# Re: [R] multiple imputation with fit.mult.impute in Hmisc - how to replace NA with imputed value?

From: Frank E Harrell Jr <f.harrell_at_vanderbilt.edu>
Date: Wed, 26 Nov 2008 07:38:32 -0600

Charlie Brush wrote:
> I am doing multiple imputation with Hmisc, and
> can't figure out how to replace the NA values with
> the imputed values.
>
> Here's a general ourline of the process:
>
> > set.seed(23)
> > library("mice")
> > library("Hmisc")
> > library("Design")
> > length(d);length(d[,1])
> [1] 43
> [1] 2666
> Do for this data set, there are 43 columns and 2666 rows
>
> Here is a piece of data.frame d:
> > d[1:20,4:6]
> P01 P02 P03
> 1 0.1 0.16 0.16
> 2 NA 0.00 0.00
> 3 NA 0.60 0.04
> 4 NA 0.15 0.00
> 5 NA 0.00 0.00
> 6 0.7 0.00 0.75
> 7 NA 0.00 0.00
> 8 NA 0.00 0.00
> 9 0.0 0.00 0.00
> 10 0.0 0.00 0.00
> 11 0.0 0.00 0.00
> 12 0.0 0.00 0.00
> 13 0.0 0.00 0.00
> 14 0.0 0.00 0.00
> 15 0.0 0.00 0.03
> 16 NA 0.00 0.00
> 17 NA 0.01 0.00
> 18 0.0 0.00 0.00
> 19 0.0 0.00 0.00
> 20 0.0 0.00 0.00
>
> These are daily precipitation values at NCDC stations, and
> NA values at station P01 will be filled using multiple
> imputation and data from highly correlated stations P02 and P08:
>
> > f <- aregImpute(~ I(P01) + I(P02) + I(P08),
> n.impute=10,match='closest',data=d)
> Iteration 13
> > fmi <- fit.mult.impute( P01 ~ P02 + P08 , ols, f, d)
>
> Variance Inflation Factors Due to Imputation:
>
> Intercept P02 P08
> 1.01 1.39 1.16
>
> Rate of Missing Information:
>
> Intercept P02 P08
> 0.01 0.28 0.14
>
> d.f. for t-distribution for Tests of Single Coefficients:
>
> Intercept P02 P08
> 242291.18 116.05 454.95
> > r <- apply(f\$imputed\$P01,1,mean)
> > r
> 2 3 4 5 7 8 16 17 249 250 251
> 0.002 0.430 0.044 0.002 0.002 0.002 0.002 0.123 0.002 0.002 0.002
> 252 253 254 255 256 257 258 259 260 261 262
> 1.033 0.529 1.264 0.611 0.002 0.513 0.085 0.002 0.705 0.840 0.719
> 263 264 265 266 267 268 269 270 271 272 273
> 1.489 0.532 0.150 0.134 0.002 0.002 0.002 0.002 0.002 0.055 0.135
> 274 275 276 277 278 279 280 281 282 283 284
> 0.009 0.002 0.002 0.002 0.008 0.454 1.676 1.462 0.071 0.002 1.029
> 285 286 287 288 289 418 419 420 421 422 700
> 0.055 0.384 0.947 0.002 0.002 0.008 0.759 0.066 0.009 0.002 0.002
>
> ------------------------------------------------------------------
> So far, this is working great.
> Now, make a copy of d:
> > dnew <- d
>
> And then fill in the NA values in P01 with the values in r
>
> For example:
> > for (i in 1:length(r)){
> dnew\$P01[r[i,1]] <- r[i,2]
> }
> This doesn't work, because each 'piece' of r is two numbers:
> > r[1]
> 2
> 0.002
> > r[1,1]
> Error in r[1, 1] : incorrect number of dimensions
>
> My question: how can I separate the the two items in (for example)
> r[1] to use the first part as an index and the second as a value,
> and then use them to replace the NA values with the imputed values?
>
> Or is there a better way to replace the NA values with the imputed values?
>
> Thanks in advance for any help.
>

You didn't state your goal, and why fit.mult.impute does not do what you want. But you can look inside fit.mult.impute to see how it retrieves the imputed values. Also see the example in documentation for transcan in which the command impute(xt, imputation=1) to retrieve one of the multiple imputations.

Note that you can say library(Design) (omit the quotes) to access both Design and Hmisc.

Frank

```--
Frank E Harrell Jr   Professor and Chair           School of Medicine
Department of Biostatistics   Vanderbilt University

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