Nsryjdtyk

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!

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

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

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).

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!