Matlab中的短路运算和空矩阵

Question

我有一个关于 Matlab 中的短路运算和空矩阵的问题。我将短路操作放在我的 if 语句中。但有时我有空矩阵，因此它们不能转换为逻辑标量值。

我已阅读此文档。 http://www.mathworks.com/help/matlab/math/empty-matrices-scalars-and-vectors.html 但我无法控制空矩阵和逻辑值的顺序。

在我看来，我想要像“[] || 1 to be 1”这样的东西。

感谢您的帮助！

这是原始代码。这是在网络分析中进化选民模型的代码。

clear 
clc
load('karate.mat');


rng shuffle
n = length(A);
rho = 0.5;
alpha = 0.2; %rewiring probability
%rewire to the same

t = 1;



B{t} = A; %temperal network
D{t} = zeros(n,n); % the discordant edges
S = zeros(n,1); % the infected nodes (set to zero if not infected)
SS = rand(n,1);
S(SS<rho) = 1;




for i = 1:n
    for j = 1:n
        if abs(S(i)-S(j)) > 0.5
            if  B{t}(i,j) == 1
                D{t}(i,j) = 1; 
            end
        end
    end
end



flag=1;
t=1;   
while(flag)
    t=t+1;
    for i=1:n
        for j=1:n
            if(D{t-1}(i,j)>0.1)
                ii=i;
                jj=j;

                neighB=find(B{t-1}(ii,:));
                neighB1=neighB(S(neighB)==1);
                neighB0=neighB(S(neighB)==0);
                kB=length(neighB);
                kB1=length(neighB1);
                kB0=kB-kB1;

                neighD=find(D{t-1}(ii,:));
                neighD1=neighD(S(neighD)==1);
                neighD0=neighD(S(neighD)==0);
                kD=length(neighD);
                kD1=length(neighD1);
                kD0=kD-kD1; 

                if S(ii)==1     
                    if(rand(1)<1-alpha) %adopting
                        S(ii)=0;
                        if(kD1>0)
                            D{t}(ii,neighD1)=1;
                            D{t}(neighD1,ii)=1;
                        end
                        if(kD0>0)
                            D{t}(ii,neighD0)=0;
                            D{t}(neighD0,ii)=0;
                        end

                    elseif((max(S(find((B{t-1}(ii,1:(ii-1)))==0)))==1) || (max(S(find((B{t-1}(ii,(ii+1):end))==0)))==1)) %rewire could happen if there exists a node having the same status and is not a neighbor
                        B{t}(ii,jj)=0;
                        % have a problem to compare [] and 1 here
                        % Operands to the || and && operators must be convertible to logical scalar values.    
                        F=find(S(find((B{t-1}(ii,1:(ii-1)))==0)))==1 && find(S((find(B{t-1}(ii,(ii+1):end))==0)))==1; %not sure here
                        C=randperm(length(F)); %not sure here
                        B{t}(ii,S((C(1))))=1; %not sure here

                        D{t}(ii,neighD1)=0;
                        D{t}(neighD1,ii)=0; 
                        D{t}(ii,neighD0)=1;
                        D{t}(neighD0,ii)=1;

                    else
                        S(ii)=0; %if cant find other nodes to rewire, then change the status     
                        D{t}(ii,neighD0)=0;
                        D{t}(neighD0,ii)=0;
                        D{t}(ii,neighD1)=1;
                        D{t}(neighD1,ii)=1;

                    end        
                end

                if S(ii)==0
                    if(rand(1)<1-alpha)
                        S(ii)=1;
                        if(kD1>0)
                            D{t}(ii,neighD1)=0;
                            D{t}(neighD1,ii)=0;
                        end
                    if(kD0>0)
                        D{t}(ii,neighD0)=1;
                        D{t}(neighD0,ii)=1;
                    end

                    elseif((min(S(find(B{t-1}(ii,1:(ii-1)))==0))==0) || (min(S(find(B{t-1}(ii,(ii+1):end))==0))==0)) %rewire could happen if there exists a node having the same status and is not a neighbor
                        B{t}(ii,jj)=0;

                        F=find(S(find(B{t-1}(ii,1:(ii-1)))==0))==0 && find(S(find(B{t-1}(ii,(ii+1):end))==0))==0; %not sure here
                        C=randperm(length(F)); %not sure here
                        B{t}(ii,S((C(1))))=1; %not sure here


                        D{t}(ii,neighD0)=0;
                        D{t}(neighD0,ii)=0;
                        D{t}(ii,neighD1)=1;
                        D{t}(neighD1,ii)=1;     

                    else
                        S(ii)=1; %if cant find other nodes to rewire, then change the status
                        D{t}(ii,neighD0)=1;
                        D{t}(neighD0,ii)=1;
                        D{t}(ii,neighD1)=0;
                        D{t}(neighD1,ii)=0; 
                    end
                end
            end
        end

    end
    if(nnz(D{t})==0)
        flag=0;
        disp(t);
    end
end
nnz(S)/length(A)
disp(t)

Answer 1

我建议您创建自己的函数，以按照您的意愿行事。例如，如果您想要以下行为：

1. 如果矩阵为空，则将值设置为 true
2. 如果矩阵不为空，则将值设置为“任意”值
3. 评估所有参数的 OR

然后你写一个函数：

function r = anytrue(varargin)
r = false;
for ii=1:numel(varargin)
  if isempty(varargin{ii})
    r = true;
    return
  end
  if varargin{ii}
    r = true;
    return
  end
end

请注意，这会表现出短路行为；您显然可以修改以适应您想要施加的任何条件。