我有成交数据,我需要从它创建关联矩阵。我试图自连接,但似乎没有任何工作。下面是示例代码和所需的输出。我R中使用的数据表中寻找一个解决方案
> TP <- data.table(
Tr = c("T1","T1","T2","T2","T2", "T3", "T4"),
Pr = c("P1","P2","P3","P1","P4", "P2", "P9")
)
> TP
Tr Pr
1: T1 P1
2: T1 P2
3: T2 P3
4: T2 P1
5: T2 P4
6: T3 P2
7: T4 P9
所需的输出是:
T1 T2 T3 T4
1: 1 1 0 0
2: 1 0 1 0
3: 1 0 0 0
4: 0 1 0 0
5: 0 0 0 1
或者如果可能甚至更好地得到这样的事情。
Pr T1 T2 T3 T4
1: P1 1 1 0 0
2: P2 1 0 1 0
3: P3 1 0 0 0
4: P4 0 1 0 0
5: P9 0 0 0 1
这应该工作:
dcast(TP, Pr ~ Tr, fun.aggregate = function(x){(length(x) > 0) * 1})
Using 'Pr' as value column. Use 'value.var' to override
Pr T1 T2 T3 T4
1: P1 1 1 0 0
2: P2 1 0 1 0
3: P3 0 1 0 0
4: P4 0 1 0 0
5: P9 0 0 0 1
@大卫Arenburg的建议是干净多了,如果我们有没有重复的关联:
dcast(TP, Pr ~ Tr, length)
我会dcast()
做大卫Arenburg的建议,但这里是(快速)的替代只是为了好玩:
TP[, data.table(unclass(table(Pr, Tr)), keep.rownames = "Pr")]
Pr T1 T2 T3 T4
1: P1 1 1 0 0
2: P2 1 0 1 0
3: P3 0 1 0 0
4: P4 0 1 0 0
5: P9 0 0 0 1
基准测试:
这似乎与数百万行打交道时dcast()
更快:
TP1 <- data.table(
Tr = paste0("T", sample(1:10, size = 1e5, replace = TRUE)),
Pr = paste0("P", sample(1:1e4, size = 1e5, replace = TRUE))
)
TP_huge <- data.table(
Tr = paste0("T", sample(1:10, size = 1e7, replace = TRUE)),
Pr = paste0("P", sample(1:1e4, size = 1e7, replace = TRUE))
)
microbenchmark::microbenchmark(
table1 = TP1[, data.table(unclass(table(Pr, Tr)), keep.rownames = "Pr")],
dcast1 = dcast(TP1, Pr ~ Tr, length, value.var = "Pr"),
table_huge = TP_huge[, data.table(unclass(table(Pr, Tr)), keep.rownames = "Pr")],
dcast_huge = dcast(TP_huge, Pr ~ Tr, length, value.var = "Pr"),
times = 5
)
Unit: milliseconds
expr min lq mean median uq max neval cld
table1 92.71867 105.8366 127.4707 124.4188 150.0642 164.3155 5 a
dcast1 255.53793 271.5194 292.2005 301.4840 302.5010 329.9600 5 b
table_huge 1719.83678 1732.1086 1771.0142 1733.8847 1771.5087 1897.7325 5 d
dcast_huge 917.94755 927.1657 971.4084 986.1038 998.1780 1027.6468 5 c