我想将相同的聚合应用于多个数据表,而不重写聚合方案. 考虑 dt1 - data.table(id = c(1,2), a = rnorm(10), b = rnorm(10), c = rnorm(10))dt2 - data.table(id = c(1,2), a = rnorm(10), b = rnorm(10), c = rnorm(10))dt1_aggreg
考虑
dt1 <- data.table(id = c(1,2), a = rnorm(10), b = rnorm(10), c = rnorm(10)) dt2 <- data.table(id = c(1,2), a = rnorm(10), b = rnorm(10), c = rnorm(10)) dt1_aggregates <- dt1[, .(mean_a=mean(a), sd_a=sd(a), mean_b=mean(b), sd_b=sd(b)), by=id] dt2_aggregates <- dt2[, .(mean_a=mean(a), sd_a=sd(a), mean_b=mean(b), sd_b=sd(b)), by=id]
有没有办法为dt2重用dt1_aggregates聚合方案而不必将其写出两次？
您可以引用所需的表达式,然后在data.table中对其进行评估：my.call=quote(list(mean_a=mean(a), sd_a=sd(a), mean_b=mean(b), sd_b=sd(b))) dt1[, eval(my.call), by=id]
产生
id mean_a sd_a mean_b sd_b 1: 1 0.004165423 0.7504691 -0.05001424 1.4440434 2: 2 -0.430910188 0.9648096 0.26918995 0.8680997
和
dt2[, eval(my.call), by=id]
产生
id mean_a sd_a mean_b sd_b 1: 1 0.2974145 1.191863 -0.0588854 0.7896988 2: 2 -0.4642856 1.438937 0.3612607 1.0581702