CODEWITHSUNDEEP
regexr
Mithilesh Kumar
Mithilesh Kumar

Reputation: 263

Extract left to a pattern upto a given variable length in R [Variable length look behind]

I have a string matrix which contains sector specific revenue contribution of certain companies. I have to extract a matrix which only contains revenue from software. The matrix is as under:

revenue <- data.frame(revenue = c("79% Software, 1% Hardware, 20% Services", NA, NA, "10.5% Software, 90% Services", "1.4% Software, 98.6% Services", "17% Software, 83% Services", NA, "100% Services", "47% Services, 39% Hardware, 14.32% Software"))

I want to provide ending pattern as "software" and then extract left to get % which extract the number(whether it is decimal or numeric).

My solution is working but it's quiet lengthy. How can I extract the matrix in single line.

EDIT

As asked by by @SabDem in comment,

My code:

library("stringr")
revenue= as.matrix(revenue)
rs <- str_split_fixed(revenue,',',3)
rs1<- matrix(0,nrow(rs), ncol(rs))
for(i in 1:nrow(rs)){
  for(j in 1:ncol(rs)){
    ifelse(grep('Software',rs[i,j])==TRUE,(rs1[i,j]=rs[i,j]),(rs1[i,j]=0)) 
  }
}
rs2 <- gsub('Software|%','',rs1)
soft.revenue <- rowSums(data.matrix(data.frame(rs2, stringsAsFactors = FALSE)))

Upvotes: 0

Views: 101

Answers (1)

bluefish
bluefish

Reputation: 405

I would use the stringr library. For your example it would be:

library("stringr")
revenue <- data.frame(revenue = c("79% Software, 1% Hardware, 20% Services", NA, NA, "10.5% Software, 90% Services", "1.4% Software, 98.6% Services", "17% Software, 83% Services", NA, "100% Services", "47% Services, 39% Hardware, 14.32% Software"))
pattern <- "(([[:digit:]]|.[[:digit:]]+)*)(?=% Software)"
as.numeric(str_extract(revenue$revenue,pattern))

The core idea is the expression (?=% Software) which looks ahead until it finds the string % Software. Variable length look behind is (as far as I know) not possible in R.

Upvotes: 1

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