在向量中查找先前相同的值并应用某些条件

Question

对于myvector1 的每个值，我想知道myvector1 中前一个相同值的mycategory 值，因为mystatus 是ON，否则我会看对应下一个相同的值，直到它为 ON。

说明如下：

1. 对于“myvector”的给定位置，给我他的值。
2. 在中查找中间前一个相同值的位置 “我的向量”
3. 检查关联状态。如果它是 ON 给我他的关联 “我的类别”。如果它是关闭的，重复到第 2 点。
4. 将获得的“mycategory”分配给新的向量“mysolution”。

给定数据集mydf，我要查找的是DesiredSolution（我手动填写的）。

mydf <- structure(list(myvector1 = structure(c(1L, 2L, 3L, 4L, 5L, 1L, 
2L, 4L, 5L, 2L, 3L, 4L, 5L, 2L, 3L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 
2L, 4L, 5L, 1L, 1L, 2L, 3L, 4L, 5L, 3L), .Label = c("0", "1", 
"2", "3", "4"), class = "factor"), mystatus = structure(c(2L, 
1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 
1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 2L), .Label = c("OFF", 
"ON"), class = "factor"), mycategory = structure(c(2L, 2L, 3L, 
1L, 1L, 1L, 1L, 3L, 3L, 1L, 2L, 2L, 3L, 1L, 1L, 1L, 1L, 1L, 2L, 
2L, 3L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L), .Label = c("bye", 
"hi", "stay"), class = "factor"), DesiredSolution = structure(c(3L, 
3L, 3L, 3L, 3L, 2L, 3L, 1L, 3L, 1L, 4L, 4L, 4L, 1L, 2L, 4L, 1L, 
1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 2L, 1L, 2L), .Label = c("bye", 
"hi", "NA", "stay"), class = "factor")), .Names = c("myvector1", 
"mystatus", "mycategory", "DesiredSolution"), row.names = c(NA, 
-32L), class = "data.frame")

Answer 1

使用 data.table...

library(data.table)
setDT(mydf)
mydf[, r := .I]
mydf[, v := mydf[mystatus == "ON"][mydf, on=.(r < r, myvector1), mult="last", x.mycategory]]

给了

    myvector1 mystatus mycategory DesiredSolution  r    v
 1:         0       ON         hi              NA  1   NA
 2:         1      OFF         hi              NA  2   NA
 3:         2       ON       stay              NA  3   NA
 4:         3       ON        bye              NA  4   NA
 5:         4      OFF        bye              NA  5   NA
 6:         0       ON        bye              hi  6   hi
 7:         1       ON        bye              NA  7   NA
 8:         3       ON       stay             bye  8  bye
 9:         4       ON       stay              NA  9   NA
10:         1      OFF        bye             bye 10  bye
11:         2       ON         hi            stay 11 stay
12:         3       ON         hi            stay 12 stay
13:         4       ON       stay            stay 13 stay
14:         1      OFF        bye             bye 14  bye
15:         2       ON        bye              hi 15   hi
16:         4       ON        bye            stay 16 stay
17:         0       ON        bye             bye 17  bye
18:         1      OFF        bye             bye 18  bye
19:         2       ON         hi             bye 19  bye
20:         3       ON         hi              hi 20   hi
21:         4      OFF       stay             bye 21  bye
22:         0      OFF        bye             bye 22  bye
23:         1       ON        bye             bye 23  bye
24:         3      OFF        bye              hi 24   hi
25:         4       ON        bye             bye 25  bye
26:         0      OFF        bye             bye 26  bye
27:         0      OFF         hi             bye 27  bye
28:         1      OFF         hi             bye 28  bye
29:         2      OFF         hi              hi 29   hi
30:         3      OFF         hi              hi 30   hi
31:         4      OFF       stay             bye 31  bye
32:         2       ON       stay              hi 32   hi
    myvector1 mystatus mycategory DesiredSolution  r    v

它是如何工作的： 在mydf[mystatus == "ON"] 中查找行号r 较低且myvector1 匹配的行。返回mycategory，如果有多个匹配则取最后一个匹配行。

Answer 2

另一种可能的方式，使用 data.frame 和 zoo::na.locf 的“切片”

首先，设置数据的位置，以便之后可以取回它

mydf$pos <- seq_len(nrow(mydf))

然后split你的data.frame根据myvector1:

spl_mydf <- split(mydf, mydf$myvector1)

然后对每个“切片”（仅myvector1 的一个值）应用一个函数，当mystatus 为ON 时，该函数仅保留mycategory 的值，将其余部分作为NA 并将NA 替换为以前不是NA 值。第一项是NA，您不要保留最后一项（为了获得所需的延迟）。

my_out <- lapply(spl_mydf,
                 function(sl_df){
                    out <- sl_df$mycategory
                    out[sl_df$mystatus=="OFF"] <- NA
                    data.frame(pos=sl_df$pos, 
                               out=c(NA, head(na.locf(as.character(out), na.rm=FALSE), -1))) # as.character is to avoid getting the factors levels
 })

最后，根据位置放回所有内容并抑制pos列：

out <- do.call(rbind, my_out)
mydf$output <- out$out[order(out$pos)]
mydf$pos <- NULL

head(mydf, 10)
   myvector1 mystatus mycategory DesiredSolution output
1          0       ON         hi              NA   <NA>
2          1      OFF         hi              NA   <NA>
3          2       ON       stay              NA   <NA>
4          3       ON        bye              NA   <NA>
5          4      OFF        bye              NA   <NA>
6          0       ON        bye              hi     hi
7          1       ON        bye              NA   <NA>
8          3       ON       stay             bye    bye
9          4       ON       stay              NA   <NA>
10         1      OFF        bye             bye    bye

检查一切正常：

all(mydf$DesiredSolution==mydf$output, na.rm=TRUE) # TRUE
all((as.character(mydf$DesiredSolution)=="NA")==is.na(mydf$output)) # TRUE 

（NA 被视为您的 data.frame 中的级别之一）

Answer 3

试试这个。我相信你可以让它更简洁。

DesiredSolution = vector()
for (i in 1:length(myvector1)) {
  step_1 = myvector1[i]
  step_2 = grep(step_1, myvector1)
  step_2 = step_2[step_2 < i]
  step_2a = sort(step_2, decreasing = T)

  if (length(step_2) != 0) {
    for (d in 1:length(step_2a)) {
      k = step_2a[d]
      step_3 = mystatus[k]
      if (step_3 == 'ON') {
        step_4 = mycategory[k]
        break
      } else if (step_3 == 'OFF' & d == length(step_2a)) {
        step_4 = NA
      }
    }
  } else {
      step_4 = NA
  }
  DesiredSolution = c(DesiredSolution, step_4)

}