Better Strategy for pulling elements from string - regex

Question

I have a string that looks like this:
x <- "\r\n Ticker Symbol: RBO\r\n \t Exchange: TSX \r\n\t Assets ($mm) 36.26 \r\n\t Units Outstanding: 1,800,000 \r\n\t Mgmt. Fee** 0.25 \r\n 2013 MER* n/a \r\n\t CUSIP: 74932K103"
What I need is this:
list(Ticker = "RBO", Assets = 36.26, Shares = 1,800,000)
I've tried splitting, regex, etc. But I feel my string manipulation skills are not up to snuff.
Here's my "best" attempt so far.
x <- unlist(strsplit(unlist(strsplit(x, "\r\n\t") ),"\r\n"))
trim <- function (x) gsub("^\\s+|\\s+$", "", x)
x <- trim(x)
gsub("[A-Z]+$","\\2",x[2]) # bad attempt to get RBO

Answer 1

Update/better answer:
A look at cat(x) and readLines(x) helps a lot here
> cat(x)
#
# Ticker Symbol: RBO
# Exchange: TSX
# Assets ($mm) 36.26 #
# Units Outstanding: 1,800,000
# Mgmt. Fee** 0.25
# 2013 MER* n/a
# CUSIP: 74932K103
> readLines(textConnection(x))
# [1] "" " Ticker Symbol: RBO"
# [3] " \t Exchange: TSX " "\t Assets ($mm) 36.26 "
# [5] "\t Units Outstanding: 1,800,000 " "\t Mgmt. Fee** 0.25 "
# [7] " 2013 MER* n/a " "\t CUSIP: 74932K103"
Now we know a few things. One, we don't need the first line, and we do want the second line. That makes things easier because now the first line matches our desired first line. Next, it would be easier your list names matched the names in the string. I chose these.
> nm <- c("Symbol", "Assets", "Units")
Now all we have to do use grep with sapply, and we'll get back a named vector of matches. Setting value = TRUE in grep will return us the strings.
> (y <- sapply(nm, grep, x = readLines(textConnection(x))[-1], value = TRUE))
# b Symbol Assets
# " Ticker Symbol: RBO" "\t Assets ($mm) 36.26 "
# Units
# "\t Units Outstanding: 1,800,000 "
Then we strsplit that on "[: ]", take the last element in each split, and we're done.
> lapply(strsplit(y, "[: ]"), tail, 1)
$Symbol
[1] "RBO"
$Assets
[1] "36.26"
$Units
[1] "1,800,000
You could achieve the same result with
> g <- gsub("[[:cntrl:]]", "", capture.output(cat(x))[-1])
> m <- mapply(grep, nm, MoreArgs = list(x = g, value = TRUE))
> lapply(strsplit(m, "[: ]"), tail, 1)
Hope that helps.
Original Answer:
It looks like if you're pulling these from a large table, that they'd all be in the same element "slot" each time, so maybe this might be a little easier.
> s <- strsplit(x, "[: ]|[[:cntrl:]]")[[1]]
Explained:
- [: ] match a ":" character followed by a space character
- | or
- [[:cntrl:]] any control character, which in this case is any of \r, \t, and \n. This is probably better explained here
Then, nzchar looks in the above result for non-zero length character strings, and returns TRUE if matched, FALSE otherwise. So we can look at the result of the first line, determine where the matches are, and subset based on that.
> as.list(s[nzchar(s)][c(3, 8, 11)])
[[1]]
[1] "RBO"
[[2]]
[1] "36.26"
[[3]]
[1] "1,800,000"
You could put is into one line by assigning s as the inner call. Since functions and calls are evaluated from the inside out, s is assigned before R reaches the outside s subset. This is a bit less readable though.
s[nzchar(s <- strsplit(x, "[: ]|[[:cntrl:]]")[[1]])][c(3,8,11)]
So this would go s <- strsplit(...) -> [[ -> nzchar -> s[.. >- [c(3,8,11)]

Answer 2

Perhaps:
sub( "\\\r\\\n.+$", "", sub( "^.+Ticker Symbol: ", "", x) )
[1] "RBO"
I suppose you might do it all in one pattern with parentheses. and backreference.
> sub( "^.+Ticker Symbol: ([[:alpha:]]{1,})\\\r\\\n.+$", "\\1", x)
[1] "RBO"

Answer 3

If you just want to extract different parts of the string, you can use regexpr to find phrases and extract the contents after the phrase. For example
extr<-list(
"Ticker" = "Ticker Symbol: ",
"Assets" = "Assets ($mm) ",
"Shares" = "Units Outstanding: "
)
lines<-strsplit(x,"\r\n")[[1]]
Map(function(p) {
m <- regexpr(p, lines, fixed=TRUE)
if(length( w<- which(m!=-1))==1) {
gsub("^\\sw+|\\s$", "",
substr(lines[w], m[w] + attr(m,"match.length")[w], nchar(lines[w])))
} else {
NA
}
}, extr)
Which returns the named list as desired
$Ticker
[1] "RBO"
$Assets
[1] "36.26"
$Shares
[1] "1,800,000"
Here extr is a list where the name of the element is the name that will be used in the final list, and the element value is the exact string that will be matched in the text. I added in a gsub as well to trim off any whitespace.

Answer 4

The stringr package is good for scraping data from strings. Here are the steps I use every time. You can always make the rules as specific or robust as you see fit.
require(stringr)
## take out annoying characters
x <- gsub("\r\n", "", x)
x <- gsub("\t", "", x)
x <- gsub("\\(\\$mm\\) ", "", x)
## define character index positions of interest
tickerEnd <- str_locate(x, "Ticker Symbol: ")[[1, "end"]]
assetsEnd <- str_locate(x, "Assets ")[[1, "end"]]
unitsStart <- str_locate(x, "Units Outstanding: ")[[1, "start"]]
unitsEnd <- str_locate(x, "Units Outstanding: ")[[1, "end"]]
mgmtStart <- str_locate(x, "Mgmt")[[1, "start"]]
## get substrings based on indices
tickerTxt <- substr(x, tickerEnd + 1, tickerEnd + 4) # allows 4-character symbols
assetsTxt <- substr(x, assetsEnd + 1, unitsStart - 1)
sharesTxt <- substr(x, unitsEnd + 1, mgmtStart - 1)
## cut out extraneous characters
ticker <- gsub(" ", "", tickerTxt)
assets <- gsub(" ", "", assetsTxt)
shares <- gsub(" |,", "", sharesTxt)
## add data to data frame
df <- data.frame(ticker, as.numeric(assets), as.numeric(shares), stringsAsFactors = FALSE)