用于替换数据框中值的 R 函数组合

Question

给定以下可重现的示例

my objective 是将数据框相邻列的原始值逐行替换为NA；我知道这是一个问题（有这么多变体）已经发布但我还没有找到我试图完成的方法的解决方案：即通过应用函数组合

在可重现的示例中，驱动用 NA 替换原始值的列是列 a

这是我目前所做的

最后一个代码片段是我实际搜索内容的失败尝试...

#-----------------------------------------------------------
# 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

对这种特殊方法有什么帮助吗？我坚持如何正确构思传递给 lapply

的替代函数

谢谢

Answer 1

试试这个函数，在输入中你有原始数据集，在输出中有清理过的数据集：

输入

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

函数

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

输出

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

Answer 2

使用词法闭包

如果您使用词法闭包 - 您定义一个函数，该函数首先生成您需要的函数。然后你就可以随意使用这个功能了

# 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"))

使用正常功能

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

泛化任何谓词函数

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

用于替换数据框中值的 R 函数组合

R fuction composition for the substitution of values in dataframe

r

function

composition

使用词法闭包

使用正常功能

泛化任何谓词函数