我正在对我的图使用边缘过滤器,并希望更新边权重:
use petgraph::prelude::*;
use petgraph::graph;
use petgraph::visit::{Dfs, EdgeFiltered, IntoEdges};
fn filter_edges(edge: graph::EdgeReference<u32>) -> bool {
match edge.weight() {
0 => true,
_ => false,
}
}
fn main() {
let mut graph: graph::Graph<u32, u32> = graph::Graph::new();
let a = graph.add_node(1);
let b = graph.add_node(2);
let e = graph.add_edge(a, b, 0);
let mut filtered_graph = EdgeFiltered::from_fn(&graph, filter_edges);
let mut dfs = Dfs::new(&filtered_graph, a);
while let Some(node_index) = dfs.next(&filtered_graph) {
for edge in filtered_graph.edges(node_index) {
filtered_graph.update_edge(edge.source(), edge.target(), 1);
//graph.update_edge(edge.source(), edge.target(), 1);
}
}
}但这是错误的,因为EdgeFiltered没有update_edge函数:
error[E0599]: no method named `update_edge` found for struct `EdgeFiltered<&Graph<u32, u32>, for<'r> fn(petgraph::graph::EdgeReference<'r, u32>) -> bool {filter_edges}>` in the current scope
--> src/main.rs:22:28
|
22 | filtered_graph.update_edge(edge.source(), edge.target(), 1);
| ^^^^^^^^^^^ method not found in `EdgeFiltered<&Graph<u32, u32>, for<'r> fn(petgraph::graph::EdgeReference<'r, u32>) -> bool {filter_edges}>`如果我改为引用原始graph,它会出现借用检查器错误(与Dfs不同,不幸的是,EdgeFiltered的设计不允许您访问原始图形):
error[E0502]: cannot borrow `graph` as mutable because it is also borrowed as immutable
--> src/main.rs:21:13
|
17 | let mut filtered_graph = EdgeFiltered::from_fn(&graph, filter_edges);
| ------ immutable borrow occurs here
18 | let mut dfs = Dfs::new(&filtered_graph, a);
19 | while let Some(node_index) = dfs.next(&filtered_graph) {
| --------------- immutable borrow later used here
20 | for edge in filtered_graph.edges(node_index) {
21 | graph.update_edge(edge.source(), edge.target(), 1);
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ mutable borrow occurs hereEdgefiltered是非常小的,并且看起来没有任何用于可变图形操作的东西。有没有办法用Petgraph附带的东西来做到这一点,或者我必须编写我自己版本的update_edge showhow?
发布于 2021-03-24 17:32:55
FilteredGraph借用了Graph,因此只要Graph存在,就不能获得对FilteredGraph的可变引用。
您可以在每次dfs.next()调用时重新创建一个FilteredGraph来解决此问题,例如,这是可行的:
use petgraph::graph;
use petgraph::visit::{Dfs, EdgeFiltered};
fn filter_edges(edge: graph::EdgeReference<u32>) -> bool {
match edge.weight() {
0 => true,
_ => false,
}
}
fn main() {
let mut graph: graph::Graph<u32, u32> = graph::Graph::new();
let a = graph.add_node(1);
let b = graph.add_node(2);
let e = graph.add_edge(a, b, 0);
let filtered_graph = EdgeFiltered::from_fn(&graph, filter_edges);
let mut dfs = Dfs::new(&filtered_graph, a);
while let Some(node_index) = dfs.next(&EdgeFiltered::from_fn(&graph, filter_edges)) {
let mut neighbors = graph.neighbors(node_index).detach();
while let Some((edge_idx, _)) = neighbors.next(&graph) {
graph[edge_idx] = 1;
}
}
}注意:这将基于graph中存在的边,而不是filtered_graph中存在的边,获取给定节点的邻居。
您可以通过丢弃EdgeFiltered并在遍历中手动处理它来解决此问题,例如:
fn main() {
let mut graph: graph::Graph<u32, u32> = graph::Graph::new();
let a = graph.add_node(1);
let b = graph.add_node(2);
let e = graph.add_edge(a, b, 0);
let mut dfs = Dfs::new(&graph, a);
while let Some(node_index) = dfs.next(&graph) {
let mut neighbors = graph.neighbors(node_index).detach();
while let Some((edge_idx, _)) = neighbors.next(&graph) {
let edge_weight = &mut graph[edge_idx];
if *edge_weight == 0 {
*edge_weight = 1;
}
}
}
}https://stackoverflow.com/questions/66774121
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