Extract nonnegative number or alternatively return negative number from a vector of strings

Question

I need to extract a nonnegative number from a string or return negative number if a number was not extracted.

For extracting the number I found the following way:

> grep("^[0-9.]+","1234.1234alsk",value=TRUE)
[1] "1234.1234alsk"

If the given string is not a number, then empty string is returned:

> grep("^[0-9.]+","",value=TRUE)
character(0)

Now I would like to replace the empty string with some proxy number, like 0 or -1 using the following kind of function:

> sub("^$","-1","")
[1] "-1"

However, If I apply that function to an empty string of character(0) I do not get the desired result:

> sub("^$","-1",grep("^[0-9.]+","",value=TRUE))
character(0)

The problem is that grep returns character(0) and not "". Then because sub works differently with character(0) and "", I do not get the desired value "-1" but unchanged character(0). As a result those values will be dropped in the following query:

> v <- c("0","","1","2")
> as.numeric(sub("^$","-1",grep("^[0-9.]+",v,value=TRUE)))
[1] 0 1 2

How could I do the above kind of conversion using one-liner?

`character(0)` is not equivalent of empty string. (""). What are you trying to do? — Ronak Shah, Oct 01 '18 at 13:15
Also, you say that you found a way to extract the number, but your code returns `"1234.1234alsk"` (i.e. returns the actual value if there's a numeric match) — AntoniosK, Oct 01 '18 at 13:17
What do you expect `"-125d"` and `"125d67"` to return? Is it possible to have a case similar to the 2nd one? — AntoniosK, Oct 01 '18 at 13:36
I expect `"-125d"` return empty string and `"125d67"` return `"125"`. — Heikki, Oct 01 '18 at 13:52

score 1 · Answer 1 · answered Oct 01 '18 at 14:46

1

By preloading stringr, I found the following one liner to fetch the first occurrence of a number:

> library(stringr)
> x <- c("","1","1.23","1.23a","-123")
> as.vector(ifelse(is.na(str_match(x,"^[0-9.]+")),"-1",str_match(x,"^[0-9.]+")))
[1] "-1"   "1"    "1.23" "1.23" "-1"

and as a result, I can do the following conversions as one-liners:

> as.numeric(as.vector(ifelse(is.na(str_match(x,"^[0-9.]+")),"-1",str_match(x,"^[0-9.]+"))))
[1] -1.00  1.00  1.23  1.23 -1.00
> all.is.numeric(as.numeric(as.vector(ifelse(is.na(str_match(x,"^[0-9.]+")),"-1",str_match(x,"^[0-9.]+")))))
[1] TRUE

The conversion is stored in a column field, therefore, the one-liner requirement is a necessity here.

answered Oct 01 '18 at 14:46

Heikki

2,214
19
34

1

Any chance you'll have a number after some letters? `"da23"` will return `"-1"` and not `"23"`. Is that OK? Because `"^[0-9.]+"` needs the string to start with a number. It's not the "first occurrence of a number". Maybe you can try `"[0-9.]+"` instead? – AntoniosK Oct 01 '18 at 15:01
Good point. I need to interpret amount as a number, or incorrect amount as a proxy value. Therefore, in practice I should maybe have even more strict condition, like `"^[0-9.]+$"`. I was thinking accepting different decimal points or spaces at the end, but then I should use another regular expression. The actual problem was however `grep` returning `character(0)` for empty strings. – Heikki Oct 01 '18 at 17:07

Extract nonnegative number or alternatively return negative number from a vector of strings

1 Answers1