Creating a function to replace NAs from one data.frame with values from another

问题

I regularly have situations where I need to replace missing values from a data.frame with values from some other data.frame that is at a different level of aggregation. So, for example, if I have a data.frame full of county data I might replace NA values with state values stored in another data.frame. After writing the same merge... ifelse(is.na()) yada yada a few dozen times I decided to break down and write a function to do this.

Here's what I cooked up, along with an example of how I use it:

fillNaDf <- function(naDf, fillDf, mergeCols, fillCols){
 mergedDf <- merge(naDf, fillDf, by=mergeCols)
 for (col in fillCols){
   colWithNas <- mergedDf[[paste(col, "x", sep=".")]]
   colWithOutNas <- mergedDf[[paste(col, "y", sep=".")]]
   k <- which( is.na( colWithNas ) )
   colWithNas[k] <- colWithOutNas[k]
   mergedDf[col] <- colWithNas
   mergedDf[[paste(col, "x", sep=".")]] <- NULL
   mergedDf[[paste(col, "y", sep=".")]] <- NULL
 }
 return(mergedDf)
}

## test case
fillDf <- data.frame(a = c(1,2,1,2), b = c(3,3,4,4) ,f = c(100,200, 300, 400), g = c(11, 12, 13, 14))
naDf <- data.frame( a = sample(c(1,2), 100, rep=TRUE), b = sample(c(3,4), 100, rep=TRUE), f = sample(c(0,NA), 100, rep=TRUE), g = sample(c(0,NA), 200, rep=TRUE) )
fillNaDf(naDf, fillDf, mergeCols=c("a","b"), fillCols=c("f","g") )

So after I got this running I had this odd feeling that someone has probably solved this problem before me and in a much more elegant way. Is there a better/easier/faster solution to this problem? Also, is there a way that eliminates the loop in the middle of my function? That loop is there because I am often replacing NAs in more than one column. And, yes, the function assumes the columns we're filling from are named the same and the columns we are filling to and the same applies to the merge.

Any guidance or refactoring would be helpful.

EDIT on Dec 2 I realized I had logic flaws in my example which I fixed.

回答1:

What a great question.

Here's a data.table solution:

# Convert data.frames to data.tables (i.e. data.frames with extra powers;)
library(data.table)
fillDT <- data.table(fillDf, key=c("a", "b"))
naDT <- data.table(naDf, key=c("a", "b"))


# Merge data.tables, based on their keys (columns a & b)
outDT <- naDT[fillDT]    
#      a b  f  g f.1 g.1
# [1,] 1 3 NA  0 100  11
# [2,] 1 3 NA NA 100  11
# [3,] 1 3 NA  0 100  11
# [4,] 1 3  0  0 100  11
# [5,] 1 3  0 NA 100  11
# First 5 rows of 200 printed.

# In outDT[i, j], on the following two lines 
#   -- i is a Boolean vector indicating which rows will be operated on
#   -- j is an expression saying "(sub)assign from right column (e.g. f.1) to 
#        left column (e.g. f)
outDT[is.na(f), f:=f.1]
outDT[is.na(g), g:=g.1]

# Just keep the four columns ultimately needed   
outDT <- outDT[,list(a,b,g,f)]
#       a b  g   f
#  [1,] 1 3  0   0
#  [2,] 1 3 11   0
#  [3,] 1 3  0   0
#  [4,] 1 3 11   0
#  [5,] 1 3 11   0
# First 5 rows of 200 printed.

回答2:

Here's a slightly more concise/robust version of your approach. You could replace the for-loop with a call to lapply, but I find the loop easier to read.

This function assumes any columns not in mergeCols are fair game to have their NAs filled. I'm not really sure this helps, but I'll take my chances with the voters.

fillNaDf.ju <- function(naDf, fillDf, mergeCols) {
  mergedDf <- merge(fillDf, naDf, by=mergeCols, suffixes=c(".fill",""))
  dataCols <- setdiff(names(naDf),mergeCols)
  # loop over all columns we didn't merge by
  for(col in dataCols) {
    rows <- is.na(mergedDf[,col])
    # skip this column if it doesn't contain any NAs
    if(!any(rows)) next
    rows <- which(rows)
    # replace NAs with values from fillDf
    mergedDf[rows,col] <- mergedDf[rows,paste(col,"fill",sep=".")]
  }
  # don't return ".fill" columns
  mergedDf[,names(naDf)]
}

回答3:

My preference would be to pull out the code from merge that does the matching and do it myself so that I could keep the ordering of the original data frame intact, both row-wise and column-wise. I also use matrix indexing to avoid any loops, though to do so I create a new data frame with the revised fillCols and replace the columns of the original with it; I thought I could fill it in directly but apparently you can't use matrix ordering to replace parts of a data.frame, so I wouldn't be surprised if a loop over the names would be faster in some situations.

With matrix indexing:

fillNaDf <- function(naDf, fillDf, mergeCols, fillCols) {
  fillB <- do.call(paste, c(fillDf[, mergeCols, drop = FALSE], sep="\r"))
  naB <- do.call(paste, c(naDf[, mergeCols, drop = FALSE], sep="\r"))
  na.ind <- is.na(naDf[,fillCols])
  fill.ind <- cbind(match(naB, fillB)[row(na.ind)[na.ind]], col(na.ind)[na.ind])
  naX <- naDf[,fillCols]
  fillX <- fillDf[,fillCols]
  naX[na.ind] <- fillX[fill.ind]
  naDf[,colnames(naX)] <- naX
  naDf
}

With a loop:

fillNaDf2 <- function(naDf, fillDf, mergeCols, fillCols) {
  fillB <- do.call(paste, c(fillDf[, mergeCols, drop = FALSE], sep="\r"))
  naB <- do.call(paste, c(naDf[, mergeCols, drop = FALSE], sep="\r"))
  m <- match(naB, fillB)
  for(col in fillCols) {
    fix <- which(is.na(naDf[,col]))
    naDf[fix, col] <- fillDf[m[fix],col]
  }
  naDf
}

来源：https://stackoverflow.com/questions/8349909/creating-a-function-to-replace-nas-from-one-data-frame-with-values-from-another

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