CODEWITHSUNDEEP
rdata.tablelapply
gibbz00
gibbz00

Reputation: 1987

lapply with two column arguments

These is my dataframes

library(data.table)
df <- fread('
            Account       Date        NextDate
            A          2016-01-01     2016-02-01
            A          2016-02-01     2016-11-05
            B          2016-03-10     2016-11-05')


ab <- fread('
      Date       Amount
     2015-06-01     55
     2016-01-31     55
     2016-02-28     65
     2016-03-31     75')

I want to create a list by doing a loop as of each row in df and pick all rows from ab where ab$Date is greater than df$Date and less than df$NextDate so that the output looks like this:

[[1]]

Date       Amount
2016-01-31 55

[[2]]

Date        Amount
2016-02-28  65
2016-03-31  75

[[3]]
Date        Amount
2016-03-31  75

This is my attempt:

list<- lapply(df$Date, function(x,y) br[Date > x & Date < y ],y=df$NextDate)

Upvotes: 3

Views: 1056

Answers (3)

Marcelo
Marcelo

Reputation: 4282

You can use apply:

apply(df, 1, function(x) ab[ab$Date>x[2] & ab$Date<x[3],])

Upvotes: 3

IRTFM
IRTFM

Reputation: 263342

It's possible this is not truly in the dataframe traditions, but I think the logic is fairly clear if you are coming from a base-R background. The data.table "[" function needs to have unquoted column names in a list to delive just column values:

apply( df[, list(Date,NextDate)] , 1, 
       function(dts)  ab[ ab$Date > dts['Date'] & ab$Date < dts['NextDate'],  ])
[[1]]
         Date Amount
1: 2016-01-31     55

[[2]]
         Date Amount
1: 2016-02-28     65
2: 2016-03-31     75

[[3]]
         Date Amount
1: 2016-03-31     75

Upvotes: 2

rsoren
rsoren

Reputation: 4206

Create a vector that checks whether rows of ab meet your conditions, then use it to select rows of ab. 

lapply(1:nrow(df), function(x) {
  ab[which(ab$Date > df[x, get("Date")] & ab$Date < df[x, get("NextDate")]), ]
})

Upvotes: 1

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