遍历df列，与列表进行比较并创建新列

Question

I have a column of numbers, like social security numbers for example. I would like to compare this column to a list of unacceptable values ( like 11111111 or 12345678 for example). There also some grepl operations i would like to perform, like the first 3 digits can't be 000 . Below is a skeleton of what I think the code could look like, I prefer a for loop logic.

ssns <- c(12343210,23454321,34565432,11111111)
badssns <- c(11111111,22222222)

for( i in 1:length(ssns)) {
    if(ssns[i] %in% badssn_list) {
        ssns$newcolumn==BADSSN
      }
    else if( grepl(first 3 numbers 0){
        ssns$newcolumn==BADSSN
      }
    else{ssns$newcolumn==GOODSSN}
}

Answer 1

Just using a nested ifelse should do the job imo:

ssns$newcolumn <- ifelse(ssns$num %in% badssns, 'BADSSN', 
                         ifelse(substr(ssns$num,1,3)=='000', 'BADSSN', 'GOODSSN'))

or shorter using an OR statement ( | ):

ssns$newcolumn <- ifelse(ssns$num %in% badssns| substr(ssns$num,1,3)=='000', 'BADSSN', 'GOODSSN')

which gives:

> ssns
       num newcolumn
1 12343210   GOODSSN
2 23454321   GOODSSN
3 34565432   GOODSSN
4 11111111    BADSSN
5 00065432    BADSSN

---

Used data:

ssns <- data.frame(num = c('12343210','23454321','34565432','11111111','00065432'), stringsAsFactors = FALSE)
badssns <- c('11111111','22222222')

Answer 2

It seems like you have some experience with computer programming, but maybe are new to R. In most cases, the best R programs don't use for loops.

Here's a more R ish way to accomplish what you've described. It will be much faster when ssns and badssns are long.

ssns<-c(12343210,23454321,34565432,11111111)
badssns<-c(11111111,22222222)

good.idxs <- is.na(match(ssns, badssns))
good.ssns <- ssns[good.idxs]

You might want to work with strings rather than numbers -- maybe you are concerned the letter "oh" was used in place of the number "zero". This approach works in that case as well. Somewhat unexpectedly (for me, anyway), it even works when ssns is a vector of characters and badssns is a vector of number or vice versa!

Answer 3

If ssns and badssns are character vectors:

ssns<-c("12343210","23454321","34565432","11111111","00023456")
badssns<-c("11111111","22222222")

then you can use just one ifelse :

result <- ifelse(ssns %in% badssns | grepl("^0{3}",ssns), "BADSSNS", "GOODSSNS")
##[1] "GOODSSNS" "GOODSSNS" "GOODSSNS" "BADSSNS"  "BADSSNS"