使用R聚合数据树上的值

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")

           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")

           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

这个 Aggregate Do 似乎只适用于同一领域:

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

数据的聚合函数。如果您想要递归地对子项求和,则tree包特别有用。在您的情况下,您需要做两件事:

2. 将总和存储在单独的变量中

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")