Get the column of a list of lists in R-CodePudding

Using R, my data set L is a list of lists. My print(L) produces the following output:

[[1]]
[[1]][[1]]
[1] 0.8198689

[[1]][[2]]
[1] 0.8166747


[[2]]
[[2]][[1]]
[1] 0.5798426

[[2]][[2]]
[1] 0.5753511


[[3]]
[[3]][[1]]
[1] 0.4713508

[[3]][[2]]
[1] 0.4698621

And I want to get a vector of the second column. However unlist(L[[2]]) gives me the second row (not the second column) and L[,2] gives me the error Error in L[, 2] : incorrect number of dimensions. I tried also L$'2' and didn't work.

How can I get the vector of the second column of this data set in R?

CodePudding user response：

1) Assuming the input shown reproducibly in the Note at the end use sapply (or use lapply if you want it as a list). No packages are used.

sapply(L, `[[`, 2)
## [1] 0.8166747 0.5753511 0.4698621

2) Using purrr we have:

library(purrr)

transpose(L)[[2]]
## [[1]]
## [1] 0.8166747
##  
## [[2]]
## [1] 0.5753511
##
## [[3]]
## [1] 0.4698621

3) If we know that L is regularly shaped we could convert it to a matrix and then take the second column.

matrix(unlist(L), length(L), byrow = TRUE)[, 2]
## [1] 0.8166747 0.5753511 0.4698621

Note

# input in reproducible form
L <- list(
  list(0.8198689, 0.8166747),
  list(0.5798426, 0.5753511),
  list(0.4713508, 0.4698621))

CodePudding user response：

Can you explain how sapply do the job ?

I assume that sapply will apply a function on the list L but what is exactly the function here ? As far as I know, '[[' is not a function

CodePudding user response：

The simple way to do this using purrr is just to use map(), which returns a list.

library(purrr)

map(L, 2)

If you want a (numeric) vector, use map_dbl().

map_dbl(L, 2)
# [1] 0.8166747 0.5753511 0.4698621