Home/ Questions/How to avoid the error in x[floor(d)] + x[ceiling(d)] : non-numeric argument to binary operator?
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Tom
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Asked: 2022-07-02T15:15:59+00:00 In:

How to avoid the error in x[floor(d)] + x[ceiling(d)] : non-numeric argument to binary operator?

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 The error in x[floor(d)] + x[ceiling(d)] : non-numeric argument to binary operator appears when I run my program:

out.liar <- function(variable) { # arguments can be called whatever
remove <- boxplot(variable, plot = F)$out
variable[variable %in% remove] <- NA
return(variable)
}

When I pass a character vector to the out.liar() function, it throws an error.

out.liar(letters)
## Error in x[floor(d)] + x[ceiling(d)]: non-numeric argument to binary operator

Please tell me the cause and how to fix the above error.

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1 Answer

  1. Best Answer
    Expert

    The cause:  The error in x[floor(d)] + x[ceiling(d)] : non-numeric argument to binary operator happened was that the letter vector was sent from out.liar() to boxplot(), which then passed it on to the + function. The code failed because you cannot add letters, which broke boxplot() and out.liar().

    Solution: You could also want to add the ability to locate outliers in a data frame to the function. To accomplish that, you must first make the function recognize which columns in the provided data frame are numerical before executing the code on each of these columns. The function should then return the full data frame, with the numeric columns updated and the non-numeric columns.

    out.liar <- function(variable) { # arguments can be called whatever
if (class(variable) == "data.frame") {
# identify numeric variables by applying the is.numeric
# function to each column of the data frame
num_vars <- unlist(lapply(variable, is.numeric))
# run the out.liar function on each numeric column
# this kind of recurrence where you run a function inside of itself
# is allowed and often used!
variable[num_vars] <- lapply(variable[num_vars], out.liar)
} else {
remove <- boxplot(variable, plot = F)$out
variable[variable %in% remove] <- NA
}
return(variable)
}

    Check to see whether it functions, make a data frame first:

    df <- data.frame(id = factor(1:10), var1 = c(100, rnorm(9)),
var2 = c(rchisq(9, 4), 100), var3 = LETTERS[1:10])
df
    ## id var1 var2 var3
## 1 1 100.00000000 3.9174945 A
## 2 2 0.47461824 2.5883314 B
## 3 3 -1.76634605 1.6285662 C
## 4 4 -0.04140268 3.5953417 D
## 5 5 -0.74529598 8.3492440 E
## 6 6 0.55631124 0.2957526 F
## 7 7 0.57117284 2.8894209 G
## 8 8 2.92466663 12.4805552 H
## 9 9 0.45180224 2.5062931 I
## 10 10 -1.31348905 100.0000000 J

    Followed by feeding df to out.liar():

    out.liar(df)
    ## id var1 var2 var3
## 1 1 NA 3.9174945 A
## 2 2 0.47461824 2.5883314 B
## 3 3 -1.76634605 1.6285662 C
## 4 4 -0.04140268 3.5953417 D
## 5 5 -0.74529598 8.3492440 E
## 6 6 0.55631124 0.2957526 F
## 7 7 0.57117284 2.8894209 G
## 8 8 NA 12.4805552 H
## 9 9 0.45180224 2.5062931 I
## 10 10 -1.31348905 NA J

    Finally, verify that the adjustment does not affect the function’s ability to work on single columns:

    x <- rnorm(20, 9, 2)
x[2] <- 1520
out.liar(x)
    ## [1] 11.227530 NA 11.122428 9.046352 9.047519 9.011521 5.715875
## [8] 7.355825 8.852045 10.386154 7.597622 8.540947 6.287205 8.310314
## [15] 7.004540 11.916455 11.264865 10.769398 5.298348 7.785135

     

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