Variable names as Input in an R Function

Question

I have a dataframe with several numeric variables along with factors. I wish to run over the numeric variables and replace the negative values to missing. I couldn't do that.

My alternative idea was to write a function that gets a dataframe and a variable, and does it. It didn't work either.

My code is:

NegativeToMissing = function(df,var)
{
  df$var[df$var < 0] = NA
}

Error in $<-.data.frame(`*tmp*`, "var", value = logical(0)) : replacement has 0 rows, data has 40

what am I doing wrong ?

Thank you.

Read ?Extract. I think what you need in there is df[[var]][ df[[var]] < 0 ] <- NA. When you use df$var, it expects a column named var in the frame, which is not the case. To reference it indirectly, the only method is with [ (which should return a single-column frame) or [[ (which always returns a vector). — r2evans
– r2evans, Commented Dec 13, 2018 at 9:47
Please provide a reproducible example along with expected output. — Ronak Shah
– Ronak Shah, Commented Dec 13, 2018 at 10:08

markus · Accepted Answer · 2018-12-13 10:10:55Z

1

Here is an example with some dummy data.

df1 <- data.frame(col1 = c(-1, 1, 2, 0, -3),
                  col2 = 1:5,
                  col3 = LETTERS[1:5])
df1
#  col1 col2 col3
#1   -1    1    A
#2    1    2    B
#3    2    3    C
#4    0    4    D
#5   -3    5    E

Now find columns that are numeric

numeric_cols <- sapply(df1, is.numeric)

And replace negative values

df1[numeric_cols] <- lapply(df1[numeric_cols], function(x) replace(x, x < 0 , NA))
df1
#  col1 col2 col3
#1   NA    1    A
#2    1    2    B
#3    2    3    C
#4    0    4    D
#5   NA    5    E

You could also do

df1[df1 < 0] <- NA

answered Dec 13, 2018 at 10:10

markus

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2 Comments

user2899944 Over a year ago

I understand why you used sapply at first, but why lapply later. Doesn't it return a list ?

markus Over a year ago

@user2899944 Yes. A data.frame is a list too (with elements of equal length) so we can use lapply here to replace the values of the numeric_cols. But you could also use sapply or vapply instead.

akrun · Accepted Answer · 2018-12-13 10:16:04Z

0

With tidyverse, we can make use of mutate_if

library(tidyverse)
df1 %>%
    mutate_if(is.numeric, funs(replace(., . < 0, NA)))

answered Dec 13, 2018 at 10:16

akrun

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Comments

RLave · Accepted Answer · 2018-12-13 10:17:33Z

0

If you still want to change only one selected variable a solution withdplyr would be to use non-standard evaluation:

library(dplyr)
NegativeToMissing <- function(df, var) {
  quo_var = quo_name(var)
  df %>% 
    mutate(!!quo_var := ifelse(!!var < 0, NA, !!var))

}

NegativeToMissing(data, var=quo(val1)) # use quo() function without ""
#   val1 val2
# 1    1    1
# 2   NA    2
# 3    2    3

Data used:

data <- data.frame(val1 = c(1, -1, 2),
                   val2 = 1:3)
data
#   val1 val2
# 1    1    1
# 2   -1    2
# 3    2    3

answered Dec 13, 2018 at 10:17

RLave

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