【发布时间】:2017-02-15 04:58:16
【问题描述】:
我有一个带有两个分组变量的 data.table。我想计算关于组变量 1 的排名,同时仍然保留组的信息。
# require(data.table)
# require(dplyr)
set.seed(1)
DT <- data.table(group = c(rep(1,5), rep(2, 5)),
id = c(letters[1:5], letters[1:5]),
var1 = rnorm(10),
var2 = runif(10))
# > DT
# group id var1 var2
# 1: 1 a -0.6264538 0.93470523
# 2: 1 b 0.1836433 0.21214252
# 3: 1 c -0.8356286 0.65167377
# 4: 1 d 1.5952808 0.12555510
# 5: 1 e 0.3295078 0.26722067
# 6: 2 a -0.8204684 0.38611409
# 7: 2 b 0.4874291 0.01339033
# 8: 2 c 0.7383247 0.38238796
# 9: 2 d 0.5757814 0.86969085
# 10: 2 e -0.3053884 0.34034900
我可以使用
计算组内排名DT[, lapply(.SD, function(x) percent_rank(x)),
.SDcols = c("var1", "var2"), by = .(group)]
# group var1 var2
# 1: 1 0.25 1.00
# 2: 1 0.50 0.25
# 3: 1 0.00 0.75
# 4: 1 1.00 0.00
# 5: 1 0.75 0.50
# 6: 2 0.00 0.75
# 7: 2 0.50 0.00
# 8: 2 1.00 0.50
# 9: 2 0.75 1.00
# 10: 2 0.25 0.25
我还想在新表中保留id 列,如
# group id var1 var2
# 1: 1 A 0.25 1.00
# 2: 1 B 0.50 0.25
# 3: 1 C 0.00 0.75
# 4: 1 D 1.00 0.00
# 5: 1 E 0.75 0.50
# 6: 2 A 0.00 0.75
# 7: 2 B 0.50 0.00
# 8: 2 C 1.00 0.50
# 9: 2 D 0.75 1.00
# 10: 2 E 0.25 0.25
【问题讨论】:
-
您也可以将
id变量放在选择中,尽管这有点尴尬 -DT[, c(.(id=id), lapply(.SD, percent_rank)), .SDcols = c("var1", "var2"), by = .(group)] -
@thelatemail 谢谢!我不知道我能做到这一点!
-
或者只是将它们分配回
DT[, c("var1", "var2") := lapply(.SD, percent_rank), .SDcols = c("var1", "var2"), by = group]
标签: r data.table ranking