Convert data frame to list of lists by row

Question

I am rewriting some old code where I take a dataframe in r and convert it using the tidyverse packages to a list of lists, where each element is one row of the original dataframe - which is itself a list with each column an element.

My previous function achieved it like so:

library(tidyverse)
feat <- function(coords, height = 20, vjust = 75, fill = "orange"){
 inst <- purrr::flatten(purrr::by_row(coords, ..f = function(x) 
 rec <- list(type = "rect", 
 x = as.numeric(x[1]), 
 y = vjust, 
 width = as.numeric(x[2]-x[1]),
 height = height, 
 fill = fill), 
 .labels = FALSE))
 return(inst)
}

However by_row() has been depreciated from the purrr package and so I would like to rewrite it. This is my attempt:

feat2 <- function(coords, height = 20, vjust = 75, fill = "orange"){
 inst <- coords %>% 
 mutate(type = "rect", 
 x = as.numeric(start), 
 y = vjust, 
 width = as.numeric(end-start),
 height = height, 
 fill = fill) %>%
 select(-start, -end) %>% 
 mutate(count = 1:n()) %>%
 nest(-count) %>%
 select(-count) %>% 
 mutate(data = map(data, ~ flatten(.x))) %>% pull()
 return(inst)
}

which does the job but I feel there should be a quicker, more elegant way to achieve this. Do you have any ideas on that aspect?

Here is an example data set:

coords <- structure(list(start = c(126, 433, 603, 1604), end = c(327, 495, 
644, 1831)), .Names = c("start", "end"), row.names = c(NA, -4L
), class = "data.frame")

and the desired output:

result <- list(structure(list(type = "rect", x = 126, y = 75, width = 201, 
 height = 20, fill = "blue"), .Names = c("type", "x", "y", 
"width", "height", "fill")), structure(list(type = "rect", x = 433, 
 y = 75, width = 62, height = 20, fill = "blue"), .Names = c("type", 
"x", "y", "width", "height", "fill")), structure(list(type = "rect", 
 x = 603, y = 75, width = 41, height = 20, fill = "blue"), .Names = c("type", 
"x", "y", "width", "height", "fill")), structure(list(type = "rect", 
 x = 1604, y = 75, width = 227, height = 20, fill = "blue"), .Names = c("type", 
"x", "y", "width", "height", "fill")))

• Aurèle: Note that by_row has been migrated to purrrlyr and purrrlyr::by_row doesn't seem to be deprecated (though I'll admit the tools in purrrlyr seem to be left "aside" of the rest of the tidyverse).

Answer 1

without any external packages:

feat3 <- function(coords, height = 20, vjust = 75, fill = "blue"){
 xx <- apply(coords, 1, function(x) {
 list(type = "rect",
 x = as.numeric(x[1]), 
 y = vjust, 
 width = as.numeric(x[2] - x[1]),
 height = height,
 fill = fill)
 }
 )
 return(xx)
}
all.equal(feat3(coords), result)
# [1] TRUE

Benchmarks:

microbenchmark::microbenchmark(
 feat3(coords),
 feat2(coords), unit = "relative"
)
# Unit: relative
# expr min lq mean median uq max neval cld
# feat3(coords) 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 100 a 
# feat2(coords) 315.8261 297.2919 254.5025 258.7822 248.2652 286.9575 100 b

• flodel: It's somewhat undesirable that apply first converts all the data to characters. It also forces you to use x[1] and x[2] to access specific columns, so you lose code readability a bit... I would suggest you build a data.frame x then do unname(do.call(Map, c(list, x))) to convert to a list of list. It's not nearly as fast but it is more robust and maintainable. In fact, the op might want to define df_to_lol <- function(df) unname(do.call(Map, c(list, df))) so he can use it within tidyverse, e.g., mutate(...) %>% df_to_lol.
• minem: @flodel It sounds that you should write a separate answer...
• flodel: There are not a lot of people posting answers in the [r] group (thanks for joining and welcome). I prefer we help each other improve our answers than get into a competition for best answer.

Answer 2

I also asked this question over on the Rstudio community forums where user mgirlich provided this answer:

feat4 <- function(coords, height = 20, vjust = 75, fill = "orange") {
 coords %>%
 # transmute keeps only the columns specified, so there is no need to
 # deselect start and end afterwards
 transmute(
 type = "rect",
 x = as.numeric(start),
 y = vjust,
 width = as.numeric(end - start),
 height = height,
 fill = fill
 ) %>%
 purrr::transpose()
}

It is slower than the answer by @minem but the use of the transpose() function was new to me so I'll post it here.