Akhil Nair - 10 months ago 40

R Question

I have a list structure which represents a table being handed to me like this

`> l = list(list(1, 4), list(2, 5), list(3, 6))`

> str(l)

List of 3

$ :List of 2

..$ : num 1

..$ : num 4

$ :List of 2

..$ : num 2

..$ : num 5

$ :List of 2

..$ : num 3

..$ : num 6

And I'd like to convert it to this

`> lt = list(x = c(1, 2, 3), y = c(4, 5, 6))`

> str(lt)

List of 2

$ x: num [1:3] 1 2 3

$ y: num [1:3] 4 5 6

I've written a function that does it in a really simple manner which uses

`Reduce`

Any help appreciated,

Thanks

Thanks all! Much appreciated. Benchmarked the answers and picked the fastest for a larger test case:

`f1 = function(l) {`

k <- length(unlist(l)) / length(l)

lapply(seq_len(k), function(i) sapply(l, "[[", i))

}

f2 = function(l) {

n <- length(l[[1]])

split(unlist(l, use.names = FALSE), paste0("x", seq_len(n)))

}

f3 = function(l) {

split(do.call(cbind, lapply(l, unlist)), seq(unique(lengths(l))))

}

f4 = function(l) {

l %>%

purrr::transpose() %>%

map(unlist)

}

f5 = function(l) {

# bind lists together into a matrix (of lists)

temp <- Reduce(rbind, l)

# split unlisted values using indices of columns

split(unlist(temp), col(temp))

}

f6 = function(l) {

data.table::transpose(lapply(l, unlist))

}

microbenchmark::microbenchmark(

lapply = f1(l),

split_seq = f2(l),

unique = f3(l),

tidy = f4(l),

Reduce = f5(l),

dt = f6(l),

times = 10000

)

Unit: microseconds

expr min lq mean median uq max neval

lapply 165.057 179.6160 199.9383 186.2460 195.0005 4983.883 10000

split_seq 85.655 94.6820 107.5544 98.5725 104.1175 4609.378 10000

unique 144.908 159.6365 182.2863 165.9625 174.7485 3905.093 10000

tidy 99.547 122.8340 141.9482 129.3565 138.3005 8545.215 10000

Reduce 172.039 190.2235 216.3554 196.8965 206.8545 3652.939 10000

dt 98.072 106.6200 120.0749 110.0985 116.0950 3353.926 10000

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Answer Source

For the specific example, you can use this pretty simple approach:

```
split(unlist(l), c("x", "y"))
#$x
#[1] 1 2 3
#
#$y
#[1] 4 5 6
```

It recycles the x-y vector and splits on that.

To generalize this to "n" elements in each list, you can use:

```
l = list(list(1, 4, 5), list(2, 5, 5), list(3, 6, 5)) # larger test case
split(unlist(l, use.names = FALSE), paste0("x", seq_len(length(l[[1L]]))))
# $x1
# [1] 1 2 3
#
# $x2
# [1] 4 5 6
#
# $x3
# [1] 5 5 5
```

This assumes, that all the list elements on the top-level of `l`

have the same length, as in your example.

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