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

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!

Answer 1

[Steven Lord]

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.

[Jonathan Salazar]

can you clarify on how to construct a graph using the pairs? I tried

G = graph(real_pair) % contains all the pairs of platelets 
plot(G)
bins = conncomp(G)

but I am getting an error and do not really understand what is going on. thank you so much!

[Walter Roberson]

real_pair = [1,2;2,4;5,7;1,3;7,8;10,9];
G = graph(real_pair(:,1), real_pair(:,2));
plot(G);
bins = conncomp(G);
numcc = max(bins);
components = cell(numcc,1);
for K = 1 : numcc
  components{K} = find(bins == K);
end

celldisp(components)

[Steven Lord]

Or call conncomp and specify 'OutputForm' as 'cell'. That can replace six of the lines in your code. [Well, it replaces a call to conncomp with a slightly different call to conncomp then eliminates five other lines.]

[Jonathan Salazar]

Thank you so much! this works perfectly! In fact when i applied it to my code i was also able to count the platelets that do not stack... which is something I was planning on doing separately.

[Jan]

@Jonathan: Then the orphaned 6 is not a problem, but useful? Fine. Then I like this solution also. +1

[Kouichi C. Nakamura]

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]

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]}

Answer 2

[Jan]

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

Answer 3

[Jan]

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, []))})

Answer 4

[John BG]

Hi Jonathan

this is John BG [e-mail]

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:[e-mail] [e-mail]>