CODEWITHSUNDEEP
r
cs0815
cs0815

Reputation: 17388

combine column name with level of factors in data frame

I have this:

df1 <- data.frame(A = c('a', 'b', 'c'), B = c('d', 'e', 'c'))

and would like to transform it to this:

    A   B
1 A:a B:d
2 A:b B:e
3 A:c B:c

My current not working miserable 'loop' attempt (using apply version preferred) is this:

for (row in 1:nrow(df1)) {
    for (col in 1:ncol(df1)) {
        levels(colnames(df1)[col])[levels(colnames(df1)[col]) == df1[row, col]] <- paste0(colnames(df1)[col], ":", df1[row, col])
    }
}

Upvotes: 1

Views: 168

Answers (3)

tmfmnk
tmfmnk

Reputation: 39858

In the case you want to do it using tidyverse:

df1 %>%
 rowid_to_column() %>% #Creating row IDs
 gather(var, val, -rowid) %>% #Transforming the data from wide to long
 mutate(temp = paste(var, val, sep = ":")) %>% #Combining the column names with the level of factors
 select(-val) %>%
 spread(var, temp) %>% #Transforming the data back to wide format
 select(-rowid) #Deleting the redundant variable

    A   B
1 A:a B:d
2 A:b B:e
3 A:c B:c

Upvotes: 1

Ronak Shah
Ronak Shah

Reputation: 388807

An option using lapply where we go through each column and paste the name of the column along with column values.

df1[] <- lapply(seq_along(df1), function(x) paste0(names(df1)[x],":", df1[,x]))

df1
#    A   B
#1 A:a B:d
#2 A:b B:e
#3 A:c B:c

Upvotes: 3

LyzandeR
LyzandeR

Reputation: 37879

One way with mapply:

data.frame(mapply(function(x, y) paste0(y, ':', x), df1, c('A', 'B')))
#    A   B
#1 A:a B:d
#2 A:b B:e
#3 A:c B:c

Or you could do:

data.frame(A = paste0('A:', df1$A),
           B = paste0('B:', df1$B))

But I would go for the first option if you have multiple columns that you would like to use this logic on.

Upvotes: 4

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