Iterate over columns in a data frame to replace values ​​from the corresponding data in the data frame list

Question

I am interested in creating a function using apply/ sapplyor Mapthat will iterate over the available columns in and replace the values ​​in each column with the consistent values ​​from the data frame, available in the unnamed list of data frames, with the list item index corresponding to the column number of the data frame . dta dta

Example

Defined objects:

set.seed(1)
size <- 20

# Data set
dta <-
    data.frame(
        unitA = sample(LETTERS[1:4], size = size, replace = TRUE),
        unitB = sample(letters[16:20], size = size, replace = TRUE),
        unitC = sample(month.abb[1:4], size = size, replace = TRUE),
        someValue = sample(1:1e6, size = size, replace = TRUE)
    )

# Meta data
lstMeta <- list(
    # Unit A definitions
    data.frame(
        V1 = c("A", "B", "D"),
        V2 = c("Letter A", "Letter B", "Letter D")
    ),
    # Unit B definitions
    data.frame(
        V1 = c("t", "q"),
        V2 = c("small t", "small q")
    ),
    # Unit C definitions
    data.frame(
        V1 = c("Mar", "Jan"),
        V2 = c("March", "January")
    )
)

Desired Results

When applied to, the function should return the corresponding extraction below: dtadata.frame

unitA       unitB    unitC      someValue
Letter B    small t  Apr        912876
Letter B    small q  March      293604
       C    s        Apr        459066
Letter D    p        March      332395
Letter A    small q  March      650871
Letter D    small q  Apr        258017
Letter D    p        January    478546
C           small q  Feb        766311
C           small t  March      84247
Letter A    small q  March      875322
Letter A    r        Feb        339073
Letter A    r        Ap         839441
C           r        Feb        346684
Letter B    p        January    333775
Letter D    small t  January    476352
(...)

Existing approach

replaceLbls <- function(dataSet, lstDict) {
    sapply(seq_along(dataSet), function(i) {
        # Take corresponding metadata data frame
        dtaDict <- lstDict[[i]]

        # Replace values in selected column
        # Where matches on V1 push corrsponding values from V2
        dataSet[,i][match(dataSet[,i], dtaDict[,1])] <- dtaDict[,2][match(dtaDict[,1], dataSet[,i])]  
    })
}

# Testing -----------------------------------------------------------------

replaceLbls(dataSet = dta, lstDict = lstMeta)

Of course, the approach proposed above does not work, as it will try to use NAin appointments; but he summarizes what I want to achieve:

x[...] <- m: NAs : : [<-.factor(*tmp*, match(dataSet[, i], dtaDict[, 1]), value = c(NA,: , NA

:

• ,
• ,
• someValue , ,

• dplyr/data.table/sqldf .
• for -loops

Answer 1

:

replaceLbls <- function(dataSet, lstDict) {
  dataSet[seq_along(lstDict)] <- Map(function(x, lst) {
    x <- as.character(x)
    idx <- match(x, as.character(lst$V1))
    replace(x, !is.na(idx), as.character(lst$V2)[na.omit(idx)])
  }, dataSet[seq_along(lstDict)], lstDict)
  dataSet
}


head(replaceLbls(dta, lstMeta))
#      unitA   unitB unitC someValue
# 1 Letter B small t   Apr    912876
# 2 Letter B small q March    293604
# 3        C       s   Apr    459066
# 4 Letter D       p March    332395
# 5 Letter A small q March    650871
# 6 Letter D small q   Apr    258017

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Answer 2

, for . factors characters , .

, lstMeta, , . <<-. R, , , , .

set.seed(1)
size <- 20

# Data set
dta <-
  data.frame(
    unitA = sample(LETTERS[1:4], size = size, replace = TRUE),
    unitB = sample(letters[16:20], size = size, replace = TRUE),
    unitC = sample(month.abb[1:4], size = size, replace = TRUE),
    someValue = sample(1:1e6, size = size, replace = TRUE),
    stringsAsFactors = F
  )

# Meta data
lstMeta <- list(
  # Unit A definitions
  data.frame(
    V1 = c("A", "B", "D"),
    V2 = c("Letter A", "Letter B", "Letter D"),
    stringsAsFactors = F
  ),
  # Unit B definitions
  data.frame(
    V1 = c("t", "q"),
    V2 = c("small t", "small q"),
    stringsAsFactors = F
  ),
  # Unit C definitions
  data.frame(
    V1 = c("Mar", "Jan"),
    V2 = c("March", "January"),
    stringsAsFactors = F
  )
)

replaceLbls <- function(dataSet, lstDict) {
  sapply(1:3, function(i) {
    # Take corresponding metadata data frame
    dtaDict <- lstDict[[i]]

    # Replace values in selected column
    # Where matches on V1 push corrsponding values from V2
    myUniques <- which(dataSet[,i] %in% dtaDict[,1])

    dataSet[myUniques,i]<<- dtaDict[,2][match(dataSet[myUniques,i],dtaDict[,1])]  
  })
  return(dataSet)
}

# Testing -----------------------------------------------------------------

replaceLbls(dataSet = dta, lstDict = lstMeta)

Answer 3

You can also try the following:

mapr<-function(t,meta){
    ind<-match(t,meta$V1)
    if(!is.na(ind)){return(meta$V2[ind])}
    else{return(t)}}

Then using sapply:

dta<-as.data.frame(cbind(sapply(1:3,function(t,df,meta){sapply(df[,t],mapr,lstMeta[[t]])},dta,lstMeta,simplify = T),dta[,4]))

Answer 4

The couple mapplycan do the work

f1 <- function(df, lst){
  d1 <- setNames(data.frame(mapply(function(x, y) x$V2[match(y, x$V1)], lst, df[1:3]), 
                            df$someValue, stringsAsFactors = FALSE), 
                 names(df))
  as.data.frame(mapply(function(x, y) replace(x, is.na(x), y[is.na(x)]), d1, df))

}