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

Frank E Harrell Jr f.harrell at vanderbilt.edu
Wed Nov 26 14:38:32 CET 2008

```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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