r - 使用R在数据树上聚合值

我正在尝试计算数据树结构中的时间。我可以将小时数直接加到父节点下，但不能在树中包括分配给父节点的小时数。任何建议都很好。

这就是我得到的：

levelName hours totalhours1 Ned NA 12 °--John 1 33 °--Kate 1 34 ¦--Dan 1 15 ¦--Ron 1 16 °--Sienna 1 1

这就是我要寻找的：

levelName hours totalHours1 Ned NA 52 °--John 1 53 °--Kate 1 44 ¦--Dan 1 15 ¦--Ron 1 16 °--Sienna 1 1

这是我的代码：

# Install package
install.packages('data.tree')
library(data.tree)

# Create data frame
to <- c("Ned", "John", "Kate", "Kate", "Kate")
from <- c("John", "Kate", "Dan", "Ron", "Sienna")
hours <- c(1,1,1,1,1)
df <- data.frame(from,to,hours)

# Create data tree
tree <- FromDataFrameNetwork(df)
print(tree, "hours")

# Get running total of hours that includes all nodes and children values.
tree$Do(function(x) x$total <- Aggregate(x, "hours", sum), traversal = "post-order")
print(tree, "hours", runningtotal = tree$Get(Aggregate, "total", sum))

最佳答案

您可以简单地使用递归函数：

myApply <- function(node) {
  node$totalHours <-
    sum(c(node$hours, purrr::map_dbl(node$children, myApply)), na.rm = TRUE)
}
myApply(tree)
print(tree, "hours", "totalHours")

结果：

           levelName hours totalHours
1 Ned                   NA          5
2  °--John               1          5
3      °--Kate           1          4
4          ¦--Dan        1          1
5          ¦--Ron        1          1
6          °--Sienna     1          1

编辑：填充两个元素：

# Create data frame
to <- c("Ned", "John", "Kate", "Kate", "Kate")
from <- c("John", "Kate", "Dan", "Ron", "Sienna")
hours <- c(1,1,1,1,1)
hours2 <- 5:1
df <- data.frame(from,to,hours, hours2)

# Create data tree
tree <- FromDataFrameNetwork(df)
print(tree, "hours", "hours2")

myApply <- function(node) {
  res.ch <- purrr::map(node$children, myApply)
  a <- node$totalHours <-
    sum(c(node$hours,  purrr::map_dbl(res.ch, 1)), na.rm = TRUE)
  b <- node$totalHours2 <-
    sum(c(node$hours2, purrr::map_dbl(res.ch, 2)), na.rm = TRUE)
  list(a, b)
}
myApply(tree)
print(tree, "hours", "totalHours", "hours2", "totalHours2")

结果：

           levelName hours totalHours hours2 totalHours2
1 Ned                   NA          5     NA          15
2  °--John               1          5      5          15
3      °--Kate           1          4      4          10
4          ¦--Dan        1          1      3           3
5          ¦--Ron        1          1      2           2
6          °--Sienna     1          1      1           1