R：按列拆分列表并转换为新数据

Question

I have the following sample data in a list called data

data <-    structure(list(`1.1` = structure(list(id = structure(1, .Dim = c(1L, 
    1L)), Sample = structure("Test1", .Dim = c(1L, 1L)), Add = structure("T", .Dim = c(1L, 
    1L))), .Names = c("id", "Sample", "Add")), `2.1` = structure(list(
        id = structure(5, .Dim = c(1L, 1L)), Sample = structure("Test2", .Dim = c(1L, 
        1L)), Add = structure("A", .Dim = c(1L, 1L))), .Names = c("id", 
    "Sample", "Add")), `3.1` = structure(list(id = structure(7, .Dim = c(1L, 
    1L)), Sample = structure("Test3", .Dim = c(1L, 1L)), Add = structure("D", .Dim = c(1L, 
    1L))), .Names = c("id", "Sample", "Add")), `4.1` = structure(list(
        id = structure(12, .Dim = c(1L, 1L)), Sample = structure("Test4", .Dim = c(1L, 
        1L)), Add = structure("Z", .Dim = c(1L, 1L))), .Names = c("id", 
    "Sample", "Add")), `5.1` = structure(list(id = structure(17, .Dim = c(1L, 
    1L)), Sample = structure("Test12", .Dim = c(1L, 1L)), Add = structure("E", .Dim = c(1L, 
    1L))), .Names = c("id", "Sample", "Add"))), .Names = c("1.1", 
    "2.1", "3.1", "4.1", "5.1"), row.names = c("id", "Sample", "Add"
    ), class = "data.frame")

It looks like this:

 data
         1.1   2.1   3.1   4.1    5.1
id         1     5     7    12     17
Sample Test1 Test2 Test3 Test4 Test12
Add        T     A     D     Z      E

How could I split this list by column into several data.frames based on the ID number? Eg that a data.frame with name data.ID1 and a data.frame with name data.ID5 and a data.frame with name data.ID 7 etc. is created (see example below)? The name of the data.frame should be the same as the ID number. My list contains about 700 different IDs and data...

data.ID1
id      1
Sample  Test1
Add     T

data.ID5
id      5
Sample  Test2
Add     A

data.ID7
id      7
Sample  Test3
Add     D

and so on...

Answer 1

Here's a possible solution:

lst <- lapply(1:ncol(data),function(c) return(data[,c,drop=F]))
names(lst) <- lapply(data,function(col) return(paste0('data.ID',col$id)))

# here you have data.ID1, data.ID2 etc inside a lst, 
# you can have access to them simply using: lst$data.ID1, lst$data.ID2 etc.
# but if you REALLY want to add these variables in the environment, 
# continue to the next loop

for(nm in names(lst)){
  assign(nm,lst[[nm]])
}

Please, note that is preferred not to use assign since as stated in the comment above, you have all you need inside the list object "lst"... but maybe you need to do it for a valid reason ;)

Answer 2

Here is a solution close enough, try

coln <- lapply(data[1,],as.character)
colnames(data) <- paste(rep("data.TD",ncol(data)),coln,sep="")
attach(data)

Then whenever you need to call a certain column by the data id, you do:

data.frame(data.TD7)

Output:

  id Sample Add
1  7  Test3   D

Or you can transpose it using t(data.frame(data.TD7))

Answer 3

I think these three lines will solve your problem:

newdata <- apply(data, 2, function(x) { y = as.data.frame(x=unlist(x)) })
newname <- paste("data.ID", unlist(data[1, ]), sep="")
names(newdata) <- newname
newdata

new data is a list containing the desired data frames

Answer 4

data <- structure(list(
  `1.1` = structure(list(id = structure(1, .Dim = c(1L, 1L)),
                         Sample = structure("Test1", .Dim = c(1L, 1L)),
                         Add = structure("T", .Dim = c(1L, 1L))),
                    .Names = c("id", "Sample", "Add")),
  `2.1` = structure(list(id = structure(5, .Dim = c(1L, 1L)),
                         Sample = structure("Test2", .Dim = c(1L, 1L)),
                         Add = structure("A", .Dim = c(1L, 1L))),
                    .Names = c("id", "Sample", "Add")),
  `3.1` = structure(list(id = structure(7, .Dim = c(1L, 1L)),
                         Sample = structure("Test3", .Dim = c(1L, 1L)),
                         Add = structure("D", .Dim = c(1L, 1L))),
                    .Names = c("id", "Sample", "Add")),
  `4.1` = structure(list(id = structure(12, .Dim = c(1L, 1L)),
                         Sample = structure("Test4", .Dim = c(1L, 1L)),
                         Add = structure("Z", .Dim = c(1L, 1L))),
                    .Names = c("id", "Sample", "Add")),
  `5.1` = structure(list(id = structure(17, .Dim = c(1L, 1L)),
                         Sample = structure("Test12", .Dim = c(1L, 1L)),
                         Add = structure("E", .Dim = c(1L, 1L))),
                    .Names = c("id", "Sample", "Add"))),
  .Names = c("1.1", "2.1", "3.1", "4.1", "5.1"),
  row.names = c("id", "Sample", "Add"), class = "data.frame")

data
# 1.1   2.1   3.1   4.1    5.1
# id         1     5     7    12     17
# Sample Test1 Test2 Test3 Test4 Test12
# Add        T     A     D     Z      E

data2 <- as.data.frame(apply(data, 1, unlist))
data2
# id Sample Add
# 1.1  1  Test1   T
# 2.1  5  Test2   A
# 3.1  7  Test3   D
# 4.1 12  Test4   Z
# 5.1 17 Test12   E

data2 <- split(x = data2, f = data2$id)
data2
# $`1`
# id Sample Add
# 1.1  1  Test1   T
# 
# $`12`
# id Sample Add
# 4.1 12  Test4   Z
# 
# $`17`
# id Sample Add
# 5.1 17 Test12   E
# 
# $`5`
# id Sample Add
# 2.1  5  Test2   A
# 
# $`7`
# id Sample Add
# 3.1  7  Test3   D

data2 <- lapply(data2, function(x){
  as.data.frame(t(x))
})
data2
# $`1`
# 1.1
# id         1
# Sample Test1
# Add        T
# 
# $`12`
# 4.1
# id        12
# Sample Test4
# Add        Z
# 
# $`17`
# 5.1
# id         17
# Sample Test12
# Add         E
# 
# $`5`
# 2.1
# id         5
# Sample Test2
# Add        A
# 
# $`7`
# 3.1
# id         7
# Sample Test3
# Add        D