CODEWITHSUNDEEP
rdata.tabletidyverse
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Reputation: 1035

How to access a different part of the data while using pipe

Let say I have the following data

AgentID Position FriendID
1 1 2
2 4 1
3 10 2

What I want to do is something like following

d[, .SD , by=AgentID] %>%
(for each of the subsets - disected according to AgentID, called the current AgentID -, 
select the part of `d` where the FriendID is the current AgentID. )
(calculate the distance between Friend's position and current Agent's position)

How can I do that?

Do note that, the above is just a toy example to just to understand the concept

Edit:

The expected output would be like

AgentID Position FriendID distanceToFriend
1 1 2 3
2 4 1 3
3 10 2 6

Upvotes: 0

Views: 96

Answers (4)

Waldi
Waldi

Reputation: 41220

The code below make a self join with data.table from d on itself, linking AgentID with FriendID via the on = .(AgentID = FriendID) argument.
You might have a look at this tutorial to understand better the join logic.
In this kind of LHS[RHS,...] join, if variables have the same name on both sides (which is obviously the case on a self join), you can choose which side to use with the x. prefix (not used here) for LHS and i. prefix for RHS.
x. and i. refers to general data.table X[i,j,by] syntax.

d[d,.(AgentID, 
      Position = i.Position,
      FriendID = i.FriendID,
      distance = abs(Position-i.Position))
   ,on = .(AgentID = FriendID)]

#   AgentID Position FriendID distance
#1:       1        1        2        3
#2:       2        4        1        3
#3:       3       10        2        6

Upvotes: 3

Ronak Shah
Ronak Shah

Reputation: 388982

We can perform a self-join :

library(dplyr)

inner_join(df, df, by = c('AgentID' = 'FriendID')) %>%
  transmute(FriendID = AgentID, 
            AgentID = AgentID.y,
            distanceToFriend = abs(Position.x - Position.y)) %>%
  arrange(AgentID)

#  FriendID AgentID distanceToFriend
#1        2       1                3
#2        1       2                3
#3        2       3                6

Upvotes: 1

ThomasIsCoding
ThomasIsCoding

Reputation: 101337

Another data.table option

d[
  ,
  distanceToFriend := .SD[
    ,
    abs(diff(d[, Position][unlist(.(AgentID, FriendID))]))
  ],
  AgentID
][]

gives

   AgentID Position FriendID distanceToFriend
1:       1        1        2                3
2:       2        4        1                3
3:       3       10        2                6

Upvotes: 0

akrun
akrun

Reputation: 887088

We can use

d$distanceToFriend <- abs(d$Position - d$Position[match(d$FriendID, d$AgentID)])

Upvotes: 0

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