Question

我有一个数据表，其中许多变量已分为正面和负面组件。我想组合这些列，以便存在变量的有符号值。（这些变量在名称中总是有positive和negative，而其他变量都没有。但是，positive和negative子字符串可能出现在变量的任何位置 - - ie 只有grepl("(positive)|(negative)", names(dt))才能正确识别它们。）

例如，

library(data.table)

set.seed(1)

(DT <- data.table(x = 1:5, 
                  a_positive = sample(1:5), 
                  a_negative = sample(1:5), 
                  b_positive = sample(1:5), 
                  b_negative = sample(1:5), 
                  c_normal = sample(1:5)))

   x a_positive a_negative b_positive b_negative c_normal
1: 1          2          5          2          3        5
2: 2          5          4          1          5        1
3: 3          4          2          3          4        2
4: 4          3          3          4          1        4
5: 5          1          1          5          2        3

预期结果：

   x c_normal  a  b
1: 1        5 -3 -1
2: 2        1  1 -4
3: 3        2  2 -1
4: 4        4  0  3
5: 5        3  0  3

我的做法依赖于for循环和dplyr：

library(dplyr)
library(lazyeval)
library(magrittr) 

unite_positive_negative <- function(dt){
  signed_names <- 
    names(dt)[
      duplicated(gsub("(positive)|(negative)", "", names(dt))) | 
        duplicated(gsub("(positive)|(negative)", "", names(dt)), fromLast = TRUE)]

  unsigned_names <- 
    gsub("_*((positive)|(negative))_*", "", signed_names)

  the_names <- 
    data.table(signed_names = signed_names, 
               unsigned_names = unsigned_names) 

  for (unsigned_name in unsigned_names){
    poz <- the_names[unsigned_names == unsigned_name & grepl("positive", signed_names, fixed = TRUE)][["signed_names"]]
    neg <- the_names[unsigned_names == unsigned_name & grepl("negative", signed_names, fixed = TRUE)][["signed_names"]]

    dt %<>%
      mutate_(.dots = setNames(list(interp(~p - n, p = as.name(poz), n = as.name(neg))), unsigned_name)) 
  }

  # Unimportant
  unselect_ <- function(.data, .dots){
    all_names <- names(.data)
    keeps <- names(.data)[!names(.data) %in% .dots]
    dplyr::select_(.data, .dots = keeps)
  }

  dt %>%
    unselect_(.dots = signed_names)
}

是否有纯粹的data.table方式？（或者更直接的方式）？

Answer 1

我们可以尝试使用melt/dcast。重塑“广泛”的数据集。长期＆＃39;格式为melt，将id.var指定为＆＃39; x＆＃39;和＆＃39; c_normal＆＃39;列（如果有很多＆＃39;普通＆＃39;列，我们也可以使用grep来实现这一目标。使用tstrsplit将＆＃39;变量＆＃39;列拆分为两列。由＆＃39; x＆＃39;＆＃39; c_normal＆＃39;和＆＃39; var1＆＃39;（来自split）分组，我们将＆＃34;否定＆＃34;和＆＃34;积极＆＃34;价值＆＃39;，将它们与-1/1相乘并将它们加在一起。然后，dcast从＆＃39; long＆＃39;到＆＃39; 39;广泛的格式。

library(data.table)
dcast(melt(DT, id.var = c("x", "c_normal"))[, 
       c("var1", "var2") := tstrsplit(variable, "_")
        ][, -1*value[var2=="negative"] + value[var2=="positive"] ,
        by = .(x, c_normal, var1)],
              x + c_normal~var1, value.var="V1")
#   x c_normal  a  b
#1: 1        5 -3 -1
#2: 2        1  1 -4
#3: 3        2  2 -1
#4: 4        4  0  3
#5: 5        3  0  3

没有melt/dcast的另一个选项是将数据集子集为＆＃34; positive＆＃34;和＆＃34;否定＆＃34;列（假设它们是有序的），乘以1/-1，进行加法（+）并将这些输出分配给数据集的子集，而不使用＆＃34;正/负＆＃34;列。

DT1 <- DT[, c("x", grep("normal", names(DT), value=TRUE)), with = FALSE]
DT2 <- DT[, grep("positive", names(DT)), with = FALSE] +
          -1 * DT[, grep("negative", names(DT)), with = FALSE]
DT1[, c("a", "b") := DT2]
DT1
#    x c_normal  a  b
# 1: 1        5 -3 -1
# 2: 2        1  1 -4
# 3: 3        2  2 -1
# 4: 4        4  0  3
# 5: 5        3  0  3

按名称

1 个答案: