将列名粘贴到 R 中数据框的每个值

Question

Hoping this isn't a repeat- I've had a search around but can't find quite what I'm looking for.

I have a dataframe (df) in R

  1 2 3 4 5 
1 1 0.5 0.5 0 1
2 0.5 0.5 0.5 0 1
3 1 1 0 0 1
4 1 1 0 0 1 
5 1 1 0 0 1

(with the 1-5 indicating row and column names)

I would like to paste the column name to each cell, separated by a ":" so that it looks like this:

  1 2 3 4 5 
1 1:1 2:0.5 3:0.5 4:0 5:1 
2 1:0.5 2:0.5 3:0.5 4:0 5:1 
3 1:1 2:1 3:0 4:0 5:1 
4 1:1 2:1 3:0 4:0 5:1 
5 1:1 2:1 3:0 4:0 5:1 

However, my actual data is quite a bit larger.

I currently have

apply(df, 2, function(x) paste(colnames(df)[x], x, sep=":"))

Of course this doesn't work as colnames(df)[x] doesn't make any sense. Is there anything I can put in that first 'paste' term to get this sorted? Or another function to do a better job?

Thanks.

Answer 1

To explain my comment, Map is a multivariate version of lapply , so

df <- data.frame(`1` = c(1, 0.5, 1, 1, 1), 
                 `2` = c(0.5, 0.5, 1, 1, 1), 
                 `3` = c(0.5, 0.5, 0, 0, 0), 
                 `4` = c(0L, 0L, 0L, 0L, 0L), 
                 `5` = c(1L, 1L, 1L, 1L, 1L), 
                 check.names = FALSE)

df[] <- Map(paste, names(df), df, sep = ':')

df
##       1     2     3   4   5
## 1   1:1 2:0.5 3:0.5 4:0 5:1
## 2 1:0.5 2:0.5 3:0.5 4:0 5:1
## 3   1:1   2:1   3:0 4:0 5:1
## 4   1:1   2:1   3:0 4:0 5:1
## 5   1:1   2:1   3:0 4:0 5:1

Here Map takes the first element of names(df) , ie 1 , and paste s it to the first element of df , ie the first column. Assigning to df[] keeps the list's data.frame class, and therefore the original structure.

If your data is a matrix, you can do the same thing with sweep :

mat <- matrix(c(1, 0.5, 1, 1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0.5, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1), 
              5, 5, 
              dimnames = list(c("1", "2", "3", "4", "5"), 
                              c("1", "2", "3", "4", "5")))

mat[] <- sweep(mat, 2, colnames(df), function(x, y) paste(y, x, sep = ':'))

mat
##   1       2       3       4     5    
## 1 "1:1"   "2:0.5" "3:0.5" "4:0" "5:1"
## 2 "1:0.5" "2:0.5" "3:0.5" "4:0" "5:1"
## 3 "1:1"   "2:1"   "3:0"   "4:0" "5:1"
## 4 "1:1"   "2:1"   "3:0"   "4:0" "5:1"
## 5 "1:1"   "2:1"   "3:0"   "4:0" "5:1"

Answer 2

As an alternative to looping, you can use col(., as.factor = TRUE) to create a matrix of column names, then paste it to the data (coerced to matrix).

df[] <- paste(col(df, TRUE), as.matrix(df), sep = ":")

Resulting in:

 1 2 3 4 5 1 1:1 2:0.5 3:0.5 4:0 5:1 2 1:0.5 2:0.5 3:0.5 4:0 5:1 3 1:1 2:1 3:0 4:0 5:1 4 1:1 2:1 3:0 4:0 5:1 5 1:1 2:1 3:0 4:0 5:1

Actually, with these particular column names, as.factor = TRUE is not necessary. But it would be necessary for column names not the same as the column numbers. For this particular example, it could be

df[] <- paste(col(df), as.matrix(df), sep = ":")

PS You should really be using a matrix with 100% numeric data, instead of a data frame.

Data:

df <- structure(list(`1` = c(1, 0.5, 1, 1, 1), `2` = c(0.5, 0.5, 1, 
1, 1), `3` = c(0.5, 0.5, 0, 0, 0), `4` = c(0L, 0L, 0L, 0L, 0L
), `5` = c(1L, 1L, 1L, 1L, 1L)), .Names = c("1", "2", "3", "4", 
"5"), class = "data.frame", row.names = c("1", "2", "3", "4", 
"5"))

Answer 3

We can unlist the dataset and paste with replicated elements of the column name

df[] <- paste(names(df)[col(df)], unlist(df), sep=":")
df
#      1     2     3   4   5
#1   1:1 2:0.5 3:0.5 4:0 5:1
#2 1:0.5 2:0.5 3:0.5 4:0 5:1
#3   1:1   2:1   3:0 4:0 5:1
#4   1:1   2:1   3:0 4:0 5:1
#5   1:1   2:1   3:0 4:0 5:1

data

df <- structure(list(`1` = c(1, 0.5, 1, 1, 1), `2` = c(0.5, 0.5, 1, 
1, 1), `3` = c(0.5, 0.5, 0, 0, 0), `4` = c(0L, 0L, 0L, 0L, 0L
), `5` = c(1L, 1L, 1L, 1L, 1L)), .Names = c("1", "2", "3", "4", 
 "5"), class = "data.frame", row.names = c("1", "2", "3", "4", 
 "5"))

Answer 4

To update this for tidyverse, you can now run

df <- map2_dfc(colnames(df), df, paste, sep = ':')

map2 takes 2 inputs for the paste function, and the _dfc tag returns a dataframe as your result instead of a list.