R: Delete first and last part of string based on pattern - r

Question

This string is a ticker for a bond: OAT 3 25/32 7/17/17. I want to extract the coupon rate which is 3 25/32 and is read as 3 + 25/32 or 3.78125. Now I've been trying to delete the date and the name OAT with gsub, however I've encountered some problems.
This is the code to delete the date:
tkr.bond <- 'OAT 3 25/32 7/17/17'
tkr.ptrn <- '[0-9][[:punct:]][0-9][[:punct:]][0-9]'
gsub(tkr.ptrn, "", tkr.bond)
However it gets me the same string. When I use [0-9][[:punct:]][0-9] in the pattern I manage to delete part of the date, however it also deletes the fraction part of the coupon rate for the bond.
The tricky thing is to find a solution that doesn't involve the pattern of the coupon because the tickers have this form: Name Coupon Date, so, using a specific pattern for the coupon may limit the scope of the solution. For example, if the ticker is this way OAT 0 7/17/17, the coupon is zero.

Answer 1

Just replace first and last word with an empty string.
> tkr.bond <- 'OAT 3 25/32 7/17/17'
> gsub("^\\S+\\s*|\\s*\\S+$", "", tkr.bond)
[1] "3 25/32"
OR
Use gsubfn function in-order to use a function in the replacement part.
> gsubfn("^\\S+\\s+(\\d+)\\s+(\\d+)/(\\d+).*", ~ as.numeric(x) + as.numeric(y)/as.numeric(z), tkr.bond)
[1] "3.78125"
Update:
> tkr.bond1 <- c(tkr.bond, 'OAT 0 7/17/17')
> m <- gsub("^\\S+\\s*|\\s*\\S+$", "", tkr.bond1)
> gsubfn(".+", ~ eval(parse(text=x)), gsub("\\s+", "+", m))
[1] "3.78125" "0"

Answer 2

Try
eval(parse(text=sub('[A-Z]+ ([0-9]+ )([0-9/]+) .*', '\\1 + \\2', tkr.bond)))
#[1] 3.78125
Or you may need
sub('^[A-Z]+ ([^A-Z]+) [^ ]+$', '\\1', tkr.bond)
#[1] "3 25/32"
Update
tkr.bond1 <- c(tkr.bond, 'OAT 0 7/17/17')
v1 <- sub('^[A-Z]+ ([^A-Z]+) [^ ]+$', '\\1', tkr.bond1)
unname(sapply(sub(' ', '+', v1), function(x) eval(parse(text=x))))
#[1] 3.78125 0.00000
Or
vapply(strsplit(tkr.bond1, ' '), function(x)
eval(parse(text= paste(x[-c(1, length(x))], collapse="+"))), 0)
#[1] 3.78125 0.00000
Or without the eval(parse
vapply(strsplit(gsub('^[^ ]+ | [^ ]+$', '', tkr.bond1), '[ /]'), function(x) {
x1 <- as.numeric(x)
sum(x1[1], x1[2]/x1[3], na.rm=TRUE)}, 0)
#[1] 3.78125 0.00000

Answer 3

Similar to akrun's answer, using sub with a replacement. How it works: you put your "desired" pattern inside parentheses and leave the rest out (while still putting regex characters to match what's there and that you don't wish to keep). Then when you say replacement = "\\1" you indicate that the whole string must be substituted by only what's inside the parentheses.
sub(pattern = ".*\\s(\\d\\s\\d+\\/\\d+)\\s.*", replacement = "\\1", x = tkr.bond, perl = TRUE)
# [1] "3 25/32"
Then you can change it to numerical:
temp <- sub(pattern = ".*\\s(\\d\\s\\d+\\/\\d+)\\s.*", replacement = "\\1", x = tkr.bond, perl = TRUE)
eval(parse(text=sub(" ","+",x = temp)))
# [1] 3.78125

Answer 4

You can also use strsplit here. Then evaluate components excluding the first and the last. Like this
> tickers <- c('OAT 3 25/32 7/17/17', 'OAT 0 7/17/17')
>
> unlist(lapply(lapply(strsplit(tickers, " "),
+ function(x) {x[-length(x)][-1]}),
+ function(y) {sum(
+ sapply(y, function (z) {eval(parse(text = z))}) )} ) )
[1] 3.78125 0.00000