在截断实际数据框的同时有条件地创建数据框折叠折叠列 [英] Conditionally create a dataframe collapse a collapse a column while truncating actual dataframe
问题描述
我的数据框如下：
+------+----+----------+----------+
| from | to | priority | distance |
+------+----+----------+----------+
| 1 | 3 | 1 | 10 |
| 1 | 5 | 1 | 10 |
| 2 | 7 | 1 | 10 |
| 3 | 9 | 1 | 15 |
| 4 | 8 | 2 | 20 |
| 5 | 6 | 2 | 20 |
| 5 | 1 | 2 | 30 |
| 6 | 2 | 2 | 30 |
| 6 | 4 | 3 | 40 |
| 7 | 2 | 3 | 40 |
| 8 | 3 | 3 | 50 |
| 9 | 5 | 3 | 60 |
| 10 | | 3 | |
| 12 | 11 | 7 | 9 |
+------+----+----------+----------+
按优先级和距离排序
我想根据符合以下条件：
I want to collapse the to column based on the following criterias:
-
至中的每个唯一值都将分组以及相应的来自（例如下表中的第1行）
Every unique values in the
towill be grouped with the correspondingfrom(example row 1 in the table below)
+--------------+----------+
| from_parent | to_child |
+--------------+----------+
| 1 | 3,5 |
+--------------+----------+
如果值已经错误地分组在 to_child 中（在我们的示例中为数字3），并且该值也出现在从从主表，如果其对应的到是新值，则该值从未出现在 from_parent 或 to_child ，则该值应独立出现在from_parent
中，例如我们表中的
If a value has alredy been grouped in the to_child (in our case the number 3) and if it also appears in the from of the main table and if its corresponding to is a new value, a value that hasn't ever appeared in the from_parent or to_child, then that value should appear independently in from_parent
For instance from our table,
+------+----+----------+----------+
| from | to | priority | distance |
+------+----+----------+----------+
| 3 | 9 | 1 | 15 |
+------+----+----------+----------+
值9应该独立出现在新表中，如下所示：
the value 9 should appear independetly in the new table as under:
+--------------+----------+
| from_parent | to_child |
+--------------+----------+
| 9 | |
+--------------+----------+
，但是如果值9稍后出现在 to 列中，则应该将其添加到 to_child 群集中，并且应删除先前的值，所以我的意思是，如果9出现为 to 起为1的$ c>，对应的 to_child 的值为1应该是 3,5,9
but if the value 9 were to later appear in a to column it should be added to the to_child cluster and the previous value should be removed, so what I mean by this is that if 9 appeared as to for a from of 1 later on, the the correponsing to_child value of 1 should be 3,5,9
所以决赛桌应该是
+--------------+----------+
| from_parent | to_child |
+--------------+----------+
| 1 | 3,5 |
| 2 | 7 |
| 4 | 8 |
| 6 | |
| 9 | |
| 10 | |
| 12 | 11 |
+--------------+----------+
推荐答案
我将使用软件包 igraph 来解决问题，因为这是一个图形理论问题。
I will use package igraph to solve the problem in the question since this is a graph theory problem.
首先，从输入data.frame的前两列构造图形。
First, construct the graph from the first 2 columns of the input data.frame.
library(igraph)
g <- graph_from_data_frame(df1[1:2], directed = TRUE)
plot(g, edge.curved = TRUE, edge.arrow.size = 0.5)
现在从 df1 $ from 中的每个顶点获取路径。路径是通过广度优先搜索获得的，函数 bfs 。
Now get the paths from each of the vertices in df1$from. The paths are obtained with a breadth-first search, function bfs.
paths_list <- vector("list", length = length(V(g)))
i <- 0L
for(v in V(g)){
i <- i + 1L
ord <- bfs(g, root = v, neimode = "out",
unreachable = FALSE, dist = TRUE)$dist
ord <- ord[is.finite(ord)]
paths_list[[i]] <- ord
}
from <- lapply(paths_list, function(x) names(x)[1])
to <- lapply(paths_list, function(x) paste(names(x)[x != 0], collapse = ","))
res <- data.frame(from = unlist(from), to = unlist(to), stringsAsFactors = FALSE)
res <- res[nchar(res$from) != 0, ]
res
# from to
#1 1 3,5,6,9,2,4,7,8
#2 2 7
#3 3 9,5,1,6,2,4,7,8
#4 4 8,3,9,5,1,6,2,7
#5 5 1,6,2,3,4,7,8,9
#6 6 2,4,7,8,3,9,5,1
#7 7 2
#8 8 3,9,5,1,6,2,4,7
#9 9 5,1,6,2,3,4,7,8
#10 10
#11 12 11
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