根据组，按两列计算唯一行数

I have a data.table in r

我在r中有一个data.table

    col1 col2 col3   col4
 1:  5.1  3.5  1.4 setosa
 2:  5.1  3.5  1.4 setosa
 3:  4.7  3.2  1.3 setosa
 4:  4.6  3.1  1.5 setosa
 5:  5.0  3.6  1.4 setosa
 6:  5.1  3.5  3.4    eer
 7:  5.1  3.5  3.4    eer
 8:  5.1  3.2  1.3    eer
 9:  5.1  3.5  1.5    eer
10:  5.1  3.5  1.4    eer


DT <- structure(list(col1 = c(5.1, 5.1, 4.7, 4.6, 5, 5.1, 5.1, 5.1, 
5.1, 5.1), col2 = c(3.5, 3.5, 3.2, 3.1, 3.6, 3.5, 3.5, 3.2, 3.5, 
3.5), col3 = c(1.4, 1.4, 1.3, 1.5, 1.4, 3.4, 3.4, 1.3, 1.5, 1.4
), col4 = structure(c(1L, 1L, 1L, 1L, 1L, 4L, 4L, 4L, 4L, 4L), .Label = c("setosa", 
"versicolor", "virginica", "eer"), class = "factor")), .Names = c("col1", 
"col2", "col3", "col4"), row.names = c(NA, -10L), class = c("data.table", 
"data.frame"))

I want to count unique (distinct) combinations of col1 and col2 for each value of col4.

我想为col4的每个值计算col1和col2的唯一(不同)组合。

Expected output is

预期的产出是

   col1 col2 col3   col4 count
 1:  5.1  3.5  1.4 setosa     4
 2:  5.1  3.5  1.4 setosa     4
 3:  4.7  3.2  1.3 setosa     4
 4:  4.6  3.1  1.5 setosa     4
 5:  5.0  3.6  1.4 setosa     4
 6:  5.1  3.5  3.4    eer     2
 7:  5.1  3.5  3.4    eer     2
 8:  5.1  3.2  1.3    eer     2
 9:  5.1  3.5  1.5    eer     2
10:  5.1  3.5  1.4    eer     2

How can I do this in 1 data.table syntax only?

我怎么能只用1 data.table语法呢?

1 个解决方案

#1

I had to go through a few attempts first, and ended up with this. Any good?

我不得不先经过几次尝试,然后结束了这一切。好不好?

DT[, count:=nrow(unique(.SD)), by=col4, .SDcols=c("col1","col2")]
DT
    col1 col2 col3   col4 count
 1:  5.1  3.5  1.4 setosa     4
 2:  5.1  3.5  1.4 setosa     4
 3:  4.7  3.2  1.3 setosa     4
 4:  4.6  3.1  1.5 setosa     4
 5:  5.0  3.6  1.4 setosa     4
 6:  5.1  3.5  3.4    eer     2
 7:  5.1  3.5  3.4    eer     2
 8:  5.1  3.2  1.3    eer     2
 9:  5.1  3.5  1.5    eer     2
10:  5.1  3.5  1.4    eer     2
>

and the same but faster thanks to Procrastinatus comment below :

同样但更快,感谢Procrastinatus评论如下:

DT[, count:=uniqueN(.SD), by=col4, .SDcols=c("col1","col2")]

#1