难以回答的问题。这是我想做的一个例子。我开始的一个例子:
set.seed(0)
dt <- data.table(dr1.d=rnorm(5), dr1.p=abs(rnorm(5, sd=0.08)),
dr2.d=rnorm(5), dr2.p=abs(rnorm(5, sd=0.08)),
dr3.d=rnorm(5), dr3.p=abs(rnorm(5, sd=0.08)),
dr4.d=rnorm(5), dr4.p=abs(rnorm(5, sd=0.08)),
sym = paste("sym", c(1,1,1,2,2)))
dt
dr1.d dr1.p dr2.d dr2.p dr3.d dr3.p dr4.d dr4.p sym
1: 1.2629543 0.1231960034 0.7635935 0.03292087 -0.22426789 0.040288638 -0.2357066 0.09215294 sym 1
2: -0.3262334 0.0742853628 -0.7990092 0.02017788 0.37739565 0.086861549 -0.5428883 0.07937283 sym 1
3: 1.3297993 0.0235776357 -1.1476570 0.07135369 0.13333636 0.055276307 -0.4333103 0.03436105 sym 1
4: 1.2724293 0.0004613738 -0.2894616 0.03485466 0.80418951 0.102767948 -0.6494716 0.09906433 sym 2
5: 0.4146414 0.1923722711 -0.2992151 0.09900307 -0.05710677 0.003738094 0.7267507 0.02234770 sym 2
对于共享药物的所有列对(例如'dr1'),我想按'sym'对行进行分组,然后在每个组中选择具有最小p值(以'.p'结尾)的行。上述data.table的最终结果将是:
dr1.d dr1.p dr2.d dr2.p dr3.d dr3.p dr4.d dr4.p sym
1: 1.3297993 0.0235776357 -0.7990092 0.02017788 -0.22426789 0.040288638 -0.4333103 0.03436105 sym 1
2: 1.2724293 0.0004613738 -0.2894616 0.03485466 -0.05710677 0.003738094 0.7267507 0.02234770 sym 2
我已尝试使用.SD和lapply来实现这一目标,但我无法绕过它。谢谢!
答案 0 :(得分:13)
理解data.table
最重要(也是最强大的)是,只要j
返回列表,列表的每个元素都将成为一列在结果中。
有了这些知识和一些基本的R乐趣,我们可以直接得到这个结果:
# I'm on v1.9.7, see https://github.com/Rdatatable/data.table/wiki/Installation
cols1 = grep("d$", names(dt), value=TRUE)
cols2 = grep("p$", names(dt), value=TRUE)
dt[, Map(`[`, mget(c(cols1,cols2)), lapply(mget(cols2), which.min)), by=sym]
# sym dr1.d dr2.d dr3.d dr4.d dr1.p dr2.p
# 1: sym 1 1.329799 -0.7990092 -0.22426789 -0.4333103 0.0235776357 0.02017788
# 2: sym 2 1.272429 -0.2894616 -0.05710677 0.7267507 0.0004613738 0.03485466
# dr3.p dr4.p
# 1: 0.040288638 0.03436105
# 2: 0.003738094 0.02234770
有关详情,请参阅vignettes。
答案 1 :(得分:3)
这是一种一体化方法,但您可能希望将其拆分为单独的步骤以便于阅读:
dcast(melt(dt, measure = patterns("\\.p$", "\\.d$"), id.vars = "sym",
value.name = c("p", "d"))[, .SD[which.min(p)], by = list(sym, variable)],
sym ~ variable, value.var = c("p", "d"))
# sym p_1 p_2 p_3 p_4 d_1 d_2 d_3 d_4
#1: sym 1 0.0235776357 0.02017788 0.040288638 0.03436105 1.329799 -0.7990092 -0.22426789 -0.4333103
#2: sym 2 0.0004613738 0.03485466 0.003738094 0.02234770 1.272429 -0.2894616 -0.05710677 0.7267507
它首先基本上由两种模式融化,然后按最小p值进行子集化,然后再转换为宽格式。
答案 2 :(得分:2)
通过一些熔化和铸造,这是相当简单的
library(data.table)
set.seed(0)
dt <- data.table(dr1.d=rnorm(5), dr1.p=abs(rnorm(5, sd=0.08)),
dr2.d=rnorm(5), dr2.p=abs(rnorm(5, sd=0.08)),
dr3.d=rnorm(5), dr3.p=abs(rnorm(5, sd=0.08)),
dr4.d=rnorm(5), dr4.p=abs(rnorm(5, sd=0.08)),
sym = paste("sym", c(1,1,1,2,2)))
dt[, rowid := .I] #add a row identifier
dt <- melt(dt, id.vars = c("sym", "rowid"), variable.factor = F)
dt[, c("col","val") := tstrsplit(variable, "." , fixed = T)] #split the column so we can group
dt[, variable := NULL] #small cleanup
dt <- dcast(dt, sym + rowid + col ~ val)
dt <- dt[, .SD[which.min(p)], by = .(sym,col)] #select min row
dt[, rowid := NULL] #cleanup
dt <- dcast(melt(dt, id.vars = c("sym","col")), sym ~ col + variable)
dt
sym dr1_d dr1_p dr2_d dr2_p dr3_d dr3_p dr4_d dr4_p
1: sym 1 1.329799 0.0235776357 -0.7990092 0.02017788 -0.22426789 0.040288638 -0.4333103 0.03436105
2: sym 2 1.272429 0.0004613738 -0.2894616 0.03485466 -0.05710677 0.003738094 0.7267507 0.02234770