这可能是一个非常简单的问题,我有一个带密钥和超过1000行的data.table,其中两行可以设置为密钥。我想计算这个数据集的组数。
例如,简单数据是(ID和Act是关键)
ID ValueDate Act Volume
1 2015-01-01 EUR 21
1 2015-02-01 EUR 22
1 2015-01-01 MAD 12
1 2015-02-01 MAD 11
2 2015-01-01 EUR 5
2 2015-02-01 EUR 7
3 2015-01-01 EUR 4
3 2015-02-01 EUR 2
3 2015-03-01 EUR 6
以下是生成测试数据的代码:
dd <- data.table(ID = c(1,1,1,1,2,2,3,3,3),
ValueDate = c("2015-01-01", "2015-02-01", "2015-01- 01","2015-02-01", "2015-01-01","2015-02-01","2015-01-01","2015-02-01","2015-03-01"),
Act = c("EUR","EUR","MAD","MAD","EUR","EUR","EUR","EUR","EUR"),
Volume=c(21,22,12,11,5,7,4,2,6))
在这种情况下,我们可以看到总共有 4 子集。
我尝试将此表的密钥设置为第一个
setkey(dd, ID, Act)
然后我认为 count 的功能可能会计算群体的数量。 使用 count 的功能是对的,还是可以有一个简单的方法?
非常感谢!
答案 0 :(得分:3)
nrow(dd[, .(cnt= sum(.N)), by= c("ID", "Act")])
# or using base R
{t <- table(interaction(dd$ID, dd$Act)); length(t[t>0])}
# or for the counts:
dd[, .(cnt= sum(.N)), by= c("ID", "Act")]
ID Act cnt
1: 1 EUR 2
2: 1 MAD 2
3: 2 EUR 2
4: 3 EUR 3
答案 1 :(得分:3)
最快的方式应该是uniqueN
。
library(data.table)
dd <- data.table(ID = c(1,1,1,1,2,2,3,3,3),
ValueDate = c("2015-01-01", "2015-02-01", "2015-01-01","2015-02-01", "2015-01-01","2015-02-01","2015-01-01","2015-02-01","2015-03-01"),
Act = c("EUR","EUR","MAD","MAD","EUR","EUR","EUR","EUR","EUR"),
Volume=c(21,22,12,11,5,7,4,2,6))
uniqueN(dd, by = c("ID", "Act"))
#[1] 4