Concatenate lists with override in R

Question

How do I concatenate two lists in R in such a way that duplicate elements from the second one will override corresponding elements from the first one?

I tried c(), but it keeps duplicates:

l1 <- list(a = 1, b = 2)
l1 <- c(l1, list(b = 3, c = 4))

# $a
# [1] 1
# 
# $b
# [1] 2
# 
# $b
# [1] 3
#
# $c
# [1] 4

I want something that will produce this instead:

# $a
# [1] 1
# 
# $b
# [1] 3
#
# $c
# [1] 4

Hope that there exists something simpler than calling some *apply function?

TMS, Do none of the solutions below work for you? How about `modifyList` as the base R approach? — A5C1D2H2I1M1N2O1R2T1, May 02 '15 at 09:57

score 5 · Answer 1 · answered Mar 05 '15 at 10:13

5

This works with duplicated:

> l1[!duplicated(names(l1), fromLast=TRUE)]
$a
[1] 1

$b
[1] 3

$c
[1] 4

answered Mar 05 '15 at 10:13

Peter Diakumis

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thanks this works, but I would still hope for something simpler for such a basic thing... – Tomas Mar 05 '15 at 10:27
2

@TMS what could be possibly simpler than just use `duplicated` combined with simple subsetting? Did you ever see anything simpler in another language? This code will be also very efficient. – David Arenburg Mar 05 '15 at 10:47
perhaps something like `l1 + l2`? :-) I mean, this feels so basic that there should be an operator for that. – Tomas Mar 05 '15 at 13:04
2

@TMS, why would this be a `+` condition? If you're not happy with the other straightforward and standard ways of achieving what you want, you might just have to write the solution yourself. If you are [unhappy with how something is designed](http://stackoverflow.com/q/27943961/1270695) and others offer you solutions that are efficient and work *but you are still unhappy because it doesn't work using your expected syntax*, then you're the best one to try to make your expected syntax work by writing your own function or method or operator or something else. :-/ – A5C1D2H2I1M1N2O1R2T1 Mar 05 '15 at 16:23

A5C1D2H2I1M1N2O1R2T1 · Accepted Answer · 2015-03-06T09:35:34.147

4

I suspect that you could get the behavior you want by using the "rlist" package; specifically, list.merge seems to do what you are looking for:

l1 <- list(a = 1, b = 2)
l2 <- list(b = 3, c = 4)
library(rlist)
list.merge(l1, l2)
# $a
# [1] 1
# 
# $b
# [1] 3
# 
# $c
# [1] 4

Try list.merge(l2, l1) to get a sense of the function's behavior.

If you have just two lists, as noted by @PeterDee, the list.merge function pretty much makes modifyList accept more than two lists. Thus, for this particular example, you could skip loading any packages and directly do:

modifyList(l1, l2)

edited Mar 06 '15 at 09:35

answered Mar 06 '15 at 05:01

A5C1D2H2I1M1N2O1R2T1

190,393
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This is great - it is built on top of `utils::modifyList`. – Peter Diakumis Mar 06 '15 at 05:35
That's it! This should be in the base R :-) – Tomas Mar 06 '15 at 09:29
@TMS, it sort of *is* in base R. See the update at the end of my answer, in reference to PeterDee's comment. – A5C1D2H2I1M1N2O1R2T1 Mar 06 '15 at 09:36
`modifyList` is the way to go! – Tomas May 12 '15 at 14:32

score 1 · Answer 3 · answered Mar 05 '15 at 10:13

1

I am sure there is a quicker way than:

l1 <- list(a = 1, b = 2)
l2 <- list(b = 3, c = 4)

l1 <- c(l1[!(names(l1) %in% names(l2))],  l2)

You take the names of the l1 that are not in l2 and concatenate the two lists

answered Mar 05 '15 at 10:13

dimitris_ps

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thanks this works, but I would still hope for something simpler for such a basic thing... – Tomas Mar 05 '15 at 10:27

score 0 · Answer 4 · answered Mar 05 '15 at 10:19

0

I would suggest union:

l1 <- list(a = 1, b = 3)
l2 <- list(b = 3, c = 4)

union(l1, l2)

This works like a union in mathematics (Wiki link)

answered Mar 05 '15 at 10:19

martin

63
5

Yes -- this is the output I get: [[1]] [1] 1 [[2]] [1] 3 [[3]] [1] 4 – martin Mar 05 '15 at 10:26

Concatenate lists with override in R

4 Answers4