Question

我问过类似的问题，例如here，并且那里提到的解决方案可以解决问题，但是这个问题比较复杂。

我有这样的数据表。

   ID1 member
 1   a parent
 2   a  child
 3   a parent
 4   a  child
 5   a  child
 6   b parent
 7   b parent
 8   b  child
 9   c  child
10   c  child
11   c parent
12   c  child

我想指定一个如下所示的序列，记住 ID1 和成员列。

   ID1 member sequence
 1   a parent        1
 2   a  child        2
 3   a parent        1
 4   a  child        2
 5   a  child        3
 6   b parent        1
 7   b parent        1
 8   b  child        2
 9   c  child        2 *
10   c  child        3
11   c parent        1
12   c  child        2

即

> dt$sequence = 1, wherever dt$member == "parent"

> dt$sequence = previous_row_value + 1, wherever dt$member=="child"

但有时可能会发生新的ID1可能无法以成员=＆＃34;父母＆＃34;开头。如果它以＆＃34; child＆＃34; （例如星号标记的行）我们必须用2开始排序。截至目前，我一直在使用循环，如下所示。

dt_sequence <- dt[ ,sequencing(.SD), by="ID1"]

sequencing <- function(dt){
  for(i in 1:nrow(dt)){
    if(i == 1){
      if(dt[i,member] %in% "child")
        dt$sequence[i] = 2
      else
        dt$sequence[i] = 1
    }
    else{
      if(dt[i,member] %in% "child")
        dt$sequence[i] = as.numeric(dt$sequence[i-1]) + 1
      else
        dt$sequence[i] = 1
    }
  }
  return(dt)
}

我在4e5行的数据表上运行此代码，需要花费大量时间才能完成（大约20分钟）。任何人都可以建议更快的方法。

Answer 1

DF <- read.table(text="   ID1 member
 1   a parent
 2   a  child
 3   a parent
 4   a  child
 5   a  child
 6   b parent
 7   b parent
 8   b  child
 9   c  child
10   c  child
11   c parent
12   c  child", header=TRUE, stringsAsFactors=FALSE)

library(data.table)
setDT(DF)
DF[, sequence := seq_along(member) + (member[1] == "child"), 
   by = list(ID1, cumsum(member == "parent"))]

#    ID1 member sequence
# 1:   a parent        1
# 2:   a  child        2
# 3:   a parent        1
# 4:   a  child        2
# 5:   a  child        3
# 6:   b parent        1
# 7:   b parent        1
# 8:   b  child        2
# 9:   c  child        2
#10:   c  child        3
#11:   c parent        1
#12:   c  child        2

Answer 2

试试这个，

dt$sequence <- rep(NA, length(dt$member))
for (i in seq_along(dt$member)){
  dt$sequence[i] <- ifelse(dt$member[i]=="parent", 1, 
                           ifelse(dt$ID1[i]==dt$ID1[i-1], dt$sequence[i-1] + 1, 2)
                           )
   }

更简单的dplyr解决方案

data <- dt %>% 
  group_by(ID1) %>% 
  mutate(
    seq = ifelse(member=="parent", 1, 2),
    sequence = ifelse(seq==1, 1, lag(seq, default = 1) + 1)
  )

如果每个小组ID1至少包含一个parent，则更容易解决方案是在group = ID1中安排数据，以便parent始终位于顶部：

dt %>% 
  group_by(ID1) %>%
  arrange(desc(member))

Answer 3

确实很好的问题。所以这是我的解决方案：

数据

dd <- structure(list(ID1 = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("a", "b", "c"), class = "factor"), member = structure(c(2L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 1L), .Label = c("child", "parent"), class = "factor")), .Names = c("ID1", "member"), row.names = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12"), class = "data.frame")

<强>代码

首先，将parent的所有元素设置为1：

parent <- dd$member == "parent" dd$sequence <- 0 dd$sequence[parent] <- 1

现在，将没有父项的所有child elemetns设置为2：

dd$sequence <- ave(dd$sequence, dd$ID1, FUN = function(.) { ret <- . ret[1] <- if (ret[1] == 0) 2 else ret[1] ret} )

现在，我们希望获得0's的每个序列的长度以及每个0的位置：

rl <- rle(dd$sequence) rl.wh <- which(rl$values == 0)

最后，我们可以生成序列：

dd$sequence[dd$sequence == 0] <- unlist(mapply(function(x, r) seq(x + 1, length.out = r, by = 1), rl$values[rl.wh - 1], rl$length[rl.wh]))

根据R中的先前行值分配序列中的值

3 个答案: