R - use if statement to regroup variable - r

Question

I want to regroup a variable into a new one.
If value is 0, new one should be 0 too.
If value ist 999, then make it missing, NA.
Everything else 1
This is my try:
id <- 1:10
variable <- c(0,0,0,1,2,3,4,5,999,999)
df <- data.frame(id,variable)
df$variable2 <-
if (df$variable == 0) {
df$variable2 = 0
} else if (df$variable == 999){
df$variable2 = NA
} else {
df$variable2 = 1
}
And this the error message:
In if (df$variable == 0) { : the condition has length > 1 and only
the first element will be used
A pretty basic question but I'm a basic user. Thanks in advance!

Answer 1

Try ifelse
df$variable2 <- ifelse(df$variable == 999, NA, ifelse(df$variable > 0, 1, 0))
df
# id variable variable2
#1 1 0 0
#2 2 0 0
#3 3 0 0
#4 4 1 1
#5 5 2 1
#6 6 3 1
#7 7 4 1
#8 8 5 1
#9 9 999 NA
#10 10 999 NA
When you do df$variable == 0 the output / condition is
#[1] TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
where it should be a length-one logical vector that is not NA in if(condition), see ?"if".
You can avoid ifelse, for example, like so
df$variable2 <- df$variable
df$variable2[df$variable2 == 999] <- NA
df$variable2[df$variable2 > 0] <- 1

Answer 2

It might be easier to avoid the if/else statement all together by using conditional statements within subset notation:
when df$variable is equal to zero, change it to zero
df$variable[df$variable==0] <- 0
when df$variable is equal to 999, change it to NA
df$variable[df$variable==999] <- NA
when df$variable is greater than 0 and is not equal to NA, change it to 1
df$variable[df$variable>0 & is.na(df$variable) == 'FALSE'] <- 1

Answer 3

Looks like you want to recode your variable. You can do this (and other data/variable transformations) with the sjmisc-package, in your case with the rec()-command:
id <- 1:10
variable <- c(0,0,0,1,2,3,4,5,999,999)
df <- data.frame(id,variable)
library(sjmisc)
rec(df, variable, rec = c("0=0;999=NA;else=1"))
#> id variable variable_r
#> 1 1 0 0
#> 2 2 0 0
#> 3 3 0 0
#> 4 4 1 1
#> 5 5 2 1
#> 6 6 3 1
#> 7 7 4 1
#> 8 8 5 1
#> 9 9 999 NA
#> 10 10 999 NA
# or a single vector as input
rec(df$variable, rec = c("0=0;999=NA;else=1"))
#> [1] 0 0 0 1 1 1 1 1 NA NA
There are many examples, also in the help-file, and you can find a sjmisc-cheatsheet at the RStudio-Cheatsheet collection (or direct PDF-download here).

Answer 4

df$variable2 <- sapply(df$variable,
function(el) if (el == 0) {0} else if (el == 999) {NA} else {1})
This one-liner reflects your:
If value is 0, new one should be 0 too. If value ist 999, then make it
missing, NA. Everything else 1
Well, it is slightly slower than #markus's second or #SPJ's solutions which are most r-ish solutions.
Why one should put away the hands from ifelse
tt <- c(TRUE, FALSE, TRUE, FALSE)
a <- c("a", "b", "c", "d")
b <- 1:4
ifelse(tt, a, b) ## [1] "a" "2" "c" "4"
# totally perfect and as expected!
df <- data.frame(a=a, b=b, c=tt)
df$d <- ifelse(df$c, df$a, df$b)
## > df
## a b c d
## 1 a 1 TRUE 1
## 2 b 2 FALSE 2
## 3 c 3 TRUE 3
## 4 d 4 FALSE 4
######### This is wrong!! ##########################
## df$d is not [1] "a" "2" "c" "4"
## the problem is that
## ifelse(df$c, df$a, df$b)
## returns for each TRUE or FALSE the entire
## df$a or df$b intead of treating it like a vector.
## Since the last df$c is FALSE, df$b is returned
## Thus we get df$b for df$d.
## Quite an unintuitive behaviour.
##
## If one uses purely vectors, ifelse is fine.
## But actually df$c, df$a, df$b should be treated each like a vector.
## However, `ifelse` does not.
## No warnings that using `ifelse` with them will lead to a
## totally different behaviour.
## In my view, this is a design mistake of `ifelse`.
## Thus I decided myself to abandon `ifelse` from my set of R commands.
## To avoid that such kind of mistakes can ever happen.
#####################################################
As #Parfait pointed out correctly, it was a misinterpretation.
The problem was that df$a was treated in the data frame as a factor.
df <- data.frame(a=a, b=b, c=tt, stringsAsFactor = F)
df$d <- ifelse(df$c, df$a, df$b)
df
Gives the correct result.
a b c d
1 a 1 TRUE a
2 b 2 FALSE 2
3 c 3 TRUE c
4 d 4 FALSE 4
Thank you #Parfait to pointing that out!
Strange that I didn't recognized that in my initial trials.
But yeah, you are absolutely right!