices:
lapply(seq_along(comp.surv), function(x) plot(comp.surv[[x]], main = names(comp.surv)[x]))
Is that what you want?
ns=names(comp.surv)
lapply(ns, function(x) plot(comp.surv[[x]], main=x,ylab="y"))

How to add column to data.table with values from list based on regex

I have the following data.table:
id fShort
1 432-12 1245
2 3242-12 453543
3 324-32 45543
4 322-34 45343
5 2324-34 13543
DT <- data.table(
id=c("432-12", "3242-12", "324-32", "322-34", "2324-34"),
fShort=c("1245", "453543", "45543", "45343", "13543"))
and the following list:
filenames <- list("3242-124342345.png", "432-124343.png", "135-13434.jpeg")
I would like to create a new column "fComplete" that includes the complete filename from the list. For this the values of column "id" need to be matched with the filename-list. If the filename starts with the "id" string, the complete filename should be returned. I use the following regex
t <- grep("432-12","432-124343.png",value=T)
that return the correct filename.
This is how the final table should look like:
id fShort fComplete
1 432-12 1245 432-124343.png
2 3242-12 453543 3242-124342345.png
3 324-32 45543 NA
4 322-34 45343 NA
5 2324-34 13543 NA
DT2 <- data.table(
id=c("432-12", "3242-12", "324-32", "322-34", "2324-34"),
fshort=c("1245", "453543", "45543", "45343", "13543"),
fComplete = c("432-124343.png", "3242-124342345.png", NA, NA, NA))
I tried using apply and data.table approaches but I always get warnings like
argument 'pattern' has length > 1 and only the first element will be used
What is a simple approach to accomplish this?
Here's a data.table solution:
DT[ , fComplete := lapply(id, function(x) {
m <- grep(x, filenames, value = TRUE)
if (!length(m)) NA else m})]
id fShort fComplete
1: 432-12 1245 432-124343.png
2: 3242-12 453543 3242-124342345.png
3: 324-32 45543 NA
4: 322-34 45343 NA
5: 2324-34 13543 NA
In my experience with similar functions, sometimes the regex functions return a list, so you have to consider that in the apply - I usually do an example manually
Also apply will not always in y experience on its own return something that always works into a data.frame,sometimes I had to use lap ply, and or unlist and data.frame to modify it
Here is an answer - I am not familiar with data.tables and I was having issues with the filenames being in a list, but with some transformations this works. I worked it out by seeing what apply was outputting and adding the [1] to get the piece I needed
DT <- data.frame(
id=c("432-12", "3242-12", "324-32", "322-34", "2324-34"),
fShort=c("1245", "453543", "45543", "45343", "13543"))
filenames <- list("3242-124342345.png", "432-124343.png", "135-13434.jpeg")
filenames1 <- unlist(filenames)
x<-apply(DT[1],1,function(x) grep(x,filenames1)[1])
DT$fielname <- filenames1[x]

How to properly manipulate a string column in a data frame in R?

I have a data.frame with a string column that contains periods e.g "a.b.c.X". I want to split out the string by periods and retain the third segment e.g. "c" in the example given. Here is what I'm doing.
> df = data.frame(v=c("a.b.a.X", "a.b.b.X", "a.b.c.X"), b=seq(1,3))
> df
v b
1 a.b.a.X 1
2 a.b.b.X 2
3 a.b.c.X 3
And what I want is
> df = data.frame(v=c("a.b.a.X", "a.b.b.X", "a.b.c.X"), b=seq(1,3))
> df
v b
1 a 1
2 b 2
3 c 3
I'm attempting to use within, but I'm getting strange results. The value in the first row in the first column is being repeated.
> get = function(x) { unlist(strsplit(x, "\\."))[3] }
> within(df, v <- get(as.character(v)))
v b
1 a 1
2 a 2
3 a 3
What is the best practice for doing this? What am I doing wrong?
Update:
Here is the solution I used from #agstudy's answer:
> df = data.frame(v=c("a.b.a.X", "a.b.b.X", "a.b.c.X"), b=seq(1,3))
> get = function(x) gsub(".*?[.].*?[.](.*?)[.].*", '\\1', x)
> within(df, v <- get(v))
v b
1 a 1
2 b 2
3 c 3
Using some regular expression you can do :
gsub(".*?[.].*?[.](.*?)[.].*", '\\1', df$v)
[1] "a" "b" "c"
Or more concise:
gsub("(.*?[.]){2}(.*?)[.].*", '\\2', v)
The problem is not with within but with your get function. It returns a single character ("a") which gets recycled when added to your data.frame. Your code should look like this:
get.third <- function(x) sapply(strsplit(x, "\\."), `[[`, 3)
within(df, v <- get.third(as.character(v)))
Here is one possible solution:
df[, "v"] <- do.call(rbind, strsplit(as.character(df[, "v"]), "\\."))[, 3]
## > df
## v b
## 1 a 1
## 2 b 2
## 3 c 3
The answer to "what am I doing wrong" is that the bit of code that you thought was extracting the third element of each split string was actually putting all the elements of all your strings in a single vector, and then returning the third element of that:
get = function(x) {
splits = strsplit(x, "\\.")
print("All the elements: ")
print(unlist(splits))
print("The third element:")
print(unlist(splits)[3])
# What you actually wanted:
third_chars = sapply(splits, function (x) x[3])
}
within(df, v2 <- get(as.character(v)))