Argument Is of Length Zero in If Statement

Argument is of length zero in if statement

"argument is of length zero" is a very specific problem that comes from one of my least-liked elements of R. Let me demonstrate the problem:

> FALSE == "turnip"
[1] FALSE
> TRUE == "turnip"
[1] FALSE
> NA == "turnip"
[1] NA
> NULL == "turnip"
logical(0)

As you can see, comparisons to a NULL not only don't produce a boolean value, they don't produce a value at all - and control flows tend to expect that a check will produce some kind of output. When they produce a zero-length output... "argument is of length zero".

(I have a very long rant about why this infuriates me so much. It can wait.)

So, my question; what's the output of sum(is.null(data[[k]]))? If it's not 0, you have NULL values embedded in your dataset and will need to either remove the relevant rows, or change the check to

if(!is.null(data[[k]][[k2]]) & temp > data[[k]][[k2]]){
    #do stuff
}

Hopefully that helps; it's hard to tell without the entire dataset. If it doesn't help, and the problem is not a NULL value getting in somewhere, I'm afraid I have no idea.

Error in if/while (condition) { : argument is of length zero

See ?NULL

You have to use is.null

‘is.null’ returns ‘TRUE’ if its argument is ‘NULL’ and ‘FALSE’
otherwise.

Try this:

if ( is.null(hic.data[[z]]) ) { print("is null")}

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From section 2.1.6 of the R Language Definition

There is a special object called NULL. It is used whenever there is a need to indicate or
specify that an object is absent. It should not be confused with a vector or list of zero
length.
The NULL object has no type and no modifiable properties. There is only one NULL object
in R, to which all instances refer. To test for NULL use is.null. You cannot set attributes
on NULL.

R argument is of length zero when using names(vector) in IF statement

Your function uses the line

Rvec = as.vector(Rvec)

The help documentation for as.vector() states that "if the result is atomic, all attributes are removed." This means that element names are dropped when you run as.vector().

One solution is not to coerce Rvec to be a vector, but rather to check that it is a vector. You'll also need to change the if() statement in your third line to either an ifelse() or an all.equal(). Here's one solution with those 2 fixes (replace the ... with the rest of your code):

check.names <- function(Rvec,Data){ 
    if(!is.vector(Rvec)) stop("Rvec must be a vector")
    if(!is.data.frame(Data)) stop("Data must be a data.frame")
    
    dnames <- names(Data)[3:ncol(Data)]
    rnames <- names(Rvec)[3:length(Rvec)]
    
    if(!all.equal(dnames, rnames)) stop("Names don't match")
    ...
}

R argument of length zero in if statement

The issue is most likely due to creation of unused levels because by default drop = FALSE in interaction

df$id_year <- interaction(df$id, df$year, drop = TRUE)

In the loop, as others have mentioned the !unique( is not clear. If we are checking the length of unique elements are greater than 0 (if (length(unique(df$main[rows])) > 0), it would print "yes"

for (i in levels(df$id_year)) {
   rows <- df$id_year == i & df$usable
   if (length(unique(df$main[rows])) > 0) {
     print("yes")
   }
 }
#[1] "yes"
#[1] "yes"
#[1] "yes"

argument is of length zero in if statement

Your query can be illustrated with the following example:

grep(pattern="W","huh")
# integer(0)

No match results in a vector of length 0, hence the error. Instead use grepl, i.e. if( grepl( "W" , y ) ).

grepl has the return value TRUE or FALSE.

As a side note, eval( parse( "sometext" ) ) is variously thought of as not a good idea. You could try using the following untidy lapply statement instead (which will be better than apply because you don't have to convert to a matrix first):

data.frame( lapply( data , function(x) 
                                ifelse( grepl("W",x) , 
                                        as.integer( gsub("W","",x) ) * 2L , 
                                        x ) ) )
#  V1 V2 V3 V4
#1  2 16  4 16
#2  1 16  4  0
#3  2 16  1  0
#4  3 64  3  0

Error in if condition {: argument is of length zero

I'd suggest researching dataframe subsetting as your missing out on a substantial benefit of r.

df$Status <- 0
df$Status[df$End_week == 1426] <- 1

r if statement meet error: argument is of length zero

the error is because grep returns logical(0) if the string you are looking for is not present. So your loop fails on i=2, as you can see when you look at the value of i when the loop breaks.

If you use grepl in stead, your loop works as planned (building on @Akarsh Jain s answer):

movieID<-array() 
customerID<-array()

for (i in 1:length(testData)){

  if(grepl(':', testData[i])){
    movieID[i]<-as.integer(substr(testData[i],1,nchar(testData[i])-1)  )
    next
  } else{
    customerID[i]<-as.integer(testData[i])

  }
}

ofcourse, the question is how useful this is. I assume you want to somehow split your data on movieID, which you can do easily using dplyr and tidyr:

library(dplyr)
library(tidyr)
#put your testData in a dataframe
testDf <- data.frame(customerID = testData)

newDf <- testDf %>% 
#identify rows with :
         mutate(movieID = ifelse(grepl(":",customerID), customerID, NA)) %>%
#fill all NA values in movieID with the previous non-NA value:         
         fill(movieID) %>%
#remove lines where customerID has a ":":
         filter(!grepl(":",customerID))

output:

    customerID movieID
1    30878       1
2  2647871       1
3  1283744       1

dummy data

testData <- read.table(text='1:
30878
                                 2647871
                                 1283744
                                 2488120
                                 317050
                                 1904905
                                 1989766
                                 14756
                                 1027056
                                 1149588
                                 1394012
                                 1406595
                                 2529547
                                 1682104
                                 2625019
                                 2603381
                                 1774623
                                 470861
                                 712610
                                 1772839
                                 1059319
                                 2380848
                                 548064
                                 10:
                                 1952305
                                 1531863
                                 1000:
                                 2326571
                                 977808
                                 1010534
                                 1861759
                                 79755
                                 98259
                                 1960212
                                 97460
                                 2623506
                                 2409123', stringsAsFactors=FALSE)[[1]]

if statement get rid of: Error in if: argument is of length zero

The answer including browser() function was indeed helpful for locating the source of the problem but the real solution is to use req() or a combination of validate() and need() functions. See respective help() pages to learn how to use them.

In short what these functions do is to check whether the argument given has been specified. In reference to the example code the solution could look like this:

server <- function(input, output, session){
  output$output <- renderUI({
    # check if input$input is specified
    # if not stop executing the rest of this code block
    req(input$input)
    div(
      # show name
      data[which(input$input == data[, "name"]), "name"], br(),
      data[which(input$input == data[, "name"]), "age"], br(),
      # if "coward" is not FALSE than show its value
      if(data[which(input$input == data[, "name"]), "coward"] != FALSE){
        data[which(input$input == data[, "name"]), "coward"]
      }
    )
  })
}

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