请参考R中data.table中的上一行，并附带条件

Question

i have a new problem with this data. Because my full data has the form like this

a=data.table(A=c(1:10),B=c(1,2,0,2,0,0,3,4,0,2),C=c(2,3,1,4,5,3,6,7,2,2),D=c(1,1,1,1,1,2,2,2,2,2)) 


#     A B C D  
# 1:  1 1 2 1  
# 2:  2 2 3 1  
# 3:  3 0 1 1  
# 4:  4 2 4 1  
# 5:  5 0 5 1  
# 6:  6 0 3 2  
# 7:  7 3 6 2  
# 8:  8 4 7 2  
# 9:  9 0 2 2  
#10: 10 2 2 2  

Now, I want to create a new column, which calculates the number of values of A multiple with B/C of the closet previous row, as long as B is not 0. For example, in line 2, I can calculate D=2*(1/2). However, in line 4, it has to be 4*(2/3), it can not be 4*(0/1). I use

a[, D:= {i1 <- (NA^!B)
list( A*shift(na.locf(i1*B))/shift(na.locf(i1*C)))},by=d]

as Akrun recommended yesterday. It does not work when i calculate it by group.the result is like this

    A B C d        D
# 1:  1 1 2 1       NA
# 2:  2 2 3 1 1.000000
# 3:  3 0 1 1 2.000000
# 4:  4 2 4 1 2.666667
# 5:  5 0 5 1 2.500000
# 6:  6 0 3 2       NA
# 7:  7 3 6 2 3.500000
# 8:  8 4 7 2 4.571429
# 9:  9 0 2 2 5.142857
# 10: 10 2 2 2       NA

Anyone knows what is the problem here? The error is longer object length is not a multiple of shorter object length.

Answer 1

We can replace the elements in 'B', 'C' that corresponds to '0' value in 'B' as NA. Use na.locf from zoo to replace those NA values with the previous non-NA elements, shift the elements (by default, it gives a lag of 1), divide the modified columns 'B' with 'C' and then multiply by 'A'. Assign ( := ) the output to a new column 'D'.

 library(zoo)
 a[B==0, c('B', 'C'):=list(NA, NA)]
 a[, c('B', 'C'):= na.locf(.SD), .SDcols=B:C]
 a[,  D:= {tmp <- shift(.SD[, 2:3, with=FALSE])
           A*(tmp[[1]]/tmp[[2]])}]

---

Or we can make it compact. We get a logical vector ( !B ) that checks for '0' elements in 'B', convert that to a vector of 1s and NA ( NA^ ), multiply with columns 'B' and 'C' so that the 1s are replaced by the corresponding elements in those columns whereas NA remains as such. Do the na.locf (as before), shift and then do the multiplication/division.

a[, D:= {i1 <- (NA^!B)
   list( A*shift(na.locf(i1*B))/shift(na.locf(i1*C)))}]

Or instead of calling shift/na.locf two times

a[,  D:= {i1 <- (NA^!B)
      tmp <- shift(na.locf(i1*.SD))
      a[['A']]*(tmp[[1]]/tmp[[2]])}, .SDcols=B:C]

Answer 2

This can be done with a rolling join:

a[, row := .I]
a[, B/C, by=row][V1 != 0][a, A*shift(V1), on="row", roll=TRUE]
# [1]       NA 1.000000 2.000000 2.666667 2.500000 3.000000 3.500000 4.000000
# [9] 5.142857 5.714286