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I keep encountering a problem in my R code with generating a new variable based upon another variable. Every participant has entries for multiple different variables. Not all of these variables matter for each participant. I have a dummy coded variable which I use to tell me which variable I should use when generating my new variable. Here is what my data would look like.

data
id use v1 v2 v3
1  1   2  2  1  
2  2   NA 1  2 
3  3   1  NA 3
4  1   3  5  NA
5  2   4  4  1

I will try to create a new variable using the dummy coded variable. For this example, is use is 1, I want to use the value of v1 for x. If use is 2, then I want to use v2 for x. If use is 3, I want to use v3 for x. Here is the code I use.

data$x [data$use == 1] <- data$v1
data$x [data$use == 2] <- data$v2
data$x [data$use == 3] <- data$v3

When I try to run the code, I will then get the error message saying "number of items to replace is not a multiple of replacement length".

I did some research and I think this has something to do with data being missing (though I could be wrong). I tried to use is.na () within the [] but this does not solve the issue.

I have used ifelse to solve problems similar to this before, but I don't think that code would work in this circumstance because I have more than two situations (I am not sure if ifelse is cumulative or not).

Why does this error occur and what is the best way to resolve this?

Eric Boorman
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2 Answers2

1

Your issue is that your left and right hand sides of the <- assignment are different lengths.

## data$x[data$use == 1] <- data$v1

data$x[data$use == 1]
#[1] 2 3

data$v1
#[1]  2 NA  1  3  4

If you match them up by selecting on both sides, you're laughing:

data$x[data$use == 1] <- data$v1[data$use == 1]
data$x[data$use == 2] <- data$v2[data$use == 2]
data$x[data$use == 3] <- data$v3[data$use == 3]

#  id use v1 v2 v3 x
#1  1   1  2  2  1 2
#2  2   2 NA  1  2 1
#3  3   3  1 NA  3 3
#4  4   1  3  5 NA 3
#5  5   2  4  4  1 4

You can avoid needing to write multiple lines and make this work for any number of variables using matrix indexing however, as per this previous answer of mine: https://stackoverflow.com/a/33862219/496803

data[c("v1","v2","v3")][cbind(seq_len(nrow(data)), data$use)]
#[1] 2 1 3 3 4

This essentially uses a matrix with a row and column index to grab the right value from the v1-3 variables:

cbind(seq_len(nrow(data)), data$use)
##    row  col
#     [,1] [,2]
#[1,]    1    1
#[2,]    2    2
#[3,]    3    3
#[4,]    4    1
#[5,]    5    2


## assign it get the same result obviously:
data$x <- data[c("v1","v2","v3")][cbind(seq_len(nrow(data)), data$use)]
thelatemail
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0

You can try the code below

v <- c("v1", "v2", "v3")
list2env(
  setNames(
    lapply(v, function(x) data[[x]][data$use == gsub("\\D", "", x)]),
    v
  ),
  envir = .GlobalEnv
)

and you can check it by

> mget(ls(pattern = "v\\d+"))
$v1
[1] 2 3

$v2
[1] 1 4

$v3
[1] 3
ThomasIsCoding
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