我正试图为具有多个嵌套列的data.table找到等效的data.table:
MT <- as.data.table(mtcars)
MT_NEST_MULT <- MT[, .(data1 = .(.SD[, .(mpg, hp)]), data2 = .(.SD[, !c("mpg", "hp")])), by = .(cyl, gear)]cyl gear data1 data2
8 3 <S3: data.table> <S3: data.table>
8 5 <S3: data.table> <S3: data.table>
6 4 <S3: data.table> <S3: data.table>
6 3 <S3: data.table> <S3: data.table>
6 5 <S3: data.table> <S3: data.table>
4 4 <S3: data.table> <S3: data.table>
4 3 <S3: data.table> <S3: data.table>
4 5 <S3: data.table> <S3: data.table> 不嵌套单个列很容易:MT_NEST_MULT[, rbindlist(data1), by = .(cyl, gear)]
但我不知道如何两者兼得,也就是做相当于tidyr::unnest(..., c(data1, data2))的事。
发布于 2022-05-25 18:15:24
这里有一个选项--我们指定列来应用rbindlist和.SDcols,在.SD (Data.table的子集)上循环,应用rbindlist,用c平平输出
library(data.table)
MT_NEST_MULT[, do.call(c, unname(lapply(.SD, rbindlist))),
.SDcols = patterns('data'), by = .(cyl, gear)]-output
cyl gear mpg hp disp drat wt qsec vs am carb
<num> <num> <num> <num> <num> <num> <num> <num> <num> <num> <num>
1: 6 4 21.0 110 160.0 3.90 2.620 16.46 0 1 4
2: 6 4 21.0 110 160.0 3.90 2.875 17.02 0 1 4
3: 6 4 19.2 123 167.6 3.92 3.440 18.30 1 0 4
4: 6 4 17.8 123 167.6 3.92 3.440 18.90 1 0 4
5: 4 4 22.8 93 108.0 3.85 2.320 18.61 1 1 1
6: 4 4 24.4 62 146.7 3.69 3.190 20.00 1 0 2
7: 4 4 22.8 95 140.8 3.92 3.150 22.90 1 0 2
8: 4 4 32.4 66 78.7 4.08 2.200 19.47 1 1 1
9: 4 4 30.4 52 75.7 4.93 1.615 18.52 1 1 2
10: 4 4 33.9 65 71.1 4.22 1.835 19.90 1 1 1
...https://stackoverflow.com/questions/72382310
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