how can I group pairs of numbers that share a common number? Thank you!

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Jonathan Salazar
Jonathan Salazar 2017-8-24
评论： Steven Lord 2022-3-23
I have pairs of numbers that represent stacked platelets, I want to continue stacking these platelets by realizing that if i have pairs 1,2;2,4;5,7;1,3;7,8;10,9 that 1 2 4 3 are in the same stack and that 5 7 8 are in the same stack and 9 and 10 are in the same stack then I need to count how many of each stack size there is. Thank you!

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Steven Lord
Steven Lord 2017-8-24
You could treat this as a graph theory problem. Use A as the list of source and destination nodes, construct a graph from it, and use the conncomp function on the graph to identify which nodes are in the same group / component.
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Kouichi C. Nakamura
Kouichi C. Nakamura 2022-3-23
A = [1,2; 2,4; 5,7; 1,3; 7,8; 10,9]
G = graph(A(:,1), A(:,2))
G =
graph with properties:
Edges: [6×1 table]
Nodes: [10×0 table]
plot(G)
bins = conncomp(G)
bins =
1 1 1 1 2 3 2 2 4 4
Steven Lord
Steven Lord 2022-3-23
Ran in:
Or:
A = [1,2; 2,4; 5,7; 1,3; 7,8; 10,9];
G = graph(A(:,1), A(:,2));
components = conncomp(G, 'OutputForm', 'cell')
components = 1×4 cell array
{[1 2 3 4]} {[5 7 8]} {[6]} {[9 10]}

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更多回答（3 个）

Jan
Jan 2017-8-24
编辑：Jan 2017-8-24
Version 1: Create the stacks as cells and search in them iteratively:
P = [1,2; 2,4; 5,7; 1,3; 7,8; 10,9];
nP = size(P, 1);
S = cell(1, nP); % Pre-allocate list of stacks
cS = 0; % Number of stacks
for iP = 1:nP % Loop over pairs
aP = P(iP, :); % Current pair
found = false; % Check is any element exists in a stack already
for iS = 1:cS
if any(aP(1) == S{iS}) || any(aP(2) == S{iS})
% Or: if any(ismember(aP, S{iS}))
S{iS} = unique([S{iS}, aP]); % Insert both elements
found = true;
break;
end
end
if ~found % Create a new stack, if no matching was found
cS = cS + 1;
S{cS} = aP;
end
end
S = S(1:cS); % Crop unused cells
Jan
Jan 2017-8-24
编辑：Jan 2017-8-24
Alternative:
P = [1,2; 2,4; 5,7; 1,3; 7,8; 10,9];
nP = size(P, 1); % Number of pairs
iS = 0; % Current stack counter
G = zeros(nP, 1); % Group
for iP = 1:nP % Loop over pairs
aP = P(iP, :); % Current pair
[k, ~] = find(P(1:iP-1, :) == aP(1), 1); % 1st element occurred before?
if isempty(k) % No
[k, ~] = find(P(1:iP-1, :) == aP(2), 1); % 2nd element occurred before?
if isempty(k) % No
iS = iS + 1; % Then it is a new stack
G(iP) = iS; % Set group to new stack index
else
G(iP) = G(k); % 2nd lement found, use former group number
end
else
G(iP) = G(k); % 1st element found, use former group number
end
end
% Create cell containing the stacks:
S = splitapply(@(x) {unique(x(:))}, P, G);
% Perhaps faster but uglier:
% S = accumarray(G, (1:nP).', [], @(k) {unique(reshape(P(k, :), 1, []))})
John BG
John BG 2017-8-24
Hi Jonathan
this is John BG jgb2012@sky.com
the attached script answers your question, explanation
1.
w is going to log the pairs, here simulated with randi
w=[0 0];
2.
the patterns to search
L1=[1 2 4 3]; % patterns to search, without order
L2=[5 7 8];
L3=[9 10];
3.
logic conditions, one for each pattern
cond1=0;cond2=0;cond3=0;
4.
keep generating with randi until all 3 conditions met
while ~(cond1 && cond2 && cond3)
w=[w;randi([1 10],10,2)];
if numel(intersect(w(:,1),L1))==numel(L1) || numel(intersect(w(:,2),L1))==numel(L1)
cond1=1;
else
cond1=0;
end
if numel(intersect(w(:,1),L2))==numel(L2) || numel(intersect(w(:,2),L2))==numel(L2)
cond2=1;
else
cond2=0;
end
if numel(intersect(w(:,1),L3))==numel(L3) || numel(intersect(w(:,2),L3))==numel(L3)
cond3=1;
else
cond3=0;
end
end
w(1,:)=[]
.
the randomly generated stack stops growing when all 3 conditions met
w =
5 9
7 8
3 9
8 1
9 7
1 9
8 3
7 3
9 6
3 9
8 4
2 3
9 2
7 1
1 2
10 9
6 4
6 8
4 10
8 6
.
5.
checking conditions met
cond1
cond2
cond3
cond1 =
1
cond2 =
1
cond3 =
1
6.
the length of the stack right when all 3 conditions have 1st been met is
L_stack=size(w,1)
L_stack =
20
since your stack may be a measurement, instead of random generation, read from your stack line by line applying same conditions until all 3 met, that to stop reading it's same
~(cond1 && cond2 && cond3)
.
if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?
To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link
thanks in advance
John BG
<mailto:jgb2012@sky.com jgb2012@sky.com>

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