Split vector of strings and paste subset of resulting elements into a new vector

Question

Define

z<- as.character(c("1_xx xx xxx_xxxx_12_sep.xls","2_xx xx xxx_xxxx_15_aug.xls"))

such that

> z
[1] "1_xx xx xxx_xxxx_12_sep.xls" "2_xx xx xxx_xxxx_15_aug.xls"

I want to create a vector w such that

> w
[1] "1_12_sep" "2_15_aug"

That is, split each element of z by _ and then join elements 1,4,5, with the .xls removed from the latter.

I can manage the split part, but not sure what function to provide, e.g something like"

w <- as.character(lapply(strsplit(z,"_"), function(x) ???))

Answer 1

You can do this using a combination of strsplit, substr and lapply:

y <- strsplit(z,"_",fixed=TRUE)
lapply(y,FUN=function(x){paste(x[1],x[4],substr(x[5],1,3),sep="_")})

Answer 2

Using a bit of magic in the stringr package: I separately extract the left and right date fields, combine them, and finally remove the .xls at the end.

library(stringr)
l <- str_extract(z, "\\d+_")
r <- str_extract(z, "\\d+_\\w*\\.xls")
gsub(".xls", "", paste(l, r, sep=""))

[1] "1_12_sep" "2_15_aug"

str_extract is a wrapper around some of the base R functions which I find easier to use.

Edit Here is a short explanation of what the regex does:

• \\d+ looks for one or more digits. It is escaped to distinguish from a normal character d.
• \\w* looks for zero or more alphanumeric characters (word). Again, it's escaped.
• \\. looks for a decimal point. This needs to be escaped because otherwise the decimal point means any single character.

In theory the regex should be quite flexible. It should find single or double characters for your dates.