CODEWITHSUNDEEP
rlistloops
Shaun
Shaun

Reputation: 949

Adding lists together with a loop

I'm trying to add lists together using a loop. Here is some example data.

df <- data.frame(var1 = c(1,1,2,2,2,2,3,3,3,3,3), var2= 1:11)
> df
   var1 var2
1     1    1
2     1    2
3     2    3
4     2    4
5     2    5
6     2    6
7     3    7
8     3    8
9     3    9
10    3   10
11    3   11

I've run this loop code, and would like the items to be stored in a file that contains 3 lists

list_container <- list()
for (i in  unique(df$var1) ) {
  templist <-  df[ df$var1==i , "var2"] 
  list_container <- list(list_container, templist)
}

it doesn't work, and ends up looking like this

> list_container
[[1]]
[[1]][[1]]
[[1]][[1]][[1]]
list()

[[1]][[1]][[2]]
[1] 1 2

[[1]][[2]]
[1] 3 4 5 6

[[2]]
[1]  7  8  9 10 11

I want the 3 sets of list to sit separately, it should end up like this

list_result <- list(1:2, 3:6, 7:11)
> list_result
[[1]]
[1] 1 2

[[2]]
[1] 3 4 5 6

[[3]]
[1]  7  8  9 10 11

Is there anyway I can modify my code to get the desired result? Any help greatly appreciated. Thanks

Upvotes: 1

Views: 47

Answers (3)

Onyambu
Onyambu

Reputation: 79208

You could also use unstack:

unstack(df, var2~var1)
$`1`
[1] 1 2

$`2`
[1] 3 4 5 6

$`3`
[1]  7  8  9 10 11

if you do not want the names, you can get rid of them:

unname(unstack(df, var2~var1))
[[1]]
[1] 1 2

[[2]]
[1] 3 4 5 6

[[3]]
[1]  7  8  9 10 11

Upvotes: 2

ThomasIsCoding
ThomasIsCoding

Reputation: 101189

Another base R option using tapply

> with(df, tapply(var2, var1, c))
$`1`
[1] 1 2

$`2`
[1] 3 4 5 6

$`3`
[1]  7  8  9 10 11

or aggregate

> aggregate(var2 ~ ., df, c)$var2
[[1]]
[1] 1 2

[[2]]
[1] 3 4 5 6

[[3]]
[1]  7  8  9 10 11

Upvotes: 1

akrun
akrun

Reputation: 887008

split would be more direct and faster

with(df, unname(split(var2, var1)))

-output

[[1]]
[1] 1 2

[[2]]
[1] 3 4 5 6

[[3]]
[1]  7  8  9 10 11

If we want to use the == with unique elements, initialize with a NULL list of length same as the length of unique elements of 'var1' column. Loop over the sequence of unique elements, and assign the subset of 'var2' to the ith element of 'list_container'

un1 <- unique(df$var1)
list_container <- vector('list', length(un1))
for(i in seq_along(un1)) 
    list_container[[i]] <- df$var2[df$var1 == un1[i]]

-output

list_container
[[1]]
[1] 1 2

[[2]]
[1] 3 4 5 6

[[3]]
[1]  7  8  9 10 11

Upvotes: 1

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