在 R 中合并两个列表

Question

I have two lists

first = list(a = 1, b = 2, c = 3)
second = list(a = 2, b = 3, c = 4)

I want to merge these two lists so the final product is

$a
[1] 1 2

$b
[1] 2 3

$c
[1] 3 4

Is there a simple function to do this?

Answer 1

如果列表始终具有相同的结构，如示例中所示，那么更简单的解决方案是

mapply(c, first, second, SIMPLIFY=FALSE)

Answer 2

This is a very simple adaptation of the modifyList function by Sarkar. Because it is recursive, it will handle more complex situations than mapply would, and it will handle mismatched name situations by ignoring the items in 'second' that are not in 'first'.

appendList <- function (x, val) 
{
    stopifnot(is.list(x), is.list(val))
    xnames <- names(x)
    for (v in names(val)) {
        x[[v]] <- if (v %in% xnames && is.list(x[[v]]) && is.list(val[[v]])) 
            appendList(x[[v]], val[[v]])
        else c(x[[v]], val[[v]])
    }
    x
}

> appendList(first,second)
$a
[1] 1 2

$b
[1] 2 3

$c
[1] 3 4

Answer 3

Here are two options, the first:

both <- list(first, second)
n <- unique(unlist(lapply(both, names)))
names(n) <- n
lapply(n, function(ni) unlist(lapply(both, `[[`, ni)))

and the second, which works only if they have the same structure:

apply(cbind(first, second),1,function(x) unname(unlist(x)))

Both give the desired result.

Answer 4

Here's some code that I ended up writing, based upon @Andrei's answer but without the elegancy/simplicity. The advantage is that it allows a more complex recursive merge and also differs between elements that should be connected with rbind and those that are just connected with c :

# Decided to move this outside the mapply, not sure this is 
# that important for speed but I imagine redefining the function
# might be somewhat time-consuming
mergeLists_internal <- function(o_element, n_element){
  if (is.list(n_element)){
    # Fill in non-existant element with NA elements
    if (length(n_element) != length(o_element)){
      n_unique <- names(n_element)[! names(n_element) %in% names(o_element)]
      if (length(n_unique) > 0){
        for (n in n_unique){
          if (is.matrix(n_element[[n]])){
            o_element[[n]] <- matrix(NA, 
                                     nrow=nrow(n_element[[n]]), 
                                     ncol=ncol(n_element[[n]]))
          }else{
            o_element[[n]] <- rep(NA, 
                                  times=length(n_element[[n]]))
          }
        }
      }

      o_unique <- names(o_element)[! names(o_element) %in% names(n_element)]
      if (length(o_unique) > 0){
        for (n in o_unique){
          if (is.matrix(n_element[[n]])){
            n_element[[n]] <- matrix(NA, 
                                     nrow=nrow(o_element[[n]]), 
                                     ncol=ncol(o_element[[n]]))
          }else{
            n_element[[n]] <- rep(NA, 
                                  times=length(o_element[[n]]))
          }
        }
      }
    }  

    # Now merge the two lists
    return(mergeLists(o_element, 
                      n_element))

  }
  if(length(n_element)>1){
    new_cols <- ifelse(is.matrix(n_element), ncol(n_element), length(n_element))
    old_cols <- ifelse(is.matrix(o_element), ncol(o_element), length(o_element))
    if (new_cols != old_cols)
      stop("Your length doesn't match on the elements,",
           " new element (", new_cols , ") !=",
           " old element (", old_cols , ")")
  }

  return(rbind(o_element, 
               n_element, 
               deparse.level=0))
  return(c(o_element, 
           n_element))
}
mergeLists <- function(old, new){
  if (is.null(old))
    return (new)

  m <- mapply(mergeLists_internal, old, new, SIMPLIFY=FALSE)
  return(m)
}

Here's my example:

v1 <- list("a"=c(1,2), b="test 1", sublist=list(one=20:21, two=21:22))
v2 <- list("a"=c(3,4), b="test 2", sublist=list(one=10:11, two=11:12, three=1:2))
mergeLists(v1, v2)

This results in:

$a
     [,1] [,2]
[1,]    1    2
[2,]    3    4

$b
[1] "test 1" "test 2"

$sublist
$sublist$one
     [,1] [,2]
[1,]   20   21
[2,]   10   11

$sublist$two
     [,1] [,2]
[1,]   21   22
[2,]   11   12

$sublist$three
     [,1] [,2]
[1,]   NA   NA
[2,]    1    2

Yeah, I know - perhaps not the most logical merge but I have a complex parallel loop that I had to generate a more customized .combine function for, and therefore I wrote this monster :-)

Answer 5

In general one could,

merge_list <- function(...) by(v<-unlist(c(...)),names(v),base::c)

Note that the by() solution returns an attribute d list, so it will print differently, but will still be a list. But you can get rid of the attributes with attr(x,"_attribute.name_")<-NULL . You can probably also use aggregate() .

Answer 6

merged = map(names(first), ~c(first[[.x]], second[[.x]])
merged = set_names(merged, names(first))

Using purrr. Also solves the problem of your lists not being in order.

Answer 7

We can do a lapply with c() , and use setNames to assign the original name to the output.

setNames(lapply(1:length(first), function(x) c(first[[x]], second[[x]])), names(first))

$a
[1] 1 2

$b
[1] 2 3

$c
[1] 3 4

Answer 8

Following @Aaron left Stack Overflow and @Theo answer, the merged list's elements are in form of vector c . But if you want to bind rows and columns use rbind and cbind .

merged = map(names(first), ~rbind(first[[.x]], second[[.x]])
merged = set_names(merged, names(first))

Answer 9

Using dplyr, I found that this line works for named lists using the same names:

as.list(bind_rows(first, second))