Splitting string into unknown number of new dataframe columns - regex

Question

I have a dataframe with a character column that contains email metadata in the form of multiple strings delimited by a newline character \n:
person myString
1 John To name5#email.com by sender6 on 01-12-2014\n
2 Jane To name#email.com,name4#email.com by sender1 on 01-22-2014\nTo name3#email.com by sender2 on 02-03-2014\nTo email5#domain.com by sender1 on 06-21-2014\n
3 Tim To name2#email.com by sender2 on 05-11-2014\nTo name#email.com by sender2 on 06-03-2015\n
I want to split the different substrings of myString into different columns so that it looks like this:
person email1 email2 email3
1 John To name5#email.com by sender6 on 01-12-2014 <NA> <NA>
2 Jane To name#email.com,name4#email.com by sender1 on 01-22-2014 To name3#email.com by sender2 on 02-03-2014 To email5#domain.com by sender1 on 06-21-2014
3 Tim To name2#email.com by sender2 on 05-11-2014 To name#email.com by sender2 on 06-03-2015 <NA>
My current approach uses separate from the tidyr package:
library(dplyr)
library(tidyr)
res1 <- df %>%
separate(col = myString, into = paste(rep("email", 3), 1:3), sep = "\\n", extra = "drop")
res1[res1 == ""] <- NA
But with this approach, I have to manually specify that there are three columns to extract.
I'm looking to improve this process with either or both of:
A way to automate that calculation of the maximum number of occurrences of the delimiting character (i.e., how many new variables are needed)
Other approaches to splitting into an unknown number of columns
And if there's a good solution that returns data in a long form, instead of wide, that would be great too.
Sample data:
df <- structure(list(person = c("John", "Jane", "Tim"), myString = c("To name5#email.com by sender6 on 01-12-2014\n",
"To name#email.com,name4#email.com by sender1 on 01-22-2014\nTo name3#email.com by sender2 on 02-03-2014\nTo email5#domain.com by sender1 on 06-21-2014\n",
"To name2#email.com by sender2 on 05-11-2014\nTo name#email.com by sender2 on 06-03-2015\n"
)), .Names = c("person", "myString"), row.names = c(NA, -3L), class = "data.frame")

Answer 1

I would suggest cSplit from my "splitstackshape" package:
library(splitstackshape)
cSplit(df, "myString", "\n")
# person myString_1
# 1: John To name5#email.com by sender6 on 01-12-2014
# 2: Jane To name#email.com,name4#email.com by sender1 on 01-22-2014
# 3: Tim To name2#email.com by sender2 on 05-11-2014
# myString_2
# 1: NA
# 2: To name3#email.com by sender2 on 02-03-2014
# 3: To name#email.com by sender2 on 06-03-2015
# myString_3
# 1: NA
# 2: To email5#domain.com by sender1 on 06-21-2014
# 3: NA
You can also try stri_split_fixed from the "stringi" package with the argument simplify = TRUE (though with your sample data, this adds an extra empty column at the end). The approach would be something like:
library(stringi)
data.frame(person = df$person,
stri_split_fixed(df$myString, "\n",
simplify = TRUE))

Answer 2

Seems hacky, but here ya go...
Use strsplit to split the char vector. Get the max length, use that for your columns.
df <- data.frame(
person = c("John", "Jane", "Tim"),
myString = c("To name5#email.com by sender6 on 01-12-2014\n",
"To name#email.com,name4#email.com by sender1 on 01-22-2014\nTo name3#email.com by sender2 on 02-03-2014\nTo email5#domain.com by sender1 on 06-21-2014\n",
"To name2#email.com by sender2 on 05-11-2014\nTo name#email.com by sender2 on 06-03-2015\n"
), stringsAsFactors=FALSE
)
a <- strsplit(df$myString, "\n")
max_len <- max(sapply(a, length))
for(i in 1:max_len){
df[,paste0("email", i)] <- sapply(a, "[", i)
}

Answer 3

Here is an efficient route to a long form:
a <- strsplit(df$myString, "\n")
lens <- vapply(a, length, integer(1L)) # or lengths(a) in R 3.2
longdf <- df[rep(seq_along(a), lens),]
longdf$string <- unlist(a)
Note that stack() is often useful for these cases.
Can be simplified by using the IRanges Bioconductor package:
longdf <- df[togroup(a),]
longdf$string <- unlist(a)
Then, if really necessary, go to wide-form:
longdf$myString <- NULL
longdf$token <- sequence(lens)
widedf <- reshape(longdf, timevar="token", idvar="person", direction="wide")

Answer 4

This might suffice:
library(data.table)
dt = as.data.table(df) # or setDT to convert in place
dt[, strsplit(myString, split = "\n"), by = person]
# person V1
#1: John To name5#email.com by sender6 on 01-12-2014
#2: Jane To name#email.com,name4#email.com by sender1 on 01-22-2014
#3: Jane To name3#email.com by sender2 on 02-03-2014
#4: Jane To email5#domain.com by sender1 on 06-21-2014
#5: Tim To name2#email.com by sender2 on 05-11-2014
#6: Tim To name#email.com by sender2 on 06-03-2015
And then can trivially convert to wide format:
dcast(dt[, strsplit(myString, split = "\n"), by = person][, idx := 1:.N, by = person],
person ~ idx, value.var = 'V1')
# person 1 2 3
#1: Jane To name#email.com,name4#email.com by sender1 on 01-22-2014 To name3#email.com by sender2 on 02-03-2014 To email5#domain.com by sender1 on 06-21-2014
#2: John To name5#email.com by sender6 on 01-12-2014 NA NA
#3: Tim To name2#email.com by sender2 on 05-11-2014 To name#email.com by sender2 on 06-03-2015 NA
# (load reshape2 and use dcast.data.table instead of dcast if not using 1.9.5+)