How to convert list of list into a tibble (datafra

Question

I have the following list of list. It contains two variables: pair and genes. The contain of pair is always vector with two strings. And the variable genes is a vector which can contain more than 1 values.

lol <- list(structure(list(pair = c("BoneMarrow", "Pulmonary"), genes = "PRR11"), .Names = c("pair", 
"genes")), structure(list(pair = c("BoneMarrow", "Umbilical"), 
    genes = "GNB2L1"), .Names = c("pair", "genes")), structure(list(
    pair = c("Pulmonary", "Umbilical"), genes = "ATP1B1"), .Names = c("pair", 
"genes")))


lol
#> [[1]]
#> [[1]]$pair
#> [1] "BoneMarrow" "Pulmonary" 
#> 
#> [[1]]$genes
#> [1] "PRR11"
#> 
#> 
#> [[2]]
#> [[2]]$pair
#> [1] "BoneMarrow" "Umbilical" 
#> 
#> [[2]]$genes
#> [1] "GNB2L1"
#> 
#> 
#> [[3]]
#> [[3]]$pair
#> [1] "Pulmonary" "Umbilical"
#> 
#> [[3]]$genes
#> [1] "ATP1B1"

How can I convert it into this dataframe:

pair1         pair2        genes_vec
BoneMarrow    Pulmonary    PRR11
BoneMarrow    Umbilical    GNB2L1
Pulmonary     Umbilical    ATP1B1

Note that the genes variable is a vector not single string.

My best try is this which doesn't give what I want:

> do.call(rbind, lapply(lol, data.frame, stringsAsFactors=FALSE))
        pair  genes
1 BoneMarrow  PRR11
2  Pulmonary  PRR11
3 BoneMarrow GNB2L1
4  Umbilical GNB2L1
5  Pulmonary ATP1B1
6  Umbilical ATP1B1

---

Update:

With new example to show vector content of genes

lol2 <- list(structure(list(pair = c("BoneMarrow", "Pulmonary"), genes = c("GNB2L1", 
"PRR11")), .Names = c("pair", "genes")), structure(list(pair = c("BoneMarrow", 
"Umbilical"), genes = "GNB2L1"), .Names = c("pair", "genes")), 
    structure(list(pair = c("Pulmonary", "Umbilical"), genes = "ATP1B1"), .Names = c("pair", 
    "genes")))

lol2
#> [[1]]
#> [[1]]$pair
#> [1] "BoneMarrow" "Pulmonary" 
#> 
#> [[1]]$genes
#> [1] "GNB2L1" "PRR11" 
#> 
#> 
#> [[2]]
#> [[2]]$pair
#> [1] "BoneMarrow" "Umbilical" 
#> 
#> [[2]]$genes
#> [1] "GNB2L1"
#> 
#> 
#> [[3]]
#> [[3]]$pair
#> [1] "Pulmonary" "Umbilical"
#> 
#> [[3]]$genes
#> [1] "ATP1B1"

The expected output is:

pair1         pair2        genes_vec
BoneMarrow    Pulmonary    PRR11,GNB2L1
BoneMarrow    Umbilical    GNB2L1
Pulmonary     Umbilical    ATP1B1

Answer 1

EDIT: updated to work with vector lol2.

Maybe like this:

as.data.frame(do.call(rbind,lapply(lol2, function(x) {c(unlist(x[1]),gene=paste(unlist(x[2]),collapse=","))})),stringsAsFactors = F)




       pair1     pair2         genes
1 BoneMarrow Pulmonary GNB2L1, PRR11
2 BoneMarrow Umbilical        GNB2L1
3  Pulmonary Umbilical        ATP1B1

Answer 2

Using tidyverse, you could use purrr to help you

library(dplyr)
library(purrr)

tibble(
  pair = map(lol, "pair"),
  genes_vec = map_chr(lol, "genes")
) %>% 
  mutate(
    pair1 = map_chr(pair, 1),
    pair2 = map_chr(pair, 2) 
  ) %>%
  select(pair1, pair2, genes_vec)
#> # A tibble: 3 x 3
#>        pair1     pair2 genes_vec
#>        <chr>     <chr>     <chr>
#> 1 BoneMarrow Pulmonary     PRR11
#> 2 BoneMarrow Umbilical    GNB2L1
#> 3  Pulmonary Umbilical    ATP1B1

---

with the second example, just replace map_chr(lol, "genes") with map(lol2, "genes") as you want to keep a nested dataframe with a list column.

tibble(
  pair = map(lol2, "pair"),
  genes_vec = map(lol2, "genes")
) %>% 
  mutate(
    pair1 = map_chr(pair, 1),
    pair2 = map_chr(pair, 2) 
  ) %>%
  select(pair1, pair2, genes_vec)
#> # A tibble: 3 x 3
#>        pair1     pair2 genes_vec
#>        <chr>     <chr>    <list>
#> 1 BoneMarrow Pulmonary <chr [2]>
#> 2 BoneMarrow Umbilical <chr [1]>
#> 3  Pulmonary Umbilical <chr [1]>

---

And a more generic approach would be to work with nested tibbles and unnest them as needed

library(dplyr)
library(purrr)
library(tidyr)

tab1 <-lol %>%
  transpose() %>%
  as_tibble() %>%
  mutate(pair = map(pair, ~as_tibble(t(.x)))) %>%
  mutate(pair = map(pair, ~set_names(.x, c("pair1", "pair2"))))
tab1
#> # A tibble: 3 x 2
#>               pair     genes
#>             <list>    <list>
#> 1 <tibble [1 x 2]> <chr [1]>
#> 2 <tibble [1 x 2]> <chr [1]>
#> 3 <tibble [1 x 2]> <chr [1]>

For lol2 nothing changes unless the list lol2 instead of lol1

tab2 <- lol2 %>%
  transpose() %>%
  as_tibble() %>%
  mutate(pair = map(pair, ~as_tibble(t(.x)))) %>%
  mutate(pair = map(pair, ~set_names(.x, c("pair1", "pair2"))))
tab2
#> # A tibble: 3 x 2
#>               pair     genes
#>             <list>    <list>
#> 1 <tibble [1 x 2]> <chr [2]>
#> 2 <tibble [1 x 2]> <chr [1]>
#> 3 <tibble [1 x 2]> <chr [1]>

You can then unnest what the column you want

tab1 %>%
  unnest()
#> # A tibble: 3 x 3
#>    genes      pair1     pair2
#>    <chr>      <chr>     <chr>
#> 1  PRR11 BoneMarrow Pulmonary
#> 2 GNB2L1 BoneMarrow Umbilical
#> 3 ATP1B1  Pulmonary Umbilical

tab2 %>% 
  unnest(pair)
#> # A tibble: 3 x 3
#>       genes      pair1     pair2
#>      <list>      <chr>     <chr>
#> 1 <chr [2]> BoneMarrow Pulmonary
#> 2 <chr [1]> BoneMarrow Umbilical
#> 3 <chr [1]>  Pulmonary Umbilical

Answer 3

> lol1 <- data.frame(t(sapply(lol,c)))
> as.data.frame(t(apply(lol1, 1, unlist)))
       pair1     pair2  genes
1 BoneMarrow Pulmonary  PRR11
2 BoneMarrow Umbilical GNB2L1
3  Pulmonary Umbilical ATP1B1

Answer 4

This should work:

data.frame(do.call(rbind,lol2))
data.frame(do.call(rbind,lol2))
                   pair         genes
1 BoneMarrow, Pulmonary GNB2L1, PRR11
2 BoneMarrow, Umbilical        GNB2L1
3  Pulmonary, Umbilical        ATP1B1

The same way you treat the genes as a vector is the same way you can the same way you can treat the pairs as a vector.. instead of pair 1 and 2 you just use both of them.