Solved – KNN imputation R packages

data-imputationk nearest neighbourr

I am looking for a KNN imputation package. I have been looking at imputation package (http://cran.r-project.org/web/packages/imputation/imputation.pdf) but for some reason the KNN impute function (even when following the example from the description) only seems to impute zero values (as per below). I have been looking around but cannot find something yet, and hence was wondering if anyone has other suggestions for good KNN imputation packages?

In the code per below – the NA values are replaced by zero's – not by the Knn mean value

require(imputation)
x = matrix(rnorm(100),10,10)
x.missing = x > 1
x[x.missing] = NA
kNNImpute(x, 3)
x

Best Answer

You could also try the following package: DMwR.

It failed on the case of 3 NN, giving 'Error in knnImputation(x, k = 3) : Not sufficient complete cases for computing neighbors.'

However, trying 2 gives.

> knnImputation(x,k=2)
             [,1]       [,2]       [,3]       [,4]       [,5]        [,6]
 [1,] -0.59091360 -1.2698175  0.5556009 -0.1327224 -0.8325065  0.71664000
 [2,] -1.27255074 -0.7853602  0.7261897  0.2969900  0.2969556 -0.44612831
 [3,]  0.55473981  0.4748735  0.5158498 -0.9493917 -1.5187722 -0.99377854
 [4,] -0.47797654  0.1647818  0.6167311 -0.5149731  0.5240514 -0.46027809
 [5,] -1.08767831 -0.3785608  0.6659499 -0.7223724 -0.9512409 -1.60547053
 [6,] -0.06153279  0.9486815 -0.5464601  0.1544475  0.2835521 -0.82250221
 [7,] -0.82536029 -0.2906253 -3.0284281 -0.8473210  0.7985286 -0.09751927
 [8,] -1.15366189  0.5341000 -1.0109258 -1.5900281  0.2742328  0.29039928
 [9,] -1.49504465 -0.5419533  0.5766574 -1.2412777 -1.4089572 -0.71069839
[10,] -0.35935440 -0.2622265  0.4048126 -2.0869817  0.2682486  0.16904559
             [,7]       [,8]        [,9]      [,10]
 [1,]  0.58027159 -1.0669137  0.48670802  0.5824858
 [2,] -0.48314440 -1.0532693 -0.34030385 -1.1041681
 [3,] -2.81996446  0.3191438 -0.48117020 -0.0352633
 [4,] -0.55080515 -1.0620243 -0.51383557  0.3161907
 [5,] -0.56808769 -0.3696951  0.35549191  0.3202675
 [6,] -0.25043479 -1.0389393  0.07810902  0.5251606
 [7,] -0.41667318  0.8809541 -0.04613332 -1.1586756
 [8,] -0.06898363 -1.0736161  0.62698065 -1.0373835
 [9,]  0.30051583 -0.2936140  0.31417921 -1.4155193
[10,] -0.68180034 -1.0789745  0.58290920 -1.0197956

You can test for sufficient observations using complete.cases(x), where that value must be at least k.

One way to overcome this problem is to relax your requirements (i.e. less incomplete rows), by 1) increasing the NA threshold, or alternatively, 2) increasing your number of observations.

Here is the first:

> x = matrix(rnorm(100),10,10)
> x.missing = x > 2
> x[x.missing] = NA
> complete.cases(x)
 [1]  TRUE  TRUE  TRUE FALSE FALSE  TRUE  TRUE  TRUE  TRUE  TRUE
> knnImputation(x,k=3)
             [,1]       [,2]       [,3]       [,4]        [,5]       [,6]       [,7]        [,8]        [,9]       [,10]
 [1,]  0.86882569 -0.2409922  0.3859031  0.5818927 -1.50310330  0.8752261 -0.5173105 -2.18244988 -0.28817656 -0.63941237
 [2,]  1.54114079  0.7227511  0.7856277  0.8512048 -1.32442954 -2.1668744  0.7017532 -0.40086348 -0.41251883  0.42924986
 [3,]  0.60062917 -0.5955623  0.6192783 -0.3836310  0.06871570  1.7804657  0.5965411 -1.62625036  1.27706937  0.72860273
 [4,] -0.07328279 -0.1738157  1.4965579 -1.1686115 -0.06954318 -1.0171604 -0.3283916  0.63493884  0.72039689 -0.20889111
 [5,]  0.78747874 -0.8607320  0.4828322  0.6558960 -0.22064430  0.2001473  0.7725701  0.06155196  0.09011719 -1.01902968
 [6,]  0.17988720 -0.8520000 -0.5911523  1.8100573 -0.56108621  0.0151522 -0.2484345 -0.80695513 -0.18532984 -1.75115335
 [7,]  1.03943492  0.4880532 -2.7588922 -0.1336166 -1.28424057  1.2871333  0.7595750 -0.55615677 -1.67765572 -0.05440992
 [8,]  1.12394474  1.4890366 -1.6034648 -1.4315445 -0.23052386 -0.3536677 -0.8694188 -0.53689507 -1.11510406 -1.39108817
 [9,] -0.30393916  0.6216156  0.1559639  1.2297105 -0.29439390  1.8224512 -0.4457441 -0.32814665  0.55487894 -0.22602598
[10,]  1.18424722 -0.1816049 -2.2975095 -0.7537477  0.86647524 -0.8710603  0.3351710 -0.79632184 -0.56254688 -0.77449398
> x
             [,1]       [,2]       [,3]       [,4]       [,5]       [,6]       [,7]        [,8]        [,9]       [,10]
 [1,]  0.86882569 -0.2409922  0.3859031  0.5818927 -1.5031033  0.8752261 -0.5173105 -2.18244988 -0.28817656 -0.63941237
 [2,]  1.54114079  0.7227511  0.7856277  0.8512048 -1.3244295 -2.1668744  0.7017532 -0.40086348 -0.41251883  0.42924986
 [3,]  0.60062917 -0.5955623  0.6192783 -0.3836310  0.0687157  1.7804657  0.5965411 -1.62625036  1.27706937  0.72860273
 [4,] -0.07328279 -0.1738157  1.4965579 -1.1686115         NA -1.0171604 -0.3283916  0.63493884  0.72039689 -0.20889111
 [5,]  0.78747874 -0.8607320  0.4828322         NA -0.2206443  0.2001473  0.7725701  0.06155196  0.09011719 -1.01902968
 [6,]  0.17988720 -0.8520000 -0.5911523  1.8100573 -0.5610862  0.0151522 -0.2484345 -0.80695513 -0.18532984 -1.75115335
 [7,]  1.03943492  0.4880532 -2.7588922 -0.1336166 -1.2842406  1.2871333  0.7595750 -0.55615677 -1.67765572 -0.05440992
 [8,]  1.12394474  1.4890366 -1.6034648 -1.4315445 -0.2305239 -0.3536677 -0.8694188 -0.53689507 -1.11510406 -1.39108817
 [9,] -0.30393916  0.6216156  0.1559639  1.2297105 -0.2943939  1.8224512 -0.4457441 -0.32814665  0.55487894 -0.22602598
[10,]  1.18424722 -0.1816049 -2.2975095 -0.7537477  0.8664752 -0.8710603  0.3351710 -0.79632184 -0.56254688 -0.77449398

here is an example of the 2nd...

x = matrix(rnorm(1000),100,10)
x.missing = x > 1
x[x.missing] = NA

complete.cases(x)

  [1]  TRUE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE FALSE FALSE
 [22] FALSE FALSE  TRUE FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 [43]  TRUE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE FALSE FALSE  TRUE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 [64] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE
 [85] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE FALSE

At least k=3 complete rows are satisfied, thus it is able to impute for k=3.

> head(knnImputation(x,k=3))
            [,1]       [,2]       [,3]       [,4]       [,5]       [,6]       [,7]       [,8]        [,9]       [,10]
[1,]  0.01817557 -2.8141502  0.3929944  0.1495092 -1.7218396  0.4159133 -0.8438809  0.6599224 -0.02451113 -1.14541016
[2,]  0.51969964 -0.4976021 -0.1495392 -0.6448184 -0.6066386 -1.6210476 -0.3118440  0.2477855 -0.30986749  0.32424673
...

Related Solutions

Logistic Regression vs Multinomial Regression – Differences and Use Cases Explained

With a multinomial logit model you impose the constraint that all the predicted probabilities add up to 1. When you use separate binary logit model you can no longer impose that constraint, they are estimated in seperate models after all. So that would be the main difference between these two models.

As you can see in the example below (In Stata, as that is the program I know best), the models tend to be similar but not the same. I would be especially careful about extrapolating predicted probabilities.

// some data preparation
. sysuse nlsw88, clear                                                               
(NLSW, 1988 extract)                                                                 

.                                                                                    
. gen byte occat = cond(occupation < 3                 , 1,      ///                 
>                  cond(inlist(occupation, 5, 6, 8, 13), 2, 3))  ///                 
>                  if !missing(occupation)                                           
(9 missing values generated)                                                         

. label variable occat "occupation in categories"                                    

. label define occat 1 "high"   ///                                                  
>                    2 "middle" ///                                                  
>                    3 "low"                                                         

. label value occat occat                                                            

.                                                                                    
. gen byte middle = (occat == 2) if occat !=1 & !missing(occat)                      
(590 missing values generated)                                                       

. gen byte high   = (occat == 1) if occat !=2 & !missing(occat)                      
(781 missing values generated)                                                       


// a multinomial logit model
. mlogit occat i.race i.collgrad , base(3) nolog                                     

Multinomial logistic regression                   Number of obs   =       2237       
                                                  LR chi2(6)      =     218.82       
                                                  Prob > chi2     =     0.0000       
Log likelihood = -2315.9312                       Pseudo R2       =     0.0451       

-------------------------------------------------------------------------------      
        occat |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]      
--------------+----------------------------------------------------------------      
high          |                                                                      
         race |                                                                      
       black  |  -.4005801   .1421777    -2.82   0.005    -.6792433    -.121917      
       other  |   .4588831   .4962591     0.92   0.355    -.5137668    1.431533      
              |                                                                      
     collgrad |                                                                      
college grad  |   1.495019   .1341625    11.14   0.000     1.232065    1.757972      
        _cons |  -.7010308   .0705042    -9.94   0.000    -.8392165   -.5628451      
--------------+----------------------------------------------------------------      
middle        |                                                                      
         race |                                                                      
       black  |   .6728568   .1106792     6.08   0.000     .4559296     .889784      
       other  |   .2678372    .509735     0.53   0.599    -.7312251    1.266899      
              |                                                                      
     collgrad |                                                                      
college grad  |    .976244   .1334458     7.32   0.000      .714695    1.237793      
        _cons |   -.517313   .0662238    -7.81   0.000    -.6471092   -.3875168      
--------------+----------------------------------------------------------------      
low           |  (base outcome)                                                      
-------------------------------------------------------------------------------      

// separate logits:
. logit high   i.race i.collgrad , nolog                                             

Logistic regression                               Number of obs   =       1465       
                                                  LR chi2(3)      =     154.21       
                                                  Prob > chi2     =     0.0000       
Log likelihood = -906.79453                       Pseudo R2       =     0.0784       

-------------------------------------------------------------------------------      
         high |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]      
--------------+----------------------------------------------------------------      
         race |                                                                      
       black  |  -.5309439   .1463507    -3.63   0.000     -.817786   -.2441017      
       other  |   .2670161   .5116686     0.52   0.602     -.735836    1.269868      
              |                                                                      
     collgrad |                                                                      
college grad  |   1.525834   .1347081    11.33   0.000     1.261811    1.789857      
        _cons |  -.6808361   .0694323    -9.81   0.000     -.816921   -.5447512      
-------------------------------------------------------------------------------      

. logit middle i.race i.collgrad , nolog                                             

Logistic regression                               Number of obs   =       1656       
                                                  LR chi2(3)      =      90.13       
                                                  Prob > chi2     =     0.0000       
Log likelihood = -1098.9988                       Pseudo R2       =     0.0394       

-------------------------------------------------------------------------------      
       middle |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]      
--------------+----------------------------------------------------------------      
         race |                                                                      
       black  |   .6942945   .1114418     6.23   0.000     .4758725    .9127164      
       other  |   .3492788   .5125802     0.68   0.496    -.6553598    1.353918      
              |                                                                      
     collgrad |                                                                      
college grad  |   .9979952   .1341664     7.44   0.000     .7350339    1.260957      
        _cons |  -.5287625   .0669093    -7.90   0.000    -.6599023   -.3976226      
-------------------------------------------------------------------------------

Solved – Fast missing data imputation in R for big data that is more sophisticated than simply imputing the means

I used mice (multiple imputation by chained equation). It's fairly fast, and quite simple. I used it on 3000 obs. for c.a. 10 variables. Done in 10min on an old computer. Further, I believe it is one of the best multiple-imputation packages out there. It can use regression to impute, among other methods.

You need to create a dataframe with the variable you want to impute, and include every variable that might predict values of that variable (so every var. in your model + possibly other var. as well). The mice package will impute every missing value in that dataframe.

Simplest way of imputing. Gives you a dataframe Datimp that has five imputed data + the original data.

library(mice)
#m=5 number of multiple imputations
#maxit=10 number of iterations. 10-20 is sufficient.
imp <- mice(Dat1, m=5, maxit=10, printFlag=TRUE) 
Datimp <- complete(imp, "long", include=TRUE)
write.table(Datimp, "C:/.../impute1.txt",
            sep="\t", dec=",", row.names=FALSE)

A better way to do this is:

library(mice)
Dat1 <- subset(Dat, select=c(id, faculty, gender, age, job, salary)) #create subset
#of variables you would like to either impute or use as predictors for imputation.
ini <- mice(Dat1, maxit=0, pri=F)
pred <- ini$pred
    pred[,c("id", "faculty")] <- 0 #variables you do not want to use as predictors (but
    #want to have in the dataset, can't add them later.
    meth <- ini$meth
meth[c("id", "faculty", "gender", "age", "job")] <- "" #choose a prediction method
#for imputing your variables. Here I don't want these variables to be imputed, so I
#choose "" (empty, no mehod).
imp <- mice(Dat1, m=5, maxit=10, printFlag=TRUE, pred=pred, meth=meth, seed=2345) 
Datimp <- complete(imp, "long", include=TRUE)
write.table(Datimp, "C:/.../impute1.txt",
            sep="\t", dec=",", row.names=FALSE)

See if your imputations were any good:

library(lattice)
com <- complete(imp, "long", inc=T)
col <- rep(c("blue","red")[1+as.numeric(is.na(imp$salary))],6)
stripplot(salary~.imp, data=com, jit=TRUE, fac=0.8, col=col, pch=20,
xlab="Imputation number",cex=0.25) 
densityplot(salary~.imp, data=com, jit=TRUE, fac=0.8, col=col, pch=20,
xlab="Imputation number",cex=0.25) 

long <- complete(imp,"long")
levels(long$.imp) <- paste("Imputation",1:22)
    long <- cbind(long, salary.na=is.na(imp$data$salary))
densityplot(~salary|.imp, data=long, group=salary, plot.points=FALSE, ref=TRUE, 
xlab="Salary",scales=list(y=list(draw=F)),
par.settings=simpleTheme(col.line=rep(c("blue","red"))), auto.key =
list(columns=2,text=c("Observed","Imputed")))

Finally, and importantly. You can't just save your new dataset and use your imputed values as normal observed values. You use pooled regression or pooled lmer ...So the uncertainty of the imputed values is taken into account.

fit1 <- with(imp, lm(salary ~ gender, na.action=na.omit))
summary(est <- pool(fit1))
pool.r.squared(fit1,adjusted=FALSE)

Best Answer

Related Solutions

Logistic Regression vs Multinomial Regression – Differences and Use Cases Explained

Solved – Fast missing data imputation in R for big data that is more sophisticated than simply imputing the means

Related Question