合并列表和合并元素

Question

我有几张清单。例如：

b1 <-
  list(
    duck = list(
      day = "Monday",
      location = list("Cisco Park", "Elephant Park")
    ),
    eagle = list(day = "Saturday"),
    pigeon = list(location = "backyard")
  )

b2 <- list(
  duck = list(day = "Tuesday", location = "Valley Green"),
  goose = list(location = "Old man Johnson's Farm")
)

我想以一种聚合这些列表中每个元素的元素的方式来合并它们。只有当day或list可以是原始列表中的向量时，这才是递归的。但这是一个很深的问题。

desired <-
  list(
    duck = list(
      day = list("Monday", "Tuesday"),
      location = list("Cisco Park", "Elephant Park", "Valley Green")
    ),
    eagle = list(day = "Saturday"),
    pigeon = list(location = "backyard"),
    goose = list(location = "Old man Johnson's Farm")
  )

我写了一个有效的lapply()解决方案，但它很长而且非常慢。接下来，我尝试了Combine/merge lists by elements names：

l <-
  list(b1, b2)
keys <- unique(unlist(lapply(l, names)))
merged <-
  setNames(do.call(mapply, c(FUN = c, lapply(l, `[`, keys))), keys)

dput(merged)

这很快，合并了两个列表，但创建了多个具有相同名称的元素：

list(duck = list(day = "Monday", location = list("Cisco Park", 
    "Elephant Park"), day = "Tuesday", location = "Valley Green"), 
    eagle = list(day = "Saturday"), pigeon = list(location = "backyard"), 
    goose = list(location = "Old man Johnson's Farm"))

Answer 1

我不知道这样的事情将如何高效地完成，但这里有一个选择：

out <- c(b1, b2[setdiff(names(b2), names(b1))])
for (k in intersect(names(b2), names(b1))) {
  elements <- names(b2[[k]])
  for (v in elements) {
    if (v %in% names(b1[[k]])) {
      out[[k]][[v]] <- unique(c(b1[[k]][[v]], b2[[k]][v]))
    } else {
      out[[k]][[v]] <- b2[[k]][v]
    }
  }
}

也许列表不是手头任务的最佳数据结构？我会尝试使用列表列的data.table。下面是一个原始的例子：

library(data.table)
bDT <- rbindlist(lapply(bt, function(x) lapply(x, identity)), fill = TRUE)
bDT[, animal := rep(names(bt), sapply(bt, function(x) max(lengths(x))))]
bDT <- bDT[, .(day = list(unique(day)), location = list(unique(location))), by = animal]

bDT[animal == "duck", location]
# [[1]]
# [[1]][[1]]
# [1] "Cisco Park"
# 
# [[1]][[2]]
# [1] "Elephant Park"
# 
# [[1]][[3]]
# [1] "Valley Green"

Answer 2

多有趣的问题啊！

这种使用嵌套lapply的方法能得到您想要的结果吗？我不确定它的效率有多高--我猜它应该不会太差(保留空值应该可以提高性能)。

l <- list(b1, b2)
l1.keys <- unique(unlist(lapply(l, names)))
l2.keys <- unique(unlist(lapply(lapply(l, "[[", 1), names)))

l1 = lapply(l1.keys, function(l1.key){
    l2 = lapply(l2.keys, function(l2.key){
        as.list(do.call("c", lapply(lapply(l, "[[", l1.key), "[[", l2.key))) # This [as.list(c())]...
    })
    names(l2) = l2.keys
    l2[lengths(l2) != 0] # ...and this prevent NULL in the output.
})
names(l1) = l1.keys
l1

输出：

$duck
$duck$day
$duck$day[[1]]
[1] "Monday"

$duck$day[[2]]
[1] "Tuesday"


$duck$location
$duck$location[[1]]
[1] "Cisco Park"

$duck$location[[2]]
[1] "Elephant Park"

$duck$location[[3]]
[1] "Valley Green"



$eagle
$eagle$day
$eagle$day[[1]]
[1] "Saturday"



$pigeon
$pigeon$location
$pigeon$location[[1]]
[1] "backyard"



$goose
$goose$location
$goose$location[[1]]
[1] "Old man Johnson's Farm"

Answer 3

这基本上是列表的递归合并。我不确定这在整个数据集上的速度有多快，但这应该也适用于任意深度

library(purrr)
library(magrittr)

mergeMyLists <- function(x, y){

  # determine names missing from x
  newNames <- setdiff(names(y), names(x))

  # create new list with all names and data from x
  z <- x
  if(length(newNames) > 0) {
    z[seq(length(x)+1, length(x) + length(newNames)) ] <- list(NULL)
    names(z) <- c(names(x), newNames)
  }

  # if member is list recursively join
  # else combine and make sure it is a list as sometimes input is vector
  map(names(z),
      ~{
        if(is.list(z[[..1]]) && is.list(y[[..1]])) {
          mergeMyLists(z[[..1]], y[[..1]]) 
        } else {
          tmp <- c(z[[..1]], y[[..1]])
          if(length(tmp)> 1 && !is.list(tmp)) as.list(tmp) else tmp
        }
      }
  ) %>% 
    set_names(names(z))
}

mergedList <- mergeMyLists(b1, b2)

你可以通过使你的输入数据更加一致来加速它(这样它就不需要做太多的检查)。例如，b1$duck$location是一个列表，而b2$duck$location是一个字符向量。如果合并函数知道这两个函数都是list，那么它就不需要检查并潜在地将输出转换为list来实现您想要的结构。