Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Convert a data frame to a treeNetwork compatible list

Tags:

r

networkd3

Consider the following data frame:

   Country     Provinces          City Zone
1   Canada   Newfondland      St Johns    A
2   Canada           PEI Charlottetown    B
3   Canada   Nova Scotia       Halifax    C
4   Canada New Brunswick   Fredericton    D
5   Canada        Quebec            NA   NA
6   Canada        Quebec   Quebec City   NA
7   Canada       Ontario       Toronto    A
8   Canada       Ontario        Ottawa    B
9   Canada      Manitoba      Winnipeg    C
10  Canada  Saskatchewan        Regina    D

Would there be a clever way to convert it to a treeNetwork compatible list (from the networkD3 package) in the form of:

CanadaPC <- list(name = "Canada",
                 children = list(
                   list(name = "Newfoundland",
                        children = list(list(name = "St. John's",
                                             children = list(list(name = "A"))))),
                   list(name = "PEI",
                        children = list(list(name = "Charlottetown",
                                             children = list(list(name = "B"))))),
                   list(name = "Nova Scotia",
                        children = list(list(name = "Halifax",
                                             children = list(list(name = "C"))))),
                   list(name = "New Brunswick",
                        children = list(list(name = "Fredericton",
                                             children = list(list(name = "D"))))),
                   list(name = "Quebec",
                        children = list(list(name = "Quebec City"))),
                   list(name = "Ontario",
                        children = list(list(name = "Toronto",
                                             children = list(list(name = "A"))),
                                        list(name = "Ottawa",
                                             children = list(list(name = "B"))))),
                   list(name = "Manitoba",
                        children = list(list(name = "Winnipeg",
                                             children = list(list(name = "C"))))),
                   list(name = "Saskatchewan",
                        children = list(list(name = "Regina",
                                             children = list(list(name = "D")))))))

In order to plot a Reingold-Tilford tree that would have an arbitrary set of levels:

enter image description here

I have tried several sub-optimal routines including a messy combination of for loops but I can't get this in the desired format.

Ideally, the function would scale in order to consider the first column as the root (starting point) and the other columns would be different levels of children.


Edit

A similar question was asked on the same topic and @MrFlick provided an interesting recursive function. The original data frame had a fixed set of levels. I introduced NAs to add another level of complexity (arbitrary set of levels) that is not adressed in @MrFlick initial solution.


Data

structure(list(Country = c("Canada", "Canada", "Canada", "Canada", 
"Canada", "Canada", "Canada", "Canada", "Canada", "Canada"), 
    Provinces = c("Newfondland", "PEI", "Nova Scotia", "New Brunswick", 
    "Quebec", "Quebec", "Ontario", "Ontario", "Manitoba", "Saskatchewan"
    ), City = c("St Johns", "Charlottetown", "Halifax", "Fredericton", 
    NA, "Quebec City", "Toronto", "Ottawa", "Winnipeg", "Regina"
    ), Zone = c("A", "B", "C", "D", NA, NA, "A", "B", "C", 
    "D")), class = "data.frame", row.names = c(NA, -10L), .Names = c("Country", 
"Provinces", "City", "Zone"))
like image 425
Steven Beaupré Avatar asked Jun 09 '15 14:06

Steven Beaupré


1 Answers

A better strategy for this scenario may be a recursive split() Here's such an implementation. First, here's the sample data

dd<-structure(list(Country = c("Canada", "Canada", "Canada", "Canada", 
"Canada", "Canada", "Canada", "Canada", "Canada", "Canada"), 
    Provinces = c("Newfondland", "PEI", "Nova Scotia", "New Brunswick", 
    "Quebec", "Quebec", "Ontario", "Ontario", "Manitoba", "Saskatchewan"
    ), City = c("St Johns", "Charlottetown", "Halifax", "Fredericton", 
    NA, "Quebec City", "Toronto", "Ottawa", "Winnipeg", "Regina"
    ), Zone = c("A", "B", "C", "D", NA, NA, "A", "B", "C", 
    "D")), class = "data.frame", row.names = c(NA, -10L), .Names = c("Country", 
"Provinces", "City", "Zone"))

note that' i've replaced the "NA" strings with true NA values. Now, the function

rsplit <- function(x) {
    x <- x[!is.na(x[,1]),,drop=FALSE]
    if(nrow(x)==0) return(NULL)
    if(ncol(x)==1) return(lapply(x[,1], function(v) list(name=v)))
    s <- split(x[,-1, drop=FALSE], x[,1])
    unname(mapply(function(v,n) {if(!is.null(v)) list(name=n, children=v) else list(name=n)}, lapply(s, rsplit), names(s), SIMPLIFY=FALSE))
}

Then we can run

rsplit(dd)

It seems to work with the test data. The only difference is the order in which the children nodes are arranged.

like image 134
MrFlick Avatar answered Nov 15 '22 22:11

MrFlick