StackOverflow2013

Note that there are some explanatory texts on larger screens.

plurals

POAdd columns to a list of data.frames
primarykey
Id
17394372
data
AcceptedAnswerId
0
AnswerCount
0
ClosedDate
CommentCount
11
CommunityOwnedDate
CreationDate
2013-06-30T20:06:56.333
FavoriteCount
0
LastActivityDate
2017-03-21T10:35:30.963
LastEditDate
2017-05-23T12:32:11.370
LastEditorUserId
-1
OwnerUserId
2380782
ParentId
0
PostTypeId
1
Score
0
ViewCount
109
LastEditorDisplayName
text
Body
This is a similar question to an old post of mine <a href="https://stackoverflow.com/questions/17286813/add-columns-to-a-dataframe-based-on-values-from-a-list">Add columns to a dataframe based on values from a list</a>. Now, I would like to use the solution provide it, but instead of applying it to a single data.frame, I'd like to use it on a list of data.frame. Briefly, I have a list of data.frames that looks like this: <pre><code>df <- list(data.frame(A=c("a","b","c"), B=c("1","2","1"), C=c(0.1,0.7,0.4)), data.frame(A=c("d","e","f"), B=c("2","2","3"), C=c(0.5,0.1,0.5)), data.frame(A=c("g","h","i"), B=c("3","1","2"), C=c(0.2,0.1,0.5))) </code></pre> And a list with elements which names match to <code>df$B</code>, i.e, these values are permutations of values from <code>df$B</code>, here is an example: <pre><code> ll <- list('1'=c(0.1,0.1,0.4,0.2,0.1,0.4), '2'=c(0.1,0.1,0.5,0.7,0.5,0.7), '3'=c(0.1,0.1,0.2,0.2,0.2,0.5)) </code></pre> I want to create a new list of data.frames but with new columns in each dataframe of list <code>df</code> that correspond to the values of <code>df$B</code> in list <code>ll</code> but at the same time they are sampled values from <code>ll</code>? Here is a desired output for a better explanation <pre><code>> list.df [[1]] A B C P1 P2 P3 P4 P5 P6 1 a 1 0.1 0.1 0.1 0.4 0.2 0.1 0.4 2 b 2 0.7 0.1 0.5 0.7 0.1 0.5 0.1 3 c 1 0.4 0.4 0.1 0.2 0.1 0.1 0.4 [[2]] A B C P1 P2 P3 P4 P5 P6 1 d 2 0.5 0.1 0.7 0.5 0.1 0.7 0.1 2 e 2 0.1 0.7 0.5 0.1 0.7 0.1 0.5 3 f 3 0.5 0.5 0.5 0.2 0.1 0.2 0.1 [[3]] A B C P1 P2 P3 P4 P5 P6 1 g 3 0.2 0.1 0.5 0.2 0.2 0.2 0.5 2 h 1 0.1 0.2 0.1 0.4 0.2 0.2 0.4 3 i 2 0.5 0.1 0.5 0.1 0.1 0.5 0.7 </code></pre> The solution that I have for one single data.frame is this: <pre><code>sampfun <- function(i, l) sample(l[[as.character(i)]], 10000, replace=TRUE) list.df <- cbind(df, t(sapply(df$B, sampfun, l = ll))) </code></pre> The problem is that I don't know how to implement this solution for use with a list of data.frames. Many thanks for the help Note: my real list of data.frames has 9,000 elements, and I look to add more than 10,000 columns so the memory and speed up are important. 
Tags
<r><dataframe>
Title
Add columns to a list of data.frames
singulars
PostAcceptedAnswerId
1. This table or related slice is empty.
PostParentId
1. This table or related slice is empty.
PostTypePostTypeId
1. PTQuestion
UserLastEditorUserId
1. USCommunity
UserOwnerUserId
1. USuser2380782
plurals
PostLinksPostIdRelatedPostId
1. PL
 singulars
 LinkTypeLinkTypeId
 LTLinked
PostLinksRelatedPostIdPostId
1. This table or related slice is empty.
PostsAcceptedAnswerId
1. This table or related slice is empty.
PostsParentIdCreationDate
1. This table or related slice is empty.
VotesPostIdCreationDate
1. This table or related slice is empty.
CommentsPostId

Querying!

Guidance

A row detail

Detail views are divided into sections. All the information in the data section comes from columns in the selected row. The other sections display data from other, related rows.

Related data can be related in a to-one or a to-many fashion. Captions of data related in a to-many fashion link to a list view showing a filtered view of the table.

Try moving around until you find a non-empty to-many entry and click on the label to get to one. You can move back to the root by clicking on the database name in the header.