r - 合并多个 data.tables 并重命名列以反射(reflect)来源

Question

作为我的问题的一个简化示例，假设我有四个 data.tables dt1, ..., dt4，都带有相同的结构:

head(dt1)
          date          x           y
 1: 2000-10-01  0.4527087 -0.11590788
 2: 2001-10-01  0.7200252 -0.55722270
 3: 2002-10-01 -1.3804472 -1.47030087
 4: 2003-10-01 -0.1380225  2.34157766
 5: 2004-10-01 -0.9288675 -1.32993998
 6: 2005-10-01 -0.9592633  0.76316150

也就是说，它们都有名为date、x 和y 的三列。我想要的输出是一个合并的 data.table(日期)，包含五列:date，然后每个单独表中的 x 列重命名反射(reflect)其原始 data.table:

head(desired_output)
          date      x_dt1       x_dt2      x_dt3      x_dt4
 1: 2000-10-01  0.4527087 -0.11590788  1.1581946 -1.5159040
 2: 2001-10-01  0.7200252 -0.55722270 -1.6247254 -0.3325556
 3: 2002-10-01 -1.3804472 -1.47030087 -0.9766309 -0.2368857
 4: 2003-10-01 -0.1380225  2.34157766  1.1831091 -0.4399184
 5: 2004-10-01 -0.9288675 -1.32993998  0.8716144 -0.4086229
 6: 2005-10-01 -0.9592633  0.76316150 -0.8860816 -0.4299365

我假设这可以通过使用 merge.data.table 的 suffixes 参数以某种方式完成。我尝试从 this answer 修改 mergeDTs，但没有成功。成功修改 mergeDTs 的解决方案(或仅使用可应用于多个 data.tables 的列表的函数)将是非常好的。

我知道 this very slick dplyr/purrr answer 但更喜欢 data.table 解决方案。

---

示例数据

library(data.table)
dt1 <- data.table(date = seq(from = as.Date("2000-10-01"), to = as.Date("2010-10-01"), by = "years"),
                  x = rnorm(11),
                  y = rnorm(11))

dt2 <- data.table(date = seq(from = as.Date("2000-10-01"), to = as.Date("2010-10-01"), by = "years"),
                  x = rnorm(11),
                  y = rnorm(11))

dt3 <- data.table(date = seq(from = as.Date("2000-10-01"), to = as.Date("2010-10-01"), by = "years"),
                  x = rnorm(11),
                  y = rnorm(11))

dt4 <- data.table(date = seq(from = as.Date("2000-10-01"), to = as.Date("2010-10-01"), by = "years"),
                  x = rnorm(11),
                  y = rnorm(11))

---

解决方案

下面我将 B. Christian Kamgang 的答案转化为函数形式(使其更容易适应我的实际问题)并删除了对新管道的依赖(因为我的组织尚未升级):

merge_select <- function(on, vars, ..., suffix = "_") {
  dts <- list(...)
  names(dts) <- sapply(as.list(substitute(list(...)))[-1L], deparse)
  
  nv <- length(vars)
  ndt <- length(dts)
  
  old_cols <- split(rep(vars, ndt),
                    ceiling(seq_along(rep(vars, ndt))/nv))
  
  new_cols <- split(paste0(vars, suffix, rep(names(dts), each = nv)),
                    ceiling(seq_along(paste0(vars, 
                                             suffix, 
                                             rep(names(dts), each = nv)))/nv))
  
  sep_cols <- lapply(dts, function(x) subset(x, select = c(on, vars)))
  
  Reduce(f = function(x,y) merge(x, y, by = on), 
         Map(f = setnames, sep_cols, old_cols, new_cols))
}

在我的情况下转化为:

merge_select("date", "x", dt1, dt2, dt3, dt4)
          date      x_dt1       x_dt2       x_dt3       x_dt4
 1: 2000-10-01 -0.6365707  0.11804268 -0.01084163 -0.88127011
 2: 2001-10-01 -0.2533127 -3.16924568  0.45746415  0.69742537
 3: 2002-10-01  2.3069266 -0.82670409 -0.54236745 -1.49613384
 4: 2003-10-01  0.7075547 -0.91809007 -0.67888707 -0.26106146
 5: 2004-10-01 -0.7165651 -0.45711888 -0.83903416  1.45113260
 6: 2005-10-01  0.5703561  0.24587897  0.13862020  0.33928202
 7: 2006-10-01 -0.6258097 -0.77652389 -0.49252474 -0.80460241
 8: 2007-10-01 -0.4600565  0.55612959  0.86749410 -1.30850411
 9: 2008-10-01 -0.8841649 -0.48113848 -1.55858406  0.83076846
10: 2009-10-01 -0.6262272 -0.73618265  0.13350581  0.06640803
11: 2010-10-01  0.1406454  0.08994779  1.28450204 -1.18329081

此解决方案也适用于多个变量，例如。

merge_select("date", c("x","y"), dt1, dt2, dt3, dt4)

Answer 1

这是一种可能的方法，它在一个简单的 for 循环中累积合并结果:

library(data.table)

dt <- dt1[, .(date, x)]

for(i in 2:4) {
  dt <- merge(dt, get(paste0("dt", i))[, .(date, x)], by = "date", suffixes = c("", paste0("_dt", i)))
}

setnames(dt, old = "x", new = "x_dt1")

head(dt)
#>          date      x_dt1      x_dt2        x_dt3      x_dt4
#> 1: 2000-10-01 -1.5035218  2.0463775 -0.120544283 -0.5662290
#> 2: 2001-10-01  0.5977386 -0.1968421 -0.840102174  1.2412272
#> 3: 2002-10-01 -0.9100557 -0.1687148 -1.738526471  1.3685767
#> 4: 2003-10-01  0.7027232  0.9009135 -0.247273205  1.3135718
#> 5: 2004-10-01  0.5269265  0.6176381 -0.007662592 -0.2928206
#> 6: 2005-10-01 -0.8350406 -0.7343245 -0.643701996  2.3068948

或者，使用 Reduce() 累积合并结果:

Reduce(
  f = function(dt, dti) merge(dt, get(dti)[, .(date, x)], by = "date", suffixes = c("", paste0("_", dti))),
  x = paste0("dt", 1:4),
  init = dt1[, .(date, x)]
)[, x := NULL][]

---

注意:为了摆脱 get() 调用，我们可以在合并之前收集列表中的所有 data.tables 或编写一个小的函数包装器，例如

merge_dts <- function(...) {
  dts <- list(...)
  dt <- dts[[1]][, .(date, x)]
  for(i in seq_along(dts)[-1]) {
    dt <- merge(dt, dts[[i]][, .(date, x)], by = "date", suffixes = c("", paste0("_dt", i)))
  }
  setnames(dt, old = "x", new = "x_dt1")
  return(dt)
}

merge_dts(dt1, dt2, dt3, dt4)