split join data.table R












2















Objective



Join DT1 (as i in data.table) to DT2 given key(s) column(s), within each group of DT2 specified by the Date column.



I cannot run DT2[DT1, on = 'key'] as that would be incorrect since key column is repeated across the Date column, but unique within a single date.



Reproducible example with a working solution



DT3 is my expected output. Is there any way to achieve this without the split manoeuvre, which does not feel very data.table-y? 



library(data.table)

set.seed(1)

DT1 <- data.table(

  Segment  = sample(paste0('S', 1:10), 100, TRUE), 

  Activity = sample(paste0('A', 1:5), 100, TRUE), 

  Value    = runif(100)

)

dates <- seq(as.Date('2018-01-01'), as.Date('2018-11-30'), by = '1 day')

DT2 <- data.table(

  Date    = rep(dates, each = 5), 

  Segment = sample(paste0('S', 1:10), 3340, TRUE), 

  Total   = runif(3340, 1, 2)

)

rm(dates)

# To ensure that each Date Segment combination is unique

DT2 <- unique(DT2, by = c('Date', 'Segment'))

iDT2 <- split(DT2, by = 'Date')

iDT2 <- lapply(

  iDT2, 

  function(x) {

    x[DT1, on = 'Segment', nomatch = 0]

  }

)

DT3 <- rbindlist(iDT2, use.names = TRUE)





r data.table







share|improve this question























  • What about using merge function with keys 'Date' and 'Segment'?

    – Heikki
    Nov 12 '18 at 20:59













  • Date does not exist in DT1. Hence, can't be used to merge

    – Ameya
    Nov 12 '18 at 21:03











  • Sorry, I should have written merge by 'Segment' (only).

    – Heikki
    Nov 12 '18 at 21:40
















2















Objective



Join DT1 (as i in data.table) to DT2 given key(s) column(s), within each group of DT2 specified by the Date column.



I cannot run DT2[DT1, on = 'key'] as that would be incorrect since key column is repeated across the Date column, but unique within a single date.



Reproducible example with a working solution



DT3 is my expected output. Is there any way to achieve this without the split manoeuvre, which does not feel very data.table-y? 



library(data.table)

set.seed(1)

DT1 <- data.table(

  Segment  = sample(paste0('S', 1:10), 100, TRUE), 

  Activity = sample(paste0('A', 1:5), 100, TRUE), 

  Value    = runif(100)

)

dates <- seq(as.Date('2018-01-01'), as.Date('2018-11-30'), by = '1 day')

DT2 <- data.table(

  Date    = rep(dates, each = 5), 

  Segment = sample(paste0('S', 1:10), 3340, TRUE), 

  Total   = runif(3340, 1, 2)

)

rm(dates)

# To ensure that each Date Segment combination is unique

DT2 <- unique(DT2, by = c('Date', 'Segment'))

iDT2 <- split(DT2, by = 'Date')

iDT2 <- lapply(

  iDT2, 

  function(x) {

    x[DT1, on = 'Segment', nomatch = 0]

  }

)

DT3 <- rbindlist(iDT2, use.names = TRUE)





r data.table







share|improve this question























  • What about using merge function with keys 'Date' and 'Segment'?

    – Heikki
    Nov 12 '18 at 20:59













  • Date does not exist in DT1. Hence, can't be used to merge

    – Ameya
    Nov 12 '18 at 21:03











  • Sorry, I should have written merge by 'Segment' (only).

    – Heikki
    Nov 12 '18 at 21:40














2












2








2








Objective



Join DT1 (as i in data.table) to DT2 given key(s) column(s), within each group of DT2 specified by the Date column.



I cannot run DT2[DT1, on = 'key'] as that would be incorrect since key column is repeated across the Date column, but unique within a single date.



Reproducible example with a working solution



DT3 is my expected output. Is there any way to achieve this without the split manoeuvre, which does not feel very data.table-y? 



library(data.table)

set.seed(1)

DT1 <- data.table(

  Segment  = sample(paste0('S', 1:10), 100, TRUE), 

  Activity = sample(paste0('A', 1:5), 100, TRUE), 

  Value    = runif(100)

)

dates <- seq(as.Date('2018-01-01'), as.Date('2018-11-30'), by = '1 day')

DT2 <- data.table(

  Date    = rep(dates, each = 5), 

  Segment = sample(paste0('S', 1:10), 3340, TRUE), 

  Total   = runif(3340, 1, 2)

)

rm(dates)

# To ensure that each Date Segment combination is unique

DT2 <- unique(DT2, by = c('Date', 'Segment'))

iDT2 <- split(DT2, by = 'Date')

iDT2 <- lapply(

  iDT2, 

  function(x) {

    x[DT1, on = 'Segment', nomatch = 0]

  }

)

DT3 <- rbindlist(iDT2, use.names = TRUE)





r data.table







share|improve this question














Objective



Join DT1 (as i in data.table) to DT2 given key(s) column(s), within each group of DT2 specified by the Date column.



I cannot run DT2[DT1, on = 'key'] as that would be incorrect since key column is repeated across the Date column, but unique within a single date.



Reproducible example with a working solution



DT3 is my expected output. Is there any way to achieve this without the split manoeuvre, which does not feel very data.table-y? 



library(data.table)

set.seed(1)

DT1 <- data.table(

  Segment  = sample(paste0('S', 1:10), 100, TRUE), 

  Activity = sample(paste0('A', 1:5), 100, TRUE), 

  Value    = runif(100)

)

dates <- seq(as.Date('2018-01-01'), as.Date('2018-11-30'), by = '1 day')

DT2 <- data.table(

  Date    = rep(dates, each = 5), 

  Segment = sample(paste0('S', 1:10), 3340, TRUE), 

  Total   = runif(3340, 1, 2)

)

rm(dates)

# To ensure that each Date Segment combination is unique

DT2 <- unique(DT2, by = c('Date', 'Segment'))

iDT2 <- split(DT2, by = 'Date')

iDT2 <- lapply(

  iDT2, 

  function(x) {

    x[DT1, on = 'Segment', nomatch = 0]

  }

)

DT3 <- rbindlist(iDT2, use.names = TRUE)





r data.table




r data.table






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 12 '18 at 20:49









AmeyaAmeya

1,0291819




1,0291819













  • What about using merge function with keys 'Date' and 'Segment'?

    – Heikki
    Nov 12 '18 at 20:59













  • Date does not exist in DT1. Hence, can't be used to merge

    – Ameya
    Nov 12 '18 at 21:03











  • Sorry, I should have written merge by 'Segment' (only).

    – Heikki
    Nov 12 '18 at 21:40



















  • What about using merge function with keys 'Date' and 'Segment'?

    – Heikki
    Nov 12 '18 at 20:59













  • Date does not exist in DT1. Hence, can't be used to merge

    – Ameya
    Nov 12 '18 at 21:03











  • Sorry, I should have written merge by 'Segment' (only).

    – Heikki
    Nov 12 '18 at 21:40

















What about using merge function with keys 'Date' and 'Segment'?

– Heikki
Nov 12 '18 at 20:59







What about using merge function with keys 'Date' and 'Segment'?

– Heikki
Nov 12 '18 at 20:59















Date does not exist in DT1. Hence, can't be used to merge

– Ameya
Nov 12 '18 at 21:03





Date does not exist in DT1. Hence, can't be used to merge

– Ameya
Nov 12 '18 at 21:03













Sorry, I should have written merge by 'Segment' (only).

– Heikki
Nov 12 '18 at 21:40





Sorry, I should have written merge by 'Segment' (only).

– Heikki
Nov 12 '18 at 21:40












1 Answer
1






active

oldest

votes


















1














You can achieve the same result with a cartesian merge:



DT4 <- merge(DT2,DT1,by='Segment',allow.cartesian = TRUE)


Here is the proof:



> all(DT3[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')] == 

      DT4[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')])



[1] TRUE





share|improve this answer



















  • 1





    Thanks, allow.cartesian does it.

    – Ameya
    Nov 12 '18 at 22:10











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














You can achieve the same result with a cartesian merge:



DT4 <- merge(DT2,DT1,by='Segment',allow.cartesian = TRUE)


Here is the proof:



> all(DT3[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')] == 

      DT4[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')])



[1] TRUE





share|improve this answer



















  • 1





    Thanks, allow.cartesian does it.

    – Ameya
    Nov 12 '18 at 22:10
















1














You can achieve the same result with a cartesian merge:



DT4 <- merge(DT2,DT1,by='Segment',allow.cartesian = TRUE)


Here is the proof:



> all(DT3[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')] == 

      DT4[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')])



[1] TRUE





share|improve this answer



















  • 1





    Thanks, allow.cartesian does it.

    – Ameya
    Nov 12 '18 at 22:10














1












1








1







You can achieve the same result with a cartesian merge:



DT4 <- merge(DT2,DT1,by='Segment',allow.cartesian = TRUE)


Here is the proof:



> all(DT3[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')] == 

      DT4[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')])



[1] TRUE





share|improve this answer













You can achieve the same result with a cartesian merge:



DT4 <- merge(DT2,DT1,by='Segment',allow.cartesian = TRUE)


Here is the proof:



> all(DT3[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')] == 

      DT4[order(Segment,Date,Total,Activity,Value),

          c('Segment','Date','Total','Activity','Value')])



[1] TRUE






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 12 '18 at 21:39









HeikkiHeikki

1,2471017




1,2471017








  • 1





    Thanks, allow.cartesian does it.

    – Ameya
    Nov 12 '18 at 22:10














  • 1





    Thanks, allow.cartesian does it.

    – Ameya
    Nov 12 '18 at 22:10








1




1





Thanks, allow.cartesian does it.

– Ameya
Nov 12 '18 at 22:10





Thanks, allow.cartesian does it.

– Ameya
Nov 12 '18 at 22:10


















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