问题描述
Assume I have the following matrix (defined here in Julia language):
mat = [1 1 0 0 0 ; 1 1 0 0 0 ; 0 0 0 0 1 ; 0 0 0 1 1]
Considering as a "component" a group of neighbour elements that have value '1', how to identify that this matrix has 2 components and which vertices compose each one?
For the matrix mat above I would like to find the following result:
Component 1 is composed by the following elements of the matrix (row,column):
(1,1)
(1,2)
(2,1)
(2,2)
Component 2 is composed by the following elements:
(3,5)
(4,4)
(4,5)
I can use Graph algorithms like this to identify components in square matrices. However such algorithms can not be used for non-square matrices like the one I present here.
Any idea will be much appreciated.
I am open if your suggestion involves the use of a Python library + PyCall. Although I would prefer to use a pure Julia solution.
Regards
Using Image.jl
's label_components
is indeed the easiest way to solve the core problem. However, your loop over 1:maximum(labels)
may not be efficient: it's O(N*n)
, where N
is the number of elements in labels
and n
the maximum, because you visit each element of labels
n
times.
You'd be much better off just visiting each element of labels
just twice: once to determine the maximum, and once to assign each non-zero element to its proper group:
using Images
function collect_groups(labels)
groups = [Int[] for i = 1:maximum(labels)]
for (i,l) in enumerate(labels)
if l != 0
push!(groups[l], i)
end
end
groups
end
mat = [1 1 0 0 0 ; 1 1 0 0 0 ; 0 0 0 0 1 ; 0 0 0 1 1]
labels = label_components(mat)
groups = collect_groups(labels)
Output on your test matrix:
2-element Array{Array{Int64,1},1}:
[1,2,5,6]
[16,19,20]
Calling library functions like find
can occasionally be useful, but it's also a habit from slower languages that's worth leaving behind. In julia, you can write your own loops and they will be fast; better yet, often the resulting algorithm is much easier to understand. collect(zip(ind2sub(size(mat),find( x -> x == value, mat))...))
does not exactly roll off the tongue.
这篇关于如何使用 Julia 在矩阵中查找连通分量的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!