That worked perfectly, thanks!!
I'd assumed that it wasn't necessary for me to type out "...based on a MV
normal distribution that I made into 1-7 integers by blah blah blah"; the
distribution is normal-ish after I transform it around for the purposes I need
for the simulations. I'll be sure to be more accurate/specific in any future
posts ;-)
Thanks!
Mike
On Oct 17, 2011, at 6:24 PM, Rolf Turner wrote:
> On 18/10/11 10:35, Michael Parent wrote:
>> Hi, all,
>>
>> I'm running a monte carlo simulation with missing data. The data are
>> arranged such that there are k columns and n rows over a set number of
>> simulations (set to 10 right now so it runs fast while I set everything up).
>> The data are integers, numbers 1-7 only (normal distribution).
>
> For CRYING OUT LOUD. This sort of blithering nonsense makes me
> want to SCREAM!!! The normal distribution is a continuous distribution.
> It does not take on (exclusively) integer values.
>
>> The simulations are set up and run without a hitch, including imposing NA
>> missing values at a specified prevalence semi-randomly (there are not
>> allowed to be any completely empty rows).
>>
>> I'd like to replace the missing values ("NA") with the mean for the
>> non-missing items items *on that row*. I want to go through all the monte
>> carlo simulation runs that I already did (so that I'm using the same data)
>> and replace NA with the mean (e.g., if k=5 and a row has values of 3 3 NA 5
>> 5, I want to put a 4 in for NA). I also want the imputed mean values to be
>> rounded to the nearest integer.
>>
>> Does anyone have an idea for how I'd set that up? I feel like there's a
>> fairly easy way to set up searching out those NAs and replacing the the row
>> mean that is not coming to me.
>
> Let your matrix of values be "m".
>
> rv <- round(apply(m,1,mean,na.rm=TRUE))
> ij <- which(is.na(m),arr.ind=TRUE)
> m[ij] <- rv[ij[,1]]
>
> cheers,
>
> Rolf Turner
>
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