r中data.table中的组累积地识别新值

Question

How to create a new column that identifies new value appearance in Letter column cumulatively by groups of unique combs of Year + Month ?

Data sample.

require(data.table)
dt <- data.table(Letter = c(LETTERS[c(5, 1:2, 1:2, 1:4, 3:6)]),
                 Year = 2018,
                 Month = c(rep(5,5), rep(6,4), rep(7,4)))

Print.

    Letter Year Month
 1:      E 2018     5
 2:      A 2018     5
 3:      B 2018     5
 4:      A 2018     5
 5:      B 2018     5
 6:      A 2018     6
 7:      B 2018     6
 8:      C 2018     6
 9:      D 2018     6
10:      C 2018     7
11:      D 2018     7
12:      E 2018     7
13:      F 2018     7

Result I'm trying to get:

    Letter Year Month   New
 1:      E 2018     5  TRUE
 2:      A 2018     5  TRUE
 3:      B 2018     5  TRUE
 4:      A 2018     5  TRUE
 5:      B 2018     5  TRUE
 6:      A 2018     6 FALSE
 7:      B 2018     6 FALSE
 8:      C 2018     6  TRUE
 9:      D 2018     6  TRUE
10:      C 2018     7 FALSE
11:      D 2018     7 FALSE
12:      E 2018     7 FALSE
13:      F 2018     7  TRUE

Detailed Question:

1. Group1 ("E", "A", "B", "A", "B") all TRUE by default as nothing to compare with.
2. Which of the letters in group2 ("A", "B", "C", "D") is not duplicated in group1.
3. Then, which of letters in group3 ("C", "D", "E", "F") in not duplicated in both groups 1&2 ("E", "A", "B", "A", "B", "A", "B", "C", "D").

Answer 1

Initialize to FALSE; then join to first Year-Month with each Letter and update to TRUE:

dt[, v := FALSE]
dt[unique(dt, by="Letter"), on=.(Letter, Year, Month), v := TRUE][]

    Letter Year Month     v
 1:      E 2018     5  TRUE
 2:      A 2018     5  TRUE
 3:      B 2018     5  TRUE
 4:      A 2018     5  TRUE
 5:      B 2018     5  TRUE
 6:      A 2018     6 FALSE
 7:      B 2018     6 FALSE
 8:      C 2018     6  TRUE
 9:      D 2018     6  TRUE
10:      C 2018     7 FALSE
11:      D 2018     7 FALSE
12:      E 2018     7 FALSE
13:      F 2018     7  TRUE

Answer 2

Simply:

 # dt[,new := ifelse(Letter %in% dt$Letter[dt$Month<Month],F,T), by="Month"][]

 #   Letter Year Month   new
 #1:      E 2018     5  TRUE
 #2:      A 2018     5  TRUE
 #3:      B 2018     5  TRUE
 #4:      A 2018     5  TRUE
 #5:      B 2018     5  TRUE
 #6:      A 2018     6 FALSE
 #7:      B 2018     6 FALSE
 #8:      C 2018     6  TRUE
 #9:      D 2018     6  TRUE
#10:      C 2018     7 FALSE
#11:      D 2018     7 FALSE
#12:      E 2018     7 FALSE
#13:      F 2018     7  TRUE

---

With very valid comments of David A., a much faster and less verbose version: ( recommended )

dt[, new := !(Letter %in% dt$Letter[dt$Month<Month]), by=Month][]

Answer 3

Another possible approach:

dupes <- c()
dt[, New := {
    x <- !Letter %chin% dupes
    dupes <- c(dupes, unique(Letter[x]))
    x
}, by=.(Year, Month)]

Some timings for reference below:

if Letter is an integer:

library(microbenchmark)
microbenchmark(mtd0=dt0[, New := !(Letter %in% dt0$Letter[dt0$Month<Month]), by=Month],
    mtd1={
        dt1[, v := FALSE]
        dt1[unique(dt1, by="Letter"), on=.(Letter, Year, Month), v := TRUE]
    },
    mtd2={
        dupes <- c()
        dt2[, New := {
            x <- !Letter %in% dupes
            dupes <- c(dupes, unique(Letter[x]))
            x
        }, by=.(Year, Month)]        
    },
    times=3L)

integer timing output:

Unit: milliseconds
 expr       min       lq      mean    median        uq      max neval
 mtd0 1293.3100 1318.775 1331.7129 1344.2398 1350.9143 1357.589     3
 mtd1  377.1534  391.178  402.4423  405.2026  415.0868  424.971     3
 mtd2 2015.2115 2020.926 2023.7209 2026.6400 2027.9756 2029.311     3

if Letter is a character:

microbenchmark(mtd0=dt0[, New := !(Letter %chin% dt0$Letter[dt0$Month<Month]), by=Month],
    mtd1={
        dt1[, v := FALSE]
        dt1[unique(dt1, by="Letter"), on=.(Letter, Year, Month), v := TRUE]
    },
    mtd2={
        dupes <- c()
        dt2[, New := {
            x <- !Letter %chin% dupes
            dupes <- c(dupes, unique(Letter[x]))
            x
        }, by=.(Year, Month)]        
    },
    times=3L)

timing output:

Unit: milliseconds
 expr       min        lq      mean    median        uq       max neval
 mtd0 1658.5806 1689.8941 1765.9329 1721.2076 1819.6090 1918.0105     3
 mtd1  849.2361  851.1807  852.8632  853.1253  854.6768  856.2283     3
 mtd2  420.1013  426.0941  433.9202  432.0869  440.8296  449.5723     3

check:

> identical(dt2$New, dt1$v)
[1] TRUE
> identical(dt0$New, dt1$v)
[1] FALSE

data:

set.seed(0L)
nr <- 1e7
dt <- unique(data.table(Letter=sample(nr/1e2, nr, replace=TRUE),
    Year=sample(2014:2018, nr, replace=TRUE),
    Month=sample(1:12, nr, replace=TRUE)))
setorder(dt, Year, Month)#[, Letter := as.character(Letter)]
dt0 <- copy(dt)
dt1 <- copy(dt)
dt2 <- copy(dt)

#for seed=0L, dt has about 4.8mio rows