R - create new variable using if statement

Question

I am trying to create a new variable within data table under if statement: if string variable contains substring, then new variable equals to numerical value.

My data:

N X
1 aa1aa 
2 bb2bb
3 cc-1bb 
...

Dataframe contains several thousands of rows.

Result needed is new column containing numerical value which is withing string (X collumn):

N X      Y
1 aa1aa  1
2 bb2bb  2
3 cc-1bb -1 

I was trying with

for (i in 1:length(mydata)){
  if (grep('1', mydata$X) == TRUE) {
    mydata$Y <- 1  }

but I'm not sure if I'm even on correct way... Any help please?

Answer 1

This should work on more of your extended samples. Basically it takes out everything that's not a letter from the middle of the string.

X <- c("aa1aa", "bb2bb", "cc-1bb","aa+0.5b","fg-0.25h")
gsub("^[a-z]+([^a-z]*)[a-z]+$","\\1",X,perl=T)
#[1] "1"     "2"     "-1"    "+0.5"  "-0.25"

Answer 2

Using the example data from @Paulo you can use gsub from base R...

d$Y <- gsub( "[^0-9]" , "" , d$X ) 

Answer 3

something like this?

d <- data.frame(N = 1:3,
                X = c('aa1aa', 'bb2bb', 'cc-1bb'),
                stringsAsFactors = FALSE)

library(stringr)

d$Y <- as.numeric(str_extract_all(d$X,"\\(?[0-9,.]+\\)?"))

d

  N      X  Y
1 1  aa1aa  1
2 2  bb2bb  2
3 3 cc-1bb  1

EDIT - Speed test

The gsub approch provided by @Simon is much faster than stringr

library(microbenchmark)
# 30000 lines data.frame
d1 <- data.frame(N = 1:30000,
                X = rep(c('aa1aa', 'bb2bb', 'cc-1bb'), 10000),
                stringsAsFactors = FALSE)

stringr

microbenchmark(as.numeric(str_extract_all(d1$X,"\\(?[0-9,.]+\\)?")), 
               times = 10L)
Unit: seconds
                                                      expr      min      lq  median       uq      max neval
 as.numeric(str_extract_all(d1$X, "\\\\(?[0-9,.]+\\\\)?")) 2.677408 2.75283 2.76473 2.781083 2.796648    10

base gsub

microbenchmark(gsub( "[^0-9]" , "" , d1$X ), times = 10L)
Unit: milliseconds
                     expr      min       lq   median       uq      max neval
 gsub("[^0-9]", "", d1$X) 44.95564 45.05358 45.07238 45.10201 45.23645    10