如何用未知的列名替换某些数据框值？

Question

I have a large data frame with unknown column names and numeric values 1, 2, 3, or 4. Now I want to replace all 4 values with it's column name and all 1, 2 and 3's with an empty value.

Ofcourse I can make a loop of some kind, like this:

df <- data.frame(id=1:8,unknownvarname1=c(1:4,1:4),unknownvarname2=c(4:1,4:1))
for (i in 2:length(df)){
  df[,i] <- as.character(df[,i])
  df[,i] <- mgsub::mgsub(df[,i],c(1,2,3,4),c("","","",names(df)[i]))  
}

This would be the result:

  id unknownvarname1 unknownvarname2
1  1                 unknownvarname2
2  2                                
3  3                                
4  4 unknownvarname1                
5  5                 unknownvarname2
6  6                                
7  7                                
8  8 unknownvarname1 unknownvarname2

For a data frame this size that's no problem at all. But when I try this loop on large data frames with up to 30k and up to 40 uknown variables, the loop takes ages to complete.

Does anyone know of a faster way to do this? I tried functions like mutate() of dplyr package but I could not manage to make it work.

Many thanks in advance!

Answer 1

One way using base R

#Replace all the values with 1:3 with blank
df[-1][sapply(df[-1], `%in%`, 1:3)] <- ""
#Get the row/column indices where value is 4
mat <- which(df == 4, arr.ind = TRUE)
#Exclude values from first column
mat <- mat[mat[, 2] != 1, ]
#Replace remaining entries with it's corresponding column names
df[mat] <- names(df)[mat[, 2]]
df

#  id unknownvarname1 unknownvarname2
#1  1                 unknownvarname2
#2  2                                
#3  3                                
#4  4 unknownvarname1                
#5  5                 unknownvarname2
#6  6                                
#7  7                                
#8  8 unknownvarname1                

Answer 2

Just to give another option with switch (though, as this function is not vectorized, it needs a nested sapply within a lapply which doesn't make it that "pretty" and efficient...):

Basically, switch works with numeric as switch(myNumberToTest, caseIfOne, caseIfTwo, ...) .

So what you need is:

df[, 2:3] <- lapply(2:3, function(x) sapply(df[, x], switch, "", "", "", names(df)[x]))

df
#  id unknownvarname1 unknownvarname2
#1  1                 unknownvarname2
#2  2                                
#3  3                                
#4  4 unknownvarname1                
#5  5                 unknownvarname2
#6  6                                
#7  7                                
#8  8 unknownvarname1                

Answer 3

Yet another base R option, using ifelse within lapply (still looping on the columns, but vectorized approach by column):

df <- data.frame(id=1:8,unknownvarname1=c(1:4,1:4),unknownvarname2=c(4:1,4:1))
df[,2:3] <- lapply(2:3, function(x) { ifelse(df[,x] < 4, "", colnames(df)[x]) })

gives

  id unknownvarname1 unknownvarname2
1  1                 unknownvarname2
2  2                                
3  3                                
4  4 unknownvarname1                
5  5                 unknownvarname2
6  6                                
7  7                                
8  8 unknownvarname1         

Answer 4

Another base R possibility using sweep :

idx <- df[, -1] == 4
sw <- sweep(idx, 2, 1:2, FUN = '*') + 1
df[, -1] <- c("", colnames(df[, -1]))[sw]

which gives:

 > df id unknownvarname1 unknownvarname2 1 1 unknownvarname2 2 2 3 3 4 4 unknownvarname1 5 5 unknownvarname2 6 6 7 7 8 8 unknownvarname1

---

This could be shortened to:

sw <- sweep(df[, -1] == 4, 2, 1:2, FUN = '*') + 1
df[, -1] <- c("", colnames(df[, -1]))[sw]

Answer 5

A somewhat inefficient tidyverse option. This is inefficient because we need to manually select the columns later:

to_use <- names(df)[-1]
df %>% 
  mutate_at(vars(contains("unknown")),list(~ifelse(.==4,
                                             NA,
                                             ""))) -> new_df

new_df[-1] <-map2(new_df[-1], to_use,function(x,y) replace(x,is.na(x),y))

A less manual approach that also has the disadvantage of being non specific:

 df %>% 
   map2(.,names(.), function(x, y) ifelse( x==4, y,"")) %>% 
   as.data.frame() %>% 
   mutate(id=row.names(.)) # might be a way around  with `.id`
  id unknownvarname1 unknownvarname2
1  1                 unknownvarname2
2  2                                
3  3                                
4  4 unknownvarname1                
5  5                 unknownvarname2
6  6                                
7  7                                
8  8 unknownvarname1 

Result for approach 1:

new_df
     id unknownvarname1 unknownvarname2
    1  1                 unknownvarname2
    2  2                                
    3  3                                
    4  4 unknownvarname1                
    5  5                 unknownvarname2
    6  6                                
    7  7                                
    8  8 unknownvarname1 

Answer 6

Yet another option using col to line up the names and values:

sel <- df[-1] == 4
df[-1] <- ""
df[-1][sel] <- names(df[-1])[col(df[-1])[sel]]

#  id unknownvarname1 unknownvarname2
#1  1                 unknownvarname2
#2  2                                
#3  3                                
#4  4 unknownvarname1                
#5  5                 unknownvarname2
#6  6                                
#7  7                                
#8  8 unknownvarname1