Merging data frames stored in lists

Question

I have two lists. Every component in the lists is a data frame. The two lists are symmetric. They both contain data frames for years 2006-2012, just on different themes. I would like to merge the data frames ' horizontally' (that is the one of 2006 in the first list with that of 2006 in the second list, and so on) obtaining a third list of data frames. I tried to figure out how to do that with lapply, but there must be something I didn't understand about that function.

Could you please help?

Thank you.

Answer 1

Something like l3 in this code, you mean?

DT1 = data.frame(A=1:3,B=letters[1:3])
DT2 = data.frame(A=4:5,B=letters[4:5])
l1 = list(DT1,DT2)
DT1 = data.frame(A=1:3,C=letters[7:9])
DT2 = data.frame(A=4:5,C=letters[11:12])
l2 = list(DT1,DT2)

l3 <- vector(mode = "list", length = length(l1))
for ( i in 1:length(l1))
{
l3[[i]]   <- merge(l2[[i]],l1[[i]], by = "A")
}

Answer 2

Perhaps something like this is what you're after?

df1 <- data.frame(year = 2006, x = 1:3)
df2 <- data.frame(year = 2007, x = 4:6)
df3 <- data.frame(year = 2006, x = 7:9)
df4 <- data.frame(year = 2007, x = 10:12)

l1 <- list(x2006 = df1, x2007 = df2)
l2 <- list(x2006 = df3, x2007 = df4)

lapply(names(l1), function(x) cbind(l1[[x]], l2[[x]]))
####
[[1]]
  year x year x
1 2006 1 2006 7
2 2006 2 2006 8
3 2006 3 2006 9

[[2]]
  year x year  x
1 2007 4 2007 10
2 2007 5 2007 11
3 2007 6 2007 12

There may be other functions that would be more appropriate than cbind() such as merge(), but this should get you on the right path. This obviously assumes that you have named your lists and those names are consistent between l1 and l2.

EDITED TO ADD SOME MORE CONTEXT

There are a few key assumptions that make this work. Those assumptions are:

1. Your list objects have names
2. The names in each list are consistent between lists

So, what are the names I'm referring to? If you look at the code about where I define l1, you'll see x2006 = df1 and x2007 = df2. I'm defining two objects in that list, df1 and df2 with two names x2006 and x2007.

You can check the names of the list by asking for the names():

names(l1)
####
[1] "x2006" "x2007"

The other key assumption is that you can index objects in a list by their name, using the [[ function. For example:

l1[["x2006"]]
####
  year x
1 2006 1
2 2006 2
3 2006 3

So what we're doing with the lapply function is that we're iterating over the names of l1, defining an anonymous function, and then using the [[ function to index the two list objects l1 and l2. We're currently using cbind as the function, but you can replace cbind with almost any other function.

As I mentioned above, this assumes that the names are the same between the two or more list objects. For example, this does not work:

#change the names of the l2 list
names(l2) <- c("foo", "bar")
lapply(names(l1), function(x) cbind(l1[[x]], l2[[x]]))
####
Error in data.frame(..., check.names = FALSE) : 
  arguments imply differing number of rows: 3, 0

The names however do not have to be in the same order. That's where the benefit of the [[ function comes in. To wit:

#Fix names on l2 again
names(l2) <- c("x2006", "x2007")
l2reverse <- list(x2007 = df4, x2006 = df3)

all.equal(
  lapply(names(l1), function(x) cbind(l1[[x]], l2[[x]])),  
  lapply(names(l1), function(x) cbind(l1[[x]], l2reverse[[x]]))
)
####
[1] TRUE

Answer 3

mapply might be of use here too.

Here's a third interpretation of what you might be asking for:

Some sample data:

DT1 <- data.frame(A=1:3, B=letters[1:3])
DT2 <- data.frame(A=4:5, C=letters[4:5])
l1 <- list(DT1,DT2)
DT1 <- data.frame(A=1:3, B=letters[7:9])
DT2 <- data.frame(A=4:5, C=letters[11:12])
l2 = list(DT1,DT2)

merge with mapply:

mapply(FUN=function(x, y) merge(x, y, by="A"), 
       l1, l2, SIMPLIFY=FALSE)
# [[1]]
#   A B.x B.y
# 1 1   a   g
# 2 2   b   h
# 3 3   c   i
# 
# [[2]]
#   A C.x C.y
# 1 4   d   k
# 2 5   e   l

---

For reference....

Here's @Chase's interpretation of your question done with mapply:

mapply(cbind, l1, l2, SIMPLIFY=FALSE)
# $x2006
#   year x year x
# 1 2006 1 2006 7
# 2 2006 2 2006 8
# 3 2006 3 2006 9
# 
# $x2007
#   year x year  x
# 1 2007 4 2007 10
# 2 2007 5 2007 11
# 3 2007 6 2007 12

Here's @Codoremifa's interpretation of your question done with mapply:

mapply(FUN=function(x, y) merge(x, y), 
       l1, l2, SIMPLIFY=FALSE)
# [[1]]
#   A B C
# 1 1 a g
# 2 2 b h
# 3 3 c i
# 
# [[2]]
#   A B C
# 1 4 d k
# 2 5 e l

What would be more helpful is if you post some sample data and your expected output so that there is less guessing about what you're trying to do :-)