Question

我想将每个节点的“上游”节点加起来。与上面问题中的答案不同，该答案在最短路径中从父母到孩子计算，我想将所有孩子的所有值加到父母身上。在河流环境中：从下游集水区到集水区上游的所有集水区。

我的输入数据

input <- structure(list(ZHYD = c("B030000156", "B030000159", "B030000165", 
"B030000167", "B030000170", "B030000171", "B030000175", "B030000177", 
"B030000181", "B030000183", "B030000184", "B030000190", "B030000192", 
"B030000193", "B030000195", "B030000196", "B030000197", "B030000198", 
"B030000199", "B030000201", "B030000202", "B030000133", "B030000191"
), NextDown = c("B030000133", "B030000133", "B030000159", "B030000159", 
"B030000167", "B030000167", "B030000170", "B030000175", "B030000175", 
"B030000171", "B030000170", "B030000171", "B030000184", "B030000191", 
"B030000197", "B030000197", "B030000191", "B030000190", "B030000190", 
"B030000199", "B030000199", "OUTLET", "B030000184"), count = c(2, 
0, 2, 0, 0, 0, 2, 3, 0, 0, 1, 0, 1, 2, 1, 0, 7, 0, 0, 0, 0, 5, 
0), Exutoire = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L), Outlet = c("BSO0000016", 
"BSO0000016", "BSO0000016", "BSO0000016", "BSO0000016", "BSO0000016", 
"BSO0000016", "BSO0000016", "BSO0000016", "BSO0000016", "BSO0000016", 
"BSO0000016", "BSO0000016", "BSO0000016", "BSO0000016", "BSO0000016", 
"BSO0000016", "BSO0000016", "BSO0000016", "BSO0000016", "BSO0000016", 
"BSO0000016", "BSO0000016"), EcrRiv_km = c(54.91, 5.14, 37.71, 
8.28, 17.22, 5.6, 45.87, 84.1, 26.22, 43.29, 32.49, 43.85, 35.1, 
11.09, 67.88, 32.66, 102.71, 18.21, 0.81, 14.05, 16.27, 45.44, 
3.47), EcrRivCoun = c(20, 3, 18, 5, 9, 3, 29, 44, 16, 18, 18, 
19, 16, 10, 30, 19, 56, 12, 3, 11, 10, 13, 5), DFLS = c(0.5, 
1, 0.5, 1, 1, 1, 0.5, 0.333333333333333, 1, 1, 1, 1, 1, 0.5, 
1, 1, 0.142857142857143, 1, 1, 1, 1, 0.2, 1), density = c(27.455, 
0, 18.855, 0, 0, 0, 22.935, 28.0333333333333, 0, 0, 32.49, 0, 
35.1, 5.545, 67.88, 0, 14.6728571428571, 0, 0, 0, 0, 9.088, 0
), dendritic_r = c(2.7455, 1.71333333333333, 2.095, 1.656, 1.91333333333333, 
1.86666666666667, 1.58172413793103, 1.91136363636364, 1.63875, 
2.405, 1.805, 2.30789473684211, 2.19375, 1.109, 2.26266666666667, 
1.71894736842105, 1.83410714285714, 1.5175, 0.27, 1.27727272727273, 
1.627, 3.49538461538462, 0.694)), class = c("tbl_df", "tbl", 
"data.frame"), row.names = c(NA, -23L), .Names = c("ZHYD", "NextDown", 
"count", "Exutoire", "Outlet", "EcrRiv_km", "EcrRivCoun", "DFLS", 
"density", "dendritic_r"))

此代码在最短的“下游”路径上累积计数

df <- data.frame(input$ZHYD, input$NextDown, input$count)
colnames(df) <- c('parent_id', 'id', 'count')
g <- graph_from_data_frame(df)
plot(g)

df <- get.data.frame(g, what = "edges")
dtr <- FromDataFrameNetwork(df)
dtr$countcum <- 0
dtr$Do(function(node) node$countcum <- node$parent$countcum + node$count, filterFun = isNotRoot)
print(dtr, "count", "countcum")

答案

这answer完成了这项工作

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

Answer 1

我想你可能在寻找subcomponent：

> subcomponent(g,"B030000156","out")
+ 3/24 vertices, named, from 8540f89:
[1] B030000156 B030000133 OUTLET    

> subcomponent(g,"B030000196","out")
+ 9/24 vertices, named, from 8540f89:
[1] B030000196 B030000197 B030000191 B030000184 B030000170 B030000167 B030000159 B030000133 OUTLET

如果您想进入其他（或两个）方向，也可以使用in或all作为修饰符。如果您使用sapply，则可以遍历所有节点：

> sapply(V(g),subcomponent,graph=g,mode="out")
$B030000156
+ 3/24 vertices, named, from 8540f89:
[1] B030000156 B030000133 OUTLET    

$B030000159
+ 3/24 vertices, named, from 8540f89:
[1] B030000159 B030000133 OUTLET    

$B030000165
+ 4/24 vertices, named, from 8540f89:
[1] B030000165 B030000159 B030000133 OUTLET
... the rest are truncated

您可以沿着这样的路径总结所有权重：

> E(g)$weight=as.numeric(df[,3])
> sum(E(g,path=c("B030000159","B030000133","OUTLET"))$weight)
[1] 5

这是从igraph对象中提取节点名称后沿路径获取权重总和的迂回方式：

library(stringr)
paths <- sapply(V(g),subcomponent,graph=g,mode="out")
z <- capture.output(paths)  # forcefully yank output from igraph object
pathlist <- z[which(str_detect(z,"[1] "))]

对于路径列表中的第一个和最后一个顶点，总长度为：

> sum(E(g,path=unlist(strsplit(pathlist[1],"\\s+"))[2:length(unlist(strsplit(pathlist[1],"\\s+")))])$weight)
[1] 5     

> sum(E(g,path=unlist(strsplit(pathlist[13],"\\s+"))[2:length(unlist(strsplit(pathlist[13],"\\s+")))])$weight)
[1] 6

您还可以将所有下游路径提取到数据框中：

> library(stringi)
> paths.df <- as.data.frame(stri_extract_all_words(pathlist, simplify = TRUE))
> head(paths.df)
  V1         V2         V3         V4         V5         V6         V7         V8         V9    V10
1  1 B030000171 B030000167 B030000159 B030000133     OUTLET                                        
2  1 B030000181 B030000175 B030000170 B030000167 B030000159 B030000133 OUTLET                  
3  1 B030000183 B030000171 B030000167 B030000159 B030000133     OUTLET                             
4  1 B030000190 B030000171 B030000167 B030000159 B030000133     OUTLET                             
5  1 B030000193 B030000191 B030000184 B030000170 B030000167 B030000159 B030000133     OUTLET       
6  1 B030000195 B030000197 B030000191 B030000184 B030000170 B030000167 B030000159 B030000133 OUTLET

提升igraph对象时边或节点属性的累积值

1 个答案: