Comparing elements of cell arrays
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My given cell array (p) and existing cell arrays (routes_log) with whom this given one is to be compared are as below.
p={[12 11 13 9]} ; %denotes a route with links 12-11,11-13 &13-9.
routes_log = {[13 9 10]} {[6 11 13 2]} {[12 14 6]} %a 1*3 cell array. Each one denotes a route
In my code, these elements denote graph nodes and so, p cell array denotes a route with links 12-11,11-13 &13-9.
So I want to find if 'ANY' of the given links in array p are a part of 'ANY' of the links of the existing routes. For e.g. in the above case, I want the code to output {[13 9 10]} and {[6 11 13 2]}. As they have the elements 13-9 & 11-13 (consecutively) and my p array also has 13-9 & 11-13. I am confused on how to do this as I cannot use ismember() so straightforwardly.
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DGM
am 2 Dez. 2021
Bearbeitet: DGM
am 2 Dez. 2021
Well, I bet there's a more elegant way than this, but this seems to work.
p = {[12 11 13 9]}; % denotes a route with links 12-11,11-13 &13-9.
routes_log = {[13 9 10] [6 11 13 2] [12 14 6]}; % a 1*3 cell array. Each one denotes a route
% build link list as node pairs
lklistp = repelem(p{:},2);
lklistp = reshape(lklistp(2:end-1),2,[]).'
% compare link lists in logs, build logical mask
shareslinks = false(size(routes_log));
for lg = 1:numel(routes_log)
lklistlog = repelem(routes_log{lg},2);
lklistlog = reshape(lklistlog(2:end-1),2,[]).';
shareslinks(lg) = any(all(ismember(lklistlog,lklistp),2));
end
% apply mask
routeswithcommonlinks = routes_log(shareslinks);
celldisp(routeswithcommonlinks)
4 Kommentare
DGM
am 3 Dez. 2021
Oof. Yeah, you got it. See this is what happens when I try to write code in my head.
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Voss
am 2 Dez. 2021
I don't know whether it's feasible to change your code in the way I'm about to suggest, but it may be worthwhile to store the links as n-by-2 matrices with each row denoting the two endpoints of the links, e.g.:
p = {[12 11; 11 13; 13 9]};
routes_log = {[13 9; 9 10] [6 11; 11 13; 13 2] [12 14; 14 6]};
Then you could use ismember with the 'rows' option to check for existing links.
2 Kommentare
dpb
am 2 Dez. 2021
Agree w/ @Benjamin. The storage scheme you have is memory-efficient, but exceedingly difficult to use for the stated purpose. Redesign the storage around the need is also my recommendation.
Otherwise you're into nested cellfun calls with sequential sections of the cell to be compared -- ugly at best.
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