# Order of edges in FeaturedGraph
**URL:** https://discourse.julialang.org/t/order-of-edges-in-featuredgraph/88052
**Category:** Machine Learning
**Created:** [September 30, 2022, 6:03pm UTC](https://discourse.julialang.org/t/order-of-edges-in-featuredgraph/88052 "2022-09-30T18:03:31Z")
**Posts on this page:** 5
**Page:** 1
### Author:  [@Tomas\_Pevny](https://discourse.julialang.org/u/Tomas_Pevny)
#### Post date: [September 30, 2022, 6:03pm UTC](https://discourse.julialang.org/t/order-of-edges-in-featuredgraph/88052/1 "2022-09-30T18:03:31Z")
Hi,
I am trying to understand `FeaturedGraph` from `GraphSignals.jl`. I am puzzled with the documentation, which says that features of edges are stored in matrix, where each column corresponds to a feature of one edge, but the documentation does not say, how edges are ordereded. I have created a workarround, where I first create a FeaturedGraph without any features and then iterate over edges to figure out the order.
```julia
using GraphSignals, Graphs
N = 4
ug = Graph(N)
add_edge!(ug, 1, 2); add_edge!(ug, 1, 3); add_edge!(ug, 1, 4);
add_edge!(ug, 2, 3); add_edge!(ug, 3, 4);
epmap = Dict([e => i for (i,e) in enumerate(edges(ug))])
fg = FeaturedGraph(ug)
ef = zeros(Float32, 5, ne(ug));
for (i, (vi, vj)) in edges(fg)
e = (vi < vj) ? Edge(vi, vj) : Edge(Int64(vj), Int64(vi))
ef[epmap[e], i] = 1
end
fg = FeaturedGraph(ug; ef)
```
But this construction seems to me a bit bizarre and I would consider it to be more like a work-around rather than a proper construction. Does someone knows the answer?
Thanks a lot in advance.
Tomas
---
### Author:  [@yuehhua](https://discourse.julialang.org/u/yuehhua)
#### Post date: [October 4, 2022, 6:37am UTC](https://discourse.julialang.org/t/order-of-edges-in-featuredgraph/88052/2 "2022-10-04T06:37:15Z")
```julia
julia> using GraphSignals, Graphs
julia> N = 4
4
julia> ug = Graph(N)
{4, 0} undirected simple Int64 graph
julia> add_edge!(ug, 1, 2); add_edge!(ug, 1, 3); add_edge!(ug, 1, 4);
julia> add_edge!(ug, 2, 3); add_edge!(ug, 3, 4);
julia> fg = FeaturedGraph(ug)
FeaturedGraph:
Undirected graph with (#V=4, #E=5) in adjacency matrix
julia> edgeidx, dst, src = collect(edges(fg))
(UInt32[0x00000001, 0x00000002, 0x00000004, 0x00000001, 0x00000003, 0x00000002, 0x00000003, 0x00000005, 0x00000004, 0x00000005], UInt32[0x00000002, 0x00000003, 0x00000004, 0x00000001, 0x00000003, 0x00000001, 0x00000002, 0x00000004, 0x00000001, 0x00000003], UInt32[0x00000001, 0x00000001, 0x00000001, 0x00000002, 0x00000002, 0x00000003, 0x00000003, 0x00000003, 0x00000004, 0x00000004])
julia> edge_index(graph(fg), 1, 2)
0x00000001
```
`collect(edges(fg))` could collect all the index of edges and edges in pair of vertices. You can use to index edge features.
For specific edge, there is `edge_index` available for `SparseGraph` in order to get the edge index from a pair of vertex indices.
---
### Author:  [@Tomas\_Pevny](https://discourse.julialang.org/u/Tomas_Pevny)
#### Post date: [October 4, 2022, 7:21am UTC](https://discourse.julialang.org/t/order-of-edges-in-featuredgraph/88052/3 "2022-10-04T07:21:40Z")
This is similar to solution we did. I was wondering if we can get the indicies of edges without constructing the FeaturedGraph, because we need those indices to construct feature matrix for edges. Or, is construction of FeaturedGraph so cheap, that it does not matter?
Thanks for reply.
---
### Author:  [@yuehhua](https://discourse.julialang.org/u/yuehhua)
#### Post date: [October 10, 2022, 5:26am UTC](https://discourse.julialang.org/t/order-of-edges-in-featuredgraph/88052/4 "2022-10-10T05:26:18Z")
I think it’s cheap to construct `FeaturedGraph` for taking indices for edge features.
It costs O(N) time, where N is number of nodes, for constructing `FeaturedGraph` from graph objects from JuliaGraphs, so maybe not cost much time when graph size is lower than or around thousnad.
---
### Author:  [@Tomas\_Pevny](https://discourse.julialang.org/u/Tomas_Pevny)
#### Post date: [October 10, 2022, 5:56am UTC](https://discourse.julialang.org/t/order-of-edges-in-featuredgraph/88052/5 "2022-10-10T05:56:39Z")
Thanks.