Question

这是一个例子：

library("data.table")

set.seed(12)

table201901 <- data.frame(ID = as.character(paste0(sample(LETTERS,5000, replace=TRUE),sample(0:9,5000, replace=TRUE))),
                 EID = paste0(sample(LETTERS,200, replace=TRUE),sample(letters,5000, replace=TRUE),sample(1:9,5000, replace=TRUE))
                 )

set.seed(15)

table201812 <- data.frame(ID = paste0(sample(LETTERS,5000, replace=TRUE),sample(0:9,5000, replace=TRUE)),
                          EID = paste0(sample(LETTERS,5000, replace=TRUE),sample(letters,5000, replace=TRUE),sample(0:9,5000, replace=TRUE))
)

table201901 <- as.data.table(table201901)
table201812 <- as.data.table(table201812)

setkey(table201901, "ID")
setkey(table201812, "ID")


table201901[!table201812]

fsetdiff(table201901, table201812)


setkey(table201901, "ID","EID")
setkey(table201812, "ID","EID")


table201901[!table201812]

fsetdiff(table201901, table201812)

我的目标是要获得一个表，该表的行项目在table201901中，但不在table201812中。

我认为fsetdiff(table1,table2)和table1[!table2]会给我相同的输出，但是却没有。

Answer 1

以下示例说明了这两种方法。为简单起见，生成了两个小的虚拟data.tables，它们具有部分重叠的行：

library(data.table)

(dt1 <- data.table(
    col1 = rep(c(1, 3, 3), each = 2),
    col2 = rep(c("only dt1", "match col2", "match row"), each = 2)
))
#>    col1       col2
#> 1:    1   only dt1
#> 2:    1   only dt1
#> 3:    3 match col2
#> 4:    3 match col2
#> 5:    3  match row
#> 6:    3  match row

(dt2 <-  data.table(
    col1 = rep(c(3, 4, 5), each = 2),
    col2 = rep(c("match row", "match col2", "only dt2"), each = 2)    
))   
#>    col1       col2
#> 1:    3  match row
#> 2:    3  match row
#> 3:    4 match col2
#> 4:    4 match col2
#> 5:    5   only dt2
#> 6:    5   only dt2

## keep all rows in dt1 not in dt2
dt1[!dt2, on = names(dt1)]
#>    col1       col2
#> 1:    1   only dt1
#> 2:    1   only dt1
#> 3:    3 match col2
#> 4:    3 match col2

## equivalent output with fsetdiff
fsetdiff(dt1, dt2, all = TRUE)
#>    col1       col2
#> 1:    1   only dt1
#> 2:    1   only dt1
#> 3:    3 match col2
#> 4:    3 match col2

## keep unique rows in dt1 not in dt2
unique(dt1[!dt2, on = names(dt1)])
#>    col1       col2
#> 1:    1   only dt1
#> 2:    3 match col2

## equivalent output with fsetdiff
fsetdiff(dt1, dt2, all = FALSE)
#>    col1       col2
#> 1:    1   only dt1
#> 2:    3 match col2

## keep rows dt1 not matching in dt2 only by "col2" 
dt1[!dt2, on = "col2"]
#>    col1     col2
#> 1:    1 only dt1
#> 2:    1 only dt1

^{由reprex package（v0.3.0）于2019-06-28创建}

我在2条代码中获得了不同的结果，我认为它们会给出相同的结果

1 个答案: