如何解决R函数组成，用于替换数据帧中的值 使用词汇闭包概括任何谓词功能
给出以下可重复的示例
我的目标是在数据帧的相邻列中用NA行替换原始值；我知道这是一个已经发布的问题（有很多变体），但是我还没有找到我要实现的方法的解决方案：即通过应用函数组合在可重现的示例中，驱动用原始值的NA替换的列为a列。
这是我到目前为止所做的
最后一个代码段是我实际上正在搜索的内容的失败尝试...
#-----------------------------------------------------------
# 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),df$b)
df$c<-ifelse(is.na(df$a),df$c)
df
#--------------------------------------------------------
# extraction and subsitution approach
# same as above
df<-data.frame(a=c(1,6))
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(1,6))
df
fix<-function(x,y){
ifelse(is.na(x),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(1,6))
df
fix2<-function(x){
x[is.na(x[1])]<-NA
x
}
df[]<-lapply(df,fix2)
df
此特定方法有帮助吗？ 我坚持如何正确构思传递给lapply的替代函数
感谢
解决方法
尝试使用此功能，在输入中您拥有原始数据集，在输出中已清理的数据集：
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
功能
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
使用词汇闭包
如果使用词法闭包-定义一个函数，该函数首先生成所需的函数。 然后，您可以根据需要使用此功能。
# 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(1,6))
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"))
使用常规功能
df<-data.frame(a=c(1,6))
df
# define it for one column
overtake_NA <- function(df,driver_col,target_col) {
df[,target_col] <- ifelse(is.na(df[,driver_col]),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,i)
}
df
}
overtake_driver_col_NAs(df,"a")
# a b c
# 1 1 3 NA
# 2 2 NA 5
# 3 NA NA NA
概括任何谓词功能
driver_col_to_other_cols <- function(df,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,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,driver_col],target_col])
df
}
for (i in 1:ncol(df)) {
df <- overtake_(df,i,pred)
}
df
}
driver_col_to_other_cols(df,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,is.na)
# a b c
# 1 1 3 NA
# 2 2 NA 5
# 3 NA NA NA
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