CODEWITHSUNDEEP
rparsingtree
dholstius
dholstius

Reputation: 1007

How to parse a data.frame into a tree?

Here's a simple taxonomy (labels and IDs):

test_data <- data.frame(
  cat_id = c(661, 197, 228, 650, 126, 912, 949, 428),
  cat_h1 = c(rep("Animals", 5), rep("Plants", 3)),
  cat_h2 = c(rep("Mammals", 3), rep("Birds", 2), c("Wheat", "Grass", "Other")),
  cat_h3 = c("Dogs", "Dogs", "Other", "Hawks", "Other", rep(NA, 3)),
  cat_h4 = c("Big", "Little", rep(NA, 6)))

The parsed structure should match the following:

list(
  Animals = list(Mammals = list(Dogs  = list(Big = 661, Little = 197), Other = 228),
                 Birds   = list(Hawks = 650, Other = 126)),
  Plants  = list(Wheat = 912, Grass = 949, Other = 428))

Upvotes: 3

Views: 489

Answers (3)

bramtayl
bramtayl

Reputation: 4024

I would avoid list structures in preference to tidy data. Here is a way to reduce the redundancy in the data.

library(dplyr)

h1_h2 = 
  test_data %>%
  select(cat_h1, cat_h2) %>%
  distinct %>%
  filter(cat_h2 %>% is.na %>% `!`)

h2_h3 =
  test_data %>%
  select(cat_h2, cat_h3) %>%
  distinct %>%
  filter(cat_h3 %>% is.na %>% `!`)

h3_h4 = 
  test_data %>%
  select(cat_h3, cat_h4) %>%
  distinct %>%
  filter(cat_h4 %>% is.na %>% `!`)

The original can be easily reconstituted:

h1_h2 %>%
  left_join(h2_h3) %>%
  left_join(h3_h4)

Edit: And here's a way to automate the whole process.

library(dplyr)
library(lazyeval)

adjacency = function(data) {
  adjacency_table = function(data, larger_name, smaller_name)
    lazy(data %>%
           select(larger_name, smaller_name) %>%
           distinct %>%
           filter(smaller_name %>% is.na %>% `!`) ) %>%
    interp(larger_name = larger_name %>% as.name, 
           smaller_name = smaller_name %>% as.name) %>%
    lazy_eval %>%
    setNames(c("larger", "smaller"))

  data_frame(smaller_name = data %>% names) %>%
    mutate(larger_name = smaller_name %>% lag) %>%
    slice(-1) %>%
    group_by(larger_name, smaller_name) %>%
    do(adjacency_table(data, .$larger_name, .$smaller_name) )
}

result = 
  test_data %>%
  select(-cat_id) %>%
  adjacency

Upvotes: 1

Neal Fultz
Neal Fultz

Reputation: 9696

If you are OK with the order changing slightly, this is a recursive solution that processes by column:

f <- function(x, d=cbind(x,NA)) {
    c( 
       # call f by branch
       if(ncol(d) > 3) local({
         x <- d[!is.na(d[[3]]),] 
         by( x[-2], droplevels(x[2]), f, x=NA, simplify=FALSE) 
       }), 
       # leaf nodes
       setNames(as.list(d[[1]]), d[[2]])[is.na(d[[3]])] 
    )
}

which will give this:

> str(f(test_data))
List of 2
 $ Animals:List of 2
  ..$ Birds  :List of 2
  .. ..$ Hawks: num 650
  .. ..$ Other: num 126
  ..$ Mammals:List of 2
  .. ..$ Dogs :List of 2
  .. .. ..$ Big   : num 661
  .. .. ..$ Little: num 197
  .. ..$ Other: num 228
 $ Plants :List of 3
  ..$ Wheat: num 912
  ..$ Grass: num 949
  ..$ Other: num 428

Upvotes: 7

Ben Bolker
Ben Bolker

Reputation: 226881

Maybe not the most efficient, but not too hard:

Create data: 

test_data <- data.frame(
  cat_id = c(661, 197, 228, 650, 126, 912, 949, 428),
  cat_h1 = c(rep("Animals", 5), rep("Plants", 3)),
  cat_h2 = c(rep("Mammals", 3), rep("Birds", 2), c("Wheat", "Grass", "Other")),
  cat_h3 = c("Dogs", "Dogs", "Other", "Hawks", "Other", rep(NA, 3)),
  cat_h4 = c("Big", "Little", rep(NA, 6)))

Loop through the data frame and build the list/tree:

tax <- list()  ## initialize
for (i in 1:nrow(test_data)) {
    ## convert data.frame row to character vector
    taxdat <- sapply(test_data[i,-1],as.character)
    taxstr <- character(0)  ## initialize taxon string
    ntax <- length(na.omit(taxdat))
    for (j in 1:ntax) {
        taxstr <- c(taxstr,taxdat[j])  ## build string
        if (is.null(tax[[taxstr]])) {
            tax[[taxstr]] <- list()  ## initialize if necessary
        }
    }
    tax[[taxstr]] <- test_data$cat_id[i]  ## assign value to tip
}

Compare result to desired:

res <- list(
  Animals = list(Mammals = list(Dogs  = list(Big = 661, Little = 197),
                 Other = 228),
                 Birds   = list(Hawks = 650, Other = 126)),
  Plants  = list(Wheat = 912, Grass = 949, Other = 428))

all.equal(res,tax)  ## TRUE

Upvotes: 3

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