使用R在数据树上聚合值

Question

我试图从数据树结构中计算出小时数。我可以直接在父节点下添加小时数，但是我不能包括分配给树中父节点的小时数。任何建议都会很棒。

这就是我得到的：

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

这就是我要找的：

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

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))

Answer 1

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：填写两个元素：

# 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

Answer 2

Aggregate期间的Do值缓存似乎仅适用于同一字段：

tree$Do(function(node) node$totalHours = node$hours)

tree$Do(function(node) node$totalHours = sum(if(!node$isLeaf) node$totalHours else 0,
                                             Aggregate(node, "totalHours", sum)),
        traversal = "post-order")
print(tree, "hours", "totalHours")
#           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

Answer 3

如果要递归地总结子项，data.tree包的Aggregate函数特别有用。在您的情况下，您有两件事要做：

1. 总结孩子和自己的价值
2. 将金额存储在单独的变量中

执行此操作的方法是：

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 <- ifelse(is.null(x$hours), 0, x$hours) + sum(Get(x$children, "total")), traversal = "post-order")
print(tree, "hours", "total")