Hi all, 
Thank you for your responses. You are correct that it is not a matrix. I used 
the incorrect term. 
I meant I put my data in a spreadsheet with three rows and 24 columns. 

Sent from my iPhone

> On Jan 28, 2019, at 3:36 AM, Jim Lemon <drjimle...@gmail.com> wrote:
> 
> Hi Halllie,
> As Jeff noted, a data frame is not a matrix (it is a variety of list),
> so that looks like your problem.
> 
> hkdf<-data.frame(sample(3:5,4,TRUE),sample(1:3,4,TRUE),sample(2:4,4,TRUE),
> sample(3:5,4,TRUE),sample(1:3,4,TRUE),sample(2:4,4,TRUE))
> library(irr)
> kripp.alpha(hkdf)
> kripp.alpha(as.matrix(hkdf))
> 
> Jim
> 
>> On Mon, Jan 28, 2019 at 6:04 PM Hallie Kamesch <hallie.kame...@gmail.com> 
>> wrote:
>> 
>> Hi -
>> I'm trying to run Krippendorff's alpha for data consisting of 4 subjects
>> rated on 6 events each by three raters.  The ratings are interval ratio
>> scale data.
>> 
>> I've rearranged my data into a 3 x 24  of ratersXevents. (per this
>> discussion on CrossValidated: (
>> https://stats.stackexchange.com/questions/255164/inter-rater-reliability-for-binomial-repeated-ratings-from-two-or-more-raters/256144#256144)
>> ).
>> 
>> This is the code I've used:
>> library(irr)
>> dat <- read.csv(file.choose(), header = TRUE)
>> head(dat)
>> kripp.alpha(dat, method=c("ratio"))
>> #### error message: Error in sort.list(y) : 'x' must be atomic for
>> 'sort.list'
>> Have you called 'sort' on a list?
>> kripp.alpha(dat,"ratio")
>> #### error message: Error in sort.list(y) : 'x' must be atomic for
>> 'sort.list'
>> Have you called 'sort' on a list?
>> 
>> I read rhelp on sort, but I'm still confused.  Please help!
>> Thank you!
>> 
>> PS
>> I arranged my data in that matrix based upon this comment and response from
>> the CrossValidated posting forum (
>> https://stats.stackexchange.com/questions/255164/inter-rater-reliability-for-binomial-repeated-ratings-from-two-or-more-raters/256144#256144),
>> but my question above was rejected there.
>> 
>>        [[alternative HTML version deleted]]
>> 
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