使用 data.table 在 R 中进行行查找

Question

我有一个按日期排列的因素和相应值的表格，但我还想按排名而不是因素来显示数据。在第一步中，我确定每个排名的相应因子(输出中的 .fact 列)，在第二步中，我使用该因子确定每个排名的相应值(输出中的 .val2 列)。实际数据集包含更多日期和因素(名称和计数各不相同)。每一行对应于特定日期每个因素的模型预测排名，数值数据为实现值。

下面的代码可以工作，但是是否有更有效(更快)的方法来完成下面的按行查找操作？我读过的许多 data.table 建议都不鼓励使用 .SD[, with=FALSE]，但我还没有找到其他解决方案。

library(data.table)
dt = data.table(Date = c("1/31/2013", "2/28/2013", "3/31/2013", 
                         "4/30/2013", "5/31/2013"), 
                A.rnk = c(5L, 2L, 2L, 3L, 3L), 
                B.rnk = c(4L, 3L, 1L, 2L, 5L), 
                C.rnk = c(3L, 1L, 4L, 1L, 1L), 
                D.rnk = c(2L, 4L, 3L, 5L, 2L),
                E.rnk = c(1L, 5L, 5L, 4L, 4L),
                A.val = rnorm(5), B.val = rnorm(5), 
                C.val = rnorm(5), D.val = rnorm(5),
                E.val = rnorm(5))
nms = c("A", "B", "C", "D", "E")
rnks = as.character(1:5)

# determine the factor (A,B,C) for each rank, by date
dt = dt[, c(rnks):={
  cols = .SD[, paste0(nms, ".rnk"), with=FALSE]
  cols = names(cols)[order(cols)]
  as.list(stringr::str_extract(cols, stringr::perl(".{1}(?=.rnk)")))
}, by=Date]

# determine the factor value (val) for each rank, by date
dt[, paste0(rnks, ".val2"):=
     .SD[, paste0(.SD[, rnks, with=FALSE], ".val"), with=FALSE], by=Date]

Answer 1

这是一个完整的重写。根据您的要求，最好使用长格式的数据，如下所示。它使用reshape2包，其中具有melt和dcast函数分别将数据转换为长格式和宽格式。

Note that faster versions of melt and dcast are implemented (in C) in the current development version (1.8.11) of data.table. So, after the next release of data.table, you can use the same code, but you don't have to convert it back to a data.table (done using as.dat.table shown below) after the melt and dcast step + it'll be a lot faster.

现在讨论解决方案:

# loading packages
require(data.table)
require(reshape2)

# long format on just the .rnk columns
dt.m <- as.data.table(melt(dt, id="Date", measure=2:6))
setnames(dt.m, c("Date", "var1", "val1"))
dt.m[, c("var2", "val2") := as.data.table(melt(dt, id="Date", 
               measure=7:11))[, list(variable, value)]]
# sort by date column by reference
setkey(dt.m, Date)

# here you can alternatively use `order`, but `fastorder` is well, faster
oo <- data.table:::fastorder(as.list(dt.m)[c("Date","val1")])
dt.m[, val3 := rep(nms, length.out=length(oo))[oo]]
dt.m[, val4 := val2[val1], by=Date]

ans1 <- as.data.table(dcast(dt.m, Date ~ var1, value.var="val3"))[, Date := NULL]
ans2 <- as.data.table(dcast(dt.m, Date ~ var2, value.var="val4"))[, Date := NULL]
setnames(ans1, rnks)
setnames(ans2, paste(rnks, ".val2", sep=""))
cbind(dt, ans1, ans2)