CODEWITHSUNDEEP
rstringgsub
capm
capm

Reputation: 1047

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

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.

Upvotes: 1

Views: 445

Answers (4)

inscaven
inscaven

Reputation: 2584

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

Upvotes: 0

akrun
akrun

Reputation: 887971

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

Upvotes: 1

Avinash Raj
Avinash Raj

Reputation: 174874

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"

Upvotes: 2

Dominic Comtois
Dominic Comtois

Reputation: 10431

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

Upvotes: 1

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