R help - Oct 2022 - Replace values based on neither condition

Hello,
I have a data frame with a string column. All data that are neither
"POS" nor "NEG" should've replaced by an NA. How can I
implement that
(even with extra libraries)? My attempts actually wipe out POS and
NEG...
Thank you

```
df = data.frame(a = 1:5, b = c("",      "31.35",
"POS",   "20.61", "NEG"),
                stringsAsFactors = F)
df$b[!(df$b == "POS") & (df$b == "NEG")] = NA
df$b[(df$b == "POS") | (df$b == "NEG")] = NA
```

-- 
Best regards,
Luigi

?s 10:43 de 28/10/2022, Luigi Marongiu escreveu:
> Hello,
> I have a data frame with a string column. All data that are neither
> "POS" nor "NEG" should've replaced by an NA. How
can I implement that
> (even with extra libraries)? My attempts actually wipe out POS and
> NEG...
> Thank you
> 
> ```
> df = data.frame(a = 1:5, b = c("",      "31.35",
"POS",   "20.61", "NEG"),
>                  stringsAsFactors = F)
> df$b[!(df$b == "POS") & (df$b == "NEG")] = NA
> df$b[(df$b == "POS") | (df$b == "NEG")] = NA
> ```
> 
Hello,

Here is a way.
Use `%in%` to get the values equal to either "POS" or "NEG",
negate its
result and function `is.na<-` assigns NA's.


df = data.frame(a = 1:5, b = c("", "31.35", "POS",
"20.61", "NEG"),
                 stringsAsFactors = FALSE)

is.na(df$b) <- !df$b %in% c("POS", "NEG")
df
#>   a    b
#> 1 1 <NA>
#> 2 2 <NA>
#> 3 3  POS
#> 4 4 <NA>
#> 5 5  NEG


Hope this helps,

Rui Barradas