R 按一列分组，为每对列选择一列中最小的行

难以表达的问题。下面是我想做的一个例子。我从以下几个例子开始：

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

---

我已经尝试使用.SD和lapply来实现这一点，但我不能完全理解它。谢谢大家!

通过一些熔炼和铸造，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

库（data.table）
种子集（0）
dt这里有一个一体化的方法，尽管为了可读性，您可能希望将其分为单独的步骤：

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

基本上，它是由两种模式融合而成的，首先，然后按最小p值进行子集，然后再广播回宽格式。
了解
数据最重要（也是最强大）的一点是，只要
j
返回一个列表，列表中的每个元素都将成为结果中的一列

有了这些知识和一些基本的乐趣，我们可以通过以下方式直接得到这个结果：

# 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

有关更多信息，请参阅

# 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