How to create all combinations from a nested list while preserving the structure using R?

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盖世英雄少女心
盖世英雄少女心 2021-01-04 07:10

Given a nested list, how to create all possible lists from its elements, while preserving the structure of the nested list?

Nested list:



        
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  • 2021-01-04 07:30

    Putting together the great answers from Ben Nutzer and Joris Chau, we have a way to create all possible combinations from a nested list, regardless of whether some sublist components are of unequal length.

    Put together as a function:

    list.combine <- function(input) {
        # Create list skeleton.
        skeleton <- rapply(input, head, n = 1, how = "list")
    
        # Create storage for the flattened list.
        flattened = list()
    
        # Flatten the list.
        invisible(rapply(input, function(x) {
            flattened <<- c(flattened, list(x))
        }))
    
        # Create all possible combinations from list elements.
        combinations <- expand.grid(flattened, stringsAsFactors = FALSE)
    
        # Create list for storing the output.
        output <- apply(combinations, 1, relist, skeleton = skeleton)
    
        return(output)
    }
    

    Note: If a character type exists in the sublist components, then everything will be coerced to a character. For example:

    # Input list.
    l <- list(
        a = "string",
        b = list(
            c = 1:2,
            d = 3
        )
    )
    
    # Applying the function.
    o <- list.combine(l)
    
    # View the list:
    str(o)
    
    # List of 2
    #  $ :List of 2
    #   ..$ a: chr "string"
    #   ..$ b:List of 2
    #   .. ..$ c: chr "1"
    #   .. ..$ d: chr "3"
    #  $ :List of 2
    #   ..$ a: chr "string"
    #   ..$ b:List of 2
    #   .. ..$ c: chr "2"
    #   .. ..$ d: chr "3"
    

    One--slow--way around this is to relist within a loop which will maintain the data in a 1x1 dataframe. Accessing the dataframe as df[, 1] will give a vector of length 1 of the original type as the element in the input list. For example:

    Updated list.combine():

    list.combine <- function(input) {
        # Create list skeleton.
        skeleton <- rapply(input, head, n = 1, how = "list")
    
        # Create storage for the flattened list.
        flattened = list()
    
        # Flatten the list.
        invisible(rapply(input, function(x) {
            flattened <<- c(flattened, list(x))
        }))
    
        # Create all possible combinations from list elements.
        combinations <- expand.grid(flattened, stringsAsFactors = FALSE)
    
        # Create list for storing the output.
        output <- list()
    
        # Relist and preserve original data type.
        for (i in 1:nrow(combinations)) {
            output[[i]] <- retain.element.type(relist(flesh = combinations[i, ], skeleton = skeleton))
        }
    
        return(output)
    }
    

    Then the retain.element.type():

    retain.element.type <- function(input.list) {
        for (name in names(input.list)) {
            # If the element is a list, recall the function.
            if(inherits(input.list[[name]], "list")) {
                input.list[[name]] <- Recall(input.list[[name]])
    
            # Else, get the first element and preserve the type.
            } else {
                input.list[[name]] <- input.list[[name]][, 1]
            }
        }
        return(input.list)
    }
    

    Example:

    # Input list.
    l <- list(
        a = "string",
        b = list(
            c = 1:2,
            d = 3
        )
    )
    
    # Applying the updated function to preserve the data type.
    o <- list.combine(l)
    
    # View the list:
    str(o)
    
    # List of 2
    #  $ :List of 2
    #   ..$ a: chr "string"
    #   ..$ b:List of 2
    #   .. ..$ c: int 1
    #   .. ..$ d: num 3
    #  $ :List of 2
    #   ..$ a: chr "string"
    #   ..$ b:List of 2
    #   .. ..$ c: int 2
    #   .. ..$ d: num 3
    
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  • 2021-01-04 07:33

    The relist function from utils seems to be designed for this task:

    rl <- as.relistable(l)
    r <- expand.grid(data.frame(rl), KEEP.OUT.ATTRS = F)
    > head(r, 5)
       b c.d.e c.d.f g
    1  1     3     5 7
    2  2     3     5 7
    3  1     4     5 7
    4  2     4     5 7
    5  1     3     6 7
    

    It saves the structure of the list (skeleton). This means one can now manipulate the data within the nested list and re-assign it into the structure (flesh). Here with the first row of the expanded matrix.

    r <- rep(unname(unlist(r[1,])),each = 2)
    l2 <- relist(r, skeleton = rl)
    > l2
    $a
    $a$b
    [1] 1 1
    
    
    $c
    $c$d
    $c$d$e
    [1] 3 3
    
    $c$d$f
    [1] 5 5
    
    
    
    $g
    [1] 7
    
    attr(,"class")
    [1] "relistable" "list"  
    

    Note that since the structure stays the same, I need to supply the same amount of elements as in the original list. This is why used rep to repeat the element twice. One could also fill it with NA, I guess.

    For every possible combination iterate through r (expanded):

    lapply(1:nrow(r), function(x) 
              relist(rep(unname(unlist(r[x,])),each = 2), skeleton = rl))
    
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  • 2021-01-04 07:44

    Unequal sublist lengths

    Here is an approach --extending on Uwe and Ben's answers-- that also works for arbitrary sublist lengths. Instead of calling expand.grid on data.frame(l), first flatten l to a single-level list and then call expand.grid on it:

    ## skeleton
    skel <- rapply(l, head, n = 1L, how = "list")
    
    ## flatten to single level list
    l.flat <- vector("list", length = length(unlist(skel)))
    i <- 0L
    
    invisible(
        rapply(l, function(x) {
              i <<- i + 1L
              l.flat[[i]] <<- x
            })
    )
    
    ## expand all list combinations 
    l.expand <- apply(expand.grid(l.flat), 1L, relist, skeleton = skel)
    
    str(l.expand)
    #> List of 12
    #>  $ :List of 3
    #>   ..$ a:List of 1
    #>   .. ..$ b: num 1
    #>   ..$ c:List of 1
    #>   .. ..$ d:List of 2
    #>   .. .. ..$ e: num 3
    #>   .. .. ..$ f: num 5
    #>   ..$ g: num 7
    #>  ...
    #>  ...
    #>  $ :List of 3
    #>   ..$ a:List of 1
    #>   .. ..$ b: num 2
    #>   ..$ c:List of 1
    #>   .. ..$ d:List of 2
    #>   .. .. ..$ e: num 4
    #>   .. .. ..$ f: num 7
    #>   ..$ g: num 7
    

    Data

    I slightly modified the data structure, so that the sublist components e and f are of unequal length.

    l <- list(
        a = list(
            b = 1:2
        ),
        c = list(
            d = list(
                e = 3:4,
                f = 5:7
            )
        ),
        g = 7
    )
    
    ## calling data.frame on l does not work
    data.frame(l)
    #> Error in (function (..., row.names = NULL, check.rows = FALSE, check.names = TRUE, : arguments imply differing number of rows: 2, 3
    
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  • 2021-01-04 07:45

    Combining Ben Nutzer's brilliant answer and Joris Chau's brilliant comment, the answer will become a one-liner:

    apply(expand.grid(data.frame(l)), 1L, relist, skeleton = rapply(l, head, n = 1L, how = "list")) 
    

    It creates a list of lists with as many elements as rows returned by expand.grid(). The result is better visualised by the output of str():

    str(apply(expand.grid(data.frame(l)), 1L, relist, skeleton = rapply(l, head, n = 1L, how = "list")))
    
    List of 16
     $ :List of 3
      ..$ a:List of 1
      .. ..$ b: num 1
      ..$ c:List of 1
      .. ..$ d:List of 2
      .. .. ..$ e: num 3
      .. .. ..$ f: num 5
      ..$ g: num 7
     $ :List of 3
      ..$ a:List of 1
      .. ..$ b: num 2
      ..$ c:List of 1
      .. ..$ d:List of 2
      .. .. ..$ e: num 3
      .. .. ..$ f: num 5
      ..$ g: num 7
    ...
    ...
    ...
     $ :List of 3
      ..$ a:List of 1
      .. ..$ b: num 2
      ..$ c:List of 1
      .. ..$ d:List of 2
      .. .. ..$ e: num 4
      .. .. ..$ f: num 6
      ..$ g: num 7
    
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