Convert a list of character vectors into dataframe

Question

my output a looks like:

str(a)
List of 8883695
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "ACCRUALS"
 $ : chr [1:3] "20MICRONS" "A" "AE"
 $ : chr [1:3] "20MICRONS" "A" "AE"
 $ : chr [1:3] "20MICRONS" "A" "AE"

I want to convert this into a dataframe like:

ticker frequency field

20MICRONS A ACCRUALS
20MICRONS A ACCRUALS
20MICRONS A AE
20MICRONS A AE
20MICRONS A AE
20MICRONS A AE

I have tried this:

b <- as.data.frame(a)  
Error in sort.list(y) : 'x' must be atomic for 'sort.list'
Have you called 'sort' on a list?

But not working. Please help. I am new to R and stack overflow.

Step 1. Make a *smaller* example to test your code on. You've got 8 million of these. — Spacedman, Apr 01 '17 at 16:40

score 4 · Answer 1 · answered Apr 01 '17 at 16:40

4

The list elements are not of equal length. So, we need to convert it to equal length by padding with NA

mx <- max(lengths(a))
d1 <- data.frame(lapply(a, `length<-`, mx))

answered Apr 01 '17 at 16:40

akrun

874,273
37
540
662

@shivrajgondi This does correct it automatically by assigning elements that have less than the maximum length to pad with as many NAs to get all of them having the same length – akrun Apr 01 '17 at 16:50
Thanks Akrun, found certain vectors to be of length 4. – shivrajgondi Apr 01 '17 at 16:53
@shivrajgondi Your partial `str` output suggests the length as 3, may be it is 4 based on the full `str` – akrun Apr 01 '17 at 17:09
1

This is a very cool answer, it sparked for me a followup quesiton on replacement functions I posted here: https://stackoverflow.com/questions/69485776/why-dont-lambda-functions-handle-replacement-functions-in-their-intuitive-form – Bryan Shalloway Nov 22 '21 at 20:40

score 1 · Answer 2 · answered Sep 24 '17 at 19:05

1

I found a function called do.call(rbind.data.frame, your_list) - seems like this is what you're looking for. Check the comments section of this question.

answered Sep 24 '17 at 19:05

user_cm21

23
6

score 1 · Answer 3 · answered Oct 07 '21 at 00:23

library(tidyverse)

a <- list(c("20M1", "A1", "ACC1"), 
          c("20M2", "A2", "ACC2"), 
          c("20M3"), 
          c("20M4", "A3"))

if_null_na <- function(x){
  if(is.null(x)) return(NA)
  x
}

a %>% 
  transpose() %>% 
  map(~map(.x, if_null_na)) %>% 
  map(unlist) %>% 
  as_tibble(.name_repair = "unique")
#> Warning: Element 3 must be length 3, not 1
#> Warning: Element 4 must be length 3, not 2
#> New names:
#> * `` -> ...1
#> * `` -> ...2
#> * `` -> ...3
#> # A tibble: 4 x 3
#>   ...1  ...2  ...3 
#>   <chr> <chr> <chr>
#> 1 20M1  A1    ACC1 
#> 2 20M2  A2    ACC2 
#> 3 20M3  <NA>  <NA> 
#> 4 20M4  A3    <NA>

^{Created on 2021-10-06 by the reprex package (v2.0.0)}

Though this way is much slower than Akrun's approach.

score 0 · Answer 4 · answered Oct 01 '18 at 11:11

0

You could try Reduce(rbind, a), which is handy for that.

answered Oct 01 '18 at 11:11

Kevin Zarca

2,572
1
18
18

Convert a list of character vectors into dataframe

4 Answers4

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