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R fuction composition for the substitution of values in dataframe

Stack Overflow Asked by maxbre on October 23, 2020

given the following reproducible example

my objective is to row-wise substitute the original values with NA in adjacent columns of a data frame; I know it’s a problem (with so many variants) already posted but I’ve not yet found the solution with the approach I’m trying to accomplish: i.e. by applying a function composition

in the reproducible example the column driving the substitution with NA of the original values is column a

this is what I’ve done so far

the very last code snippet is a failing attempt of what I’m actually searching for…

#-----------------------------------------------------------
# ifelse approach, it works but...
# it's error prone: i.e. copy and paste for all columns can introduce a lot of troubles

df<-data.frame(a=c(1, 2, NA), b=c(3, NA, 4), c=c(NA, 5, 6))
df

df$b<-ifelse(is.na(df$a), NA, df$b)
df$c<-ifelse(is.na(df$a), NA, df$c)

df

#--------------------------------------------------------
# extraction and subsitution approach
# same as above

df<-data.frame(a=c(1, 2, NA), b=c(3, NA, 4), c=c(NA, 5, 6))
df

df$b[is.na(df$a)]<-NA
df$c[is.na(df$a)]<-NA

df

#----------------------------------------------------------
# definition of a function
# it's a bit better, but still error prone because of the copy and paste

df<-data.frame(a=c(1, 2, NA), b=c(3, NA, 4), c=c(NA, 5, 6))
df

fix<-function(x,y){
  ifelse(is.na(x), NA, y)
}

df$b<-fix(df$a, df$b)
df$c<-fix(df$a, df$c)

df

#------------------------------------------------------------
# this approach is not working as expected!
# the idea behind is of function composition;
# lapply does the fix to some columns of data frame

df<-data.frame(a=c(1, 2, NA), b=c(3, NA, 4), c=c(NA, 5, 6))
df

fix2<-function(x){
  x[is.na(x[1])]<-NA
  x
}

df[]<-lapply(df, fix2)

df

any help for this particular approach?
I’m stuck on how to properly conceive the substitute function passed to lapply

thanx

2 Answers

Using lexical closure

If you use lexical closureing - you define a function which generates first the function you need. And then you can use this function as you wish.

# given a column all other columns' values at that row should become NA
# if the driver column's value at that row is NA

# using lexical scoping of R function definitions, one can reach that.

df<-data.frame(a=c(1, 2, NA), b=c(3, NA, 4), c=c(NA, 5, 6))
df

# whatever vector given, this vector's value should be changed
# according to first column's value

na_accustomizer <- function(df, driver_col) {
  ## Returns a function which will accustomize any vector/column
  ## to driver column's NAs
  function(vec) {
    vec[is.na(df[, driver_col])] <- NA
    vec
  }
}

df[] <- lapply(df, na_accustomizer(df, "a"))

df
##    a  b  c
## 1  1  3 NA
## 2  2 NA  5
## 3 NA NA NA

# 
# na_accustomizer(df, "a") returns
# 
#   function(vec) {
#     vec[is.na(df[, "a"])] <- NA
#     vec
#   }
# 
# which then can be used like you want:
# df[] <- lapply(df, na_accustomize(df, "a"))

Using normal functions

df<-data.frame(a=c(1, 2, NA), b=c(3, NA, 4), c=c(NA, 5, 6))
df

# define it for one column
overtake_NA <- function(df, driver_col, target_col) {
  df[, target_col] <- ifelse(is.na(df[, driver_col]), NA, df[, target_col])
  df
}

# define it for all columns of df
overtake_driver_col_NAs <- function(df, driver_col) {
  for (i in 1:ncol(df)) {
    df <- overtake_NA(df, driver_col, i)
  }
  df
}

overtake_driver_col_NAs(df, "a")
#    a  b  c
# 1  1  3 NA
# 2  2 NA  5
# 3 NA NA NA

Generalize for any predicate function

driver_col_to_other_cols <- function(df, driver_col, pred) {
  ## overtake any value of the driver column to the other columns of df,
  ## whenever predicate function (pred) is fulfilled.
  # define it for one column
  overtake_ <- function(df, driver_col, target_col, pred) {
    selectors <- do.call(pred, list(df[, driver_col]))
    if (deparse(substitute(pred)) != "is.na") {
      # this is to 'recorrect' NA's which intrude into the selector vector
      # then driver_col has NAs. For sure "is.na" is not the only possible
      # way to check for NA - so this edge case is not covered fully
      selectors[is.na(selectors)] <- FALSE
    }
    df[, target_col] <- ifelse(selectors, df[, driver_col], df[, target_col])
    df
  }
  for (i in 1:ncol(df)) {
    df <- overtake_(df, driver_col, i, pred)
  }
  df
}


driver_col_to_other_cols(df, "a", function(x) x == 1)
#    a  b c
# 1  1  1 1
# 2  2 NA 5
# 3 NA  4 6

## if the "is.na" check is not done, then this would give
## (because of NA in selectorvector):
#    a  b  c
# 1  1  1  1
# 2  2 NA  5
# 3 NA NA NA
## hence in the case that pred doesn't check for NA in 'a',
## these NA vlaues have to be reverted to the original columns' value.

driver_col_to_other_cols(df, "a", is.na)
#    a  b  c
# 1  1  3 NA
# 2  2 NA  5
# 3 NA NA NA

Correct answer by Gwang-Jin Kim on October 23, 2020

Try this function, in input you have your original dataset and in output the cleaned one:

Input

df<-data.frame(a=c(1, 2, NA), b=c(3, NA, 4), c=c(NA, 5, 6))
> df
   a  b  c
1  1  3 NA
2  2 NA  5
3 NA  4  6

Function

   fix<-function(df,var_x,list_y)
{
   df[is.na(df[,var_x]),list_y]<-NA
   return(df)
}

Output

fix(df,"a",c("b","c"))
   a  b  c
1  1  3 NA
2  2 NA  5
3 NA NA NA

Answered by Terru_theTerror on October 23, 2020

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