How to convert a list consisting of vector of different lengths to a usable data frame in R?

Question

I have a (fairly long) list of vectors. The vectors consist of Russian words that I got by using the strsplit() function on sentences.

The following is what head() returns:

[[1]]
[1] "модно"     "создавать" "резюме"    "в"         "виде"     

[[2]]
[1] "ты"        "начианешь" "работать"  "с"         "этими"    

[[3]]
[1] "модно"            "называть"         "блогер-рилейшенз" "―"                "начинается"       "задолго"         

[[4]]
[1] "видел" "по"    "сыну," "что"   "он"   

[[5]]
[1] "четырнадцать," "я"             "поселился"     "на"            "улице"        

[[6]]
[1] "широко"     "продолжали" "род."

Note the vectors are of different length.

What I want is to be able to read the first words from each sentence, the second word, the third, etc.

The desired result would be something like this:

    P1              P2           P3                 P4    P5           P6
[1] "модно"         "создавать"  "резюме"           "в"   "виде"       NA
[2] "ты"            "начианешь"  "работать"         "с"   "этими"      NA
[3] "модно"         "называть"   "блогер-рилейшенз" "―"   "начинается" "задолго"         
[4] "видел"         "по"         "сыну,"            "что" "он"         NA
[5] "четырнадцать," "я"          "поселился"        "на"  "улице"      NA
[6] "широко"        "продолжали" "род."             NA    NA           NA

I have tried to just use data.frame() but that didn't work because the rows are of different length. I also tried rbind.fill() from the plyr package, but that function can only process matrices.

I found some other questions here (that's where I got the plyr help from), but those were all about combining for instance two data frames of different size.

Thanks for your help.

Theosophy

maybe sapply(1:length(your_list),function(j) yourlist[[j]][1]) ? – Scary

Answer 1

try this:

word.list <- list(letters[1:4], letters[1:5], letters[1:2], letters[1:6])
n.obs <- sapply(word.list, length)
seq.max <- seq_len(max(n.obs))
mat <- t(sapply(word.list, "[", i = seq.max))

the trick is, that,

c(1:2)[1:4]

returns the vector + two NAs

Bethought

this could be further condensed to one line by: sapply(word.list, '[', seq(max(sapply(word.list, length)))) (as shown here) – Joletta

For those who would use @Arun's one-line solution, note that there must be a transpose t() to create the appropriate columns, as in the original question. – Marche

@Bethought Awesome solution. Could you explain the function "[" ? – Patinated

Nowadays, sapply(word.list, length) can be re-written as lengths(word.list). – Rogue

Answer 2

One liner with plyr

plyr::ldply(word.list, rbind)

Labarbera

any dplyr equivalent since plyr is retired? – Frontogenesis

Answer 3

Another option is stri_list2matrix from library(stringi)

library(stringi)
stri_list2matrix(l, byrow=TRUE)
#    [,1] [,2] [,3] [,4]
#[1,] "a"  "b"  "c"  NA  
#[2,] "a2" "b2" NA   NA  
#[3,] "a3" "b3" "c3" "d3"

NOTE: Data from @juba's post.

Or as @Valentin mentioned in the comments

sapply(l, "length<-", max(lengths(l)))

Or using tidyverse

library(purrr)
library(tidyr)
library(dplyr)
tibble(V = l) %>% 
   unnest_wider(V, names_sep = "")
# A tibble: 3 × 4
  V1    V2    V3    V4   
  <chr> <chr> <chr> <chr>
1 a     b     c     <NA> 
2 a2    b2    <NA>  <NA> 
3 a3    b3    c3    d3   

Yoshida

I think your elegant base R solution given here is worth being mentioned as well: sapply(l, "length<-", max(lengths(l))) – Negligible

if i have list and inside list how to do this? – Constitutionalism

@Constitutionalism For that you may need a nested loop i.e. lapply(l, function(subl) lapply(subl, "length<-", max(lengths(subl)))) – Yoshida

#65979452 . I would request you to have a look at this. My idea was to change the list to dataframe then to data table. Is there way directly to data table? – Constitutionalism

Sir @Yoshida im definitely not one of them and many of my phd biology datanalysis code is taken from your answers from various question you have answered.. – Constitutionalism

@Yoshida do you know any dplyr equivalent? – Frontogenesis

@ÁlvaroA.Gutiérrez-Vargas updated with tidyverse – Yoshida

Answer 4

You can do something like this :

## Example data
l <- list(c("a","b","c"), c("a2","b2"), c("a3","b3","c3","d3"))
## Compute maximum length
max.length <- max(sapply(l, length))
## Add NA values to list elements
l <- lapply(l, function(v) { c(v, rep(NA, max.length-length(v)))})
## Rbind
do.call(rbind, l)

Which gives :

     [,1] [,2] [,3] [,4]
[1,] "a"  "b"  "c"  NA  
[2,] "a2" "b2" NA   NA  
[3,] "a3" "b3" "c3" "d3"

Cryostat

Aha -- what we forgot (Juba and me) is that you don't need to "fill in" the original list elements with NA values. The sapply snippet I put in a comment returns NA for list elements which are shorter than the requested index value. Ain't it nice of sapply not to crash? :-) – Scary

instead of max(sapply(l, length)), you can also use the wrapper lengths- max(lengths(l)) – Bowens

Answer 5

You could also use rbindlist() from the data.table package.

Convert vectors to data.tables or data.frames and transpose them (not sure if this reduces speed a lot) with the help of lapply(). Then bind them with rbindlist() - filling missing cells with NA.

require(data.table)

l = list(c("a","b","c"), c("a2","b2"), c("a3","b3","c3","d3"))
dt = rbindlist(
  lapply(l, function(x) data.table(t(x))),
  fill = TRUE
)

Crossbow

Answer 6

As the question was to convert a list to a data.frame, you can bring all list vectors to the maximum length max(lengths(L)) with length<- used in lapply and the use list2DF to convert this list to a data.frame.

L <- list(a=1, b=2:3, c=3:5)

list2DF(lapply(L, `length<-`, max(lengths(L))))
#   a  b c
#1  1  2 3
#2 NA  3 4
#3 NA NA 5

Melindamelinde

Answer 7

Another option could be to define a function like this (it'd mimic rbind.fill) or use it directly from rowr package:

cbind.fill <- function(...){
  nm <- list(...) 
  nm <- lapply(nm, as.matrix)
  n <- max(sapply(nm, nrow)) 
  do.call(cbind, lapply(nm, function (x) 
    rbind(x, matrix(, n-nrow(x), ncol(x))))) 
}

This response is taken from here (and there're some usage examples).

Receiptor

Not sure, how this works, you don't provide an example. Seems like an exact copy from here: #7962767 Also: rowr is not on CRAN anymore. – Crossbow