recycle vector values with data.table shift instead of padding with fill=NA











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I would like to use the shift function from data.table to lead/lag a new column, but I would like to recycle values from the lagged vector that was added to the data.table. From what I can see, fill must be a vector of length 1, and so the values that are lagged must be populated with a constant value (ie NA here).



Please see the MWE below.



dt1 is the resulting data.table using the shift function as is. The new b column has NA values where 4, 5, and 6 should be.



dt2 is the desired data.table result. If my thinking is correct, the output requires R recycling rules but with a lead/lag value specified where the vector should begin.



I could have added a new vector (see below in x_to_avoid) but that requires more manual work that I hope to avoid.



Thanks,



library(data.table)
library(magrittr)

# vector to lead/lag when updating datatable
x = c(1:6)

# leaves NA where 4, 5, 6 "should" have gone for my purposes
dt1 <- data.table(a = c(1:10)) %>%
.[, b := shift(x,
n = 3L,
fill = NA,
type = c("lag"))]
dt1

# desired output
dt2 <- dt1[, .(a)] %>%
.[, b := c(4,5,6,1,2,3,4,5,6,1)]

# could use another vector, but my actual use is more complicated and I prefer to avoid this (if possible)
x_to_avoid = c(4,5,6,1:6,1)









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  • 1




    Perhaps you could define your own function, something like shift2 <- function(x, n, type = "lag") if(type == "lag") c(tail(x, -(length(x) - n)), head(x, -n)) else c(tail(x, -n), head(x, -(length(x) - n))). Then you could do dt1 <- data.table(a = 1:10, b = shift2(x, n = 3L)). Though n here isn't vectorized like in data.table::shift
    – David Arenburg
    Nov 12 at 14:41















up vote
0
down vote

favorite












I would like to use the shift function from data.table to lead/lag a new column, but I would like to recycle values from the lagged vector that was added to the data.table. From what I can see, fill must be a vector of length 1, and so the values that are lagged must be populated with a constant value (ie NA here).



Please see the MWE below.



dt1 is the resulting data.table using the shift function as is. The new b column has NA values where 4, 5, and 6 should be.



dt2 is the desired data.table result. If my thinking is correct, the output requires R recycling rules but with a lead/lag value specified where the vector should begin.



I could have added a new vector (see below in x_to_avoid) but that requires more manual work that I hope to avoid.



Thanks,



library(data.table)
library(magrittr)

# vector to lead/lag when updating datatable
x = c(1:6)

# leaves NA where 4, 5, 6 "should" have gone for my purposes
dt1 <- data.table(a = c(1:10)) %>%
.[, b := shift(x,
n = 3L,
fill = NA,
type = c("lag"))]
dt1

# desired output
dt2 <- dt1[, .(a)] %>%
.[, b := c(4,5,6,1,2,3,4,5,6,1)]

# could use another vector, but my actual use is more complicated and I prefer to avoid this (if possible)
x_to_avoid = c(4,5,6,1:6,1)









share|improve this question


















  • 1




    Perhaps you could define your own function, something like shift2 <- function(x, n, type = "lag") if(type == "lag") c(tail(x, -(length(x) - n)), head(x, -n)) else c(tail(x, -n), head(x, -(length(x) - n))). Then you could do dt1 <- data.table(a = 1:10, b = shift2(x, n = 3L)). Though n here isn't vectorized like in data.table::shift
    – David Arenburg
    Nov 12 at 14:41













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I would like to use the shift function from data.table to lead/lag a new column, but I would like to recycle values from the lagged vector that was added to the data.table. From what I can see, fill must be a vector of length 1, and so the values that are lagged must be populated with a constant value (ie NA here).



Please see the MWE below.



dt1 is the resulting data.table using the shift function as is. The new b column has NA values where 4, 5, and 6 should be.



dt2 is the desired data.table result. If my thinking is correct, the output requires R recycling rules but with a lead/lag value specified where the vector should begin.



I could have added a new vector (see below in x_to_avoid) but that requires more manual work that I hope to avoid.



Thanks,



library(data.table)
library(magrittr)

# vector to lead/lag when updating datatable
x = c(1:6)

# leaves NA where 4, 5, 6 "should" have gone for my purposes
dt1 <- data.table(a = c(1:10)) %>%
.[, b := shift(x,
n = 3L,
fill = NA,
type = c("lag"))]
dt1

# desired output
dt2 <- dt1[, .(a)] %>%
.[, b := c(4,5,6,1,2,3,4,5,6,1)]

# could use another vector, but my actual use is more complicated and I prefer to avoid this (if possible)
x_to_avoid = c(4,5,6,1:6,1)









share|improve this question













I would like to use the shift function from data.table to lead/lag a new column, but I would like to recycle values from the lagged vector that was added to the data.table. From what I can see, fill must be a vector of length 1, and so the values that are lagged must be populated with a constant value (ie NA here).



Please see the MWE below.



dt1 is the resulting data.table using the shift function as is. The new b column has NA values where 4, 5, and 6 should be.



dt2 is the desired data.table result. If my thinking is correct, the output requires R recycling rules but with a lead/lag value specified where the vector should begin.



I could have added a new vector (see below in x_to_avoid) but that requires more manual work that I hope to avoid.



Thanks,



library(data.table)
library(magrittr)

# vector to lead/lag when updating datatable
x = c(1:6)

# leaves NA where 4, 5, 6 "should" have gone for my purposes
dt1 <- data.table(a = c(1:10)) %>%
.[, b := shift(x,
n = 3L,
fill = NA,
type = c("lag"))]
dt1

# desired output
dt2 <- dt1[, .(a)] %>%
.[, b := c(4,5,6,1,2,3,4,5,6,1)]

# could use another vector, but my actual use is more complicated and I prefer to avoid this (if possible)
x_to_avoid = c(4,5,6,1:6,1)






r data.table






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asked Nov 12 at 2:20









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  • 1




    Perhaps you could define your own function, something like shift2 <- function(x, n, type = "lag") if(type == "lag") c(tail(x, -(length(x) - n)), head(x, -n)) else c(tail(x, -n), head(x, -(length(x) - n))). Then you could do dt1 <- data.table(a = 1:10, b = shift2(x, n = 3L)). Though n here isn't vectorized like in data.table::shift
    – David Arenburg
    Nov 12 at 14:41














  • 1




    Perhaps you could define your own function, something like shift2 <- function(x, n, type = "lag") if(type == "lag") c(tail(x, -(length(x) - n)), head(x, -n)) else c(tail(x, -n), head(x, -(length(x) - n))). Then you could do dt1 <- data.table(a = 1:10, b = shift2(x, n = 3L)). Though n here isn't vectorized like in data.table::shift
    – David Arenburg
    Nov 12 at 14:41








1




1




Perhaps you could define your own function, something like shift2 <- function(x, n, type = "lag") if(type == "lag") c(tail(x, -(length(x) - n)), head(x, -n)) else c(tail(x, -n), head(x, -(length(x) - n))). Then you could do dt1 <- data.table(a = 1:10, b = shift2(x, n = 3L)). Though n here isn't vectorized like in data.table::shift
– David Arenburg
Nov 12 at 14:41




Perhaps you could define your own function, something like shift2 <- function(x, n, type = "lag") if(type == "lag") c(tail(x, -(length(x) - n)), head(x, -n)) else c(tail(x, -n), head(x, -(length(x) - n))). Then you could do dt1 <- data.table(a = 1:10, b = shift2(x, n = 3L)). Though n here isn't vectorized like in data.table::shift
– David Arenburg
Nov 12 at 14:41












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I think that binhf::shift does the job. First you will need to lengthen the vector using rep.len and then you can cycle it using binhf::shift. I have no idea about the performance though.






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    I think that binhf::shift does the job. First you will need to lengthen the vector using rep.len and then you can cycle it using binhf::shift. I have no idea about the performance though.






    share|improve this answer

























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      I think that binhf::shift does the job. First you will need to lengthen the vector using rep.len and then you can cycle it using binhf::shift. I have no idea about the performance though.






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        up vote
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        I think that binhf::shift does the job. First you will need to lengthen the vector using rep.len and then you can cycle it using binhf::shift. I have no idea about the performance though.






        share|improve this answer












        I think that binhf::shift does the job. First you will need to lengthen the vector using rep.len and then you can cycle it using binhf::shift. I have no idea about the performance though.







        share|improve this answer












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        share|improve this answer










        answered Nov 14 at 19:02









        Alp Arıbal

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