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matrix matlab中对的计数频率_Matlab_Sparse Matrix - Fatal编程技术网

matrix matlab中对的计数频率

matrix matlab中对的计数频率,matlab,sparse-matrix,Matlab,Sparse Matrix,我有矩阵X,mX2,我想得到一个大小为S的矩阵 size(unique(X(:,2),1) X size(unique(X(:,2),1) 对于每个S(i,j),我想计算i,j一起出现的次数。 例如: X = [1 11 ; 2 11; 3 11; 5 23; 6 23; 1 23; 9 24; 9 25; 3 23; 10 23] unique(X(:,2)) 11 23 24 25

我有矩阵X,mX2,我想得到一个大小为S的矩阵

size(unique(X(:,2),1) X size(unique(X(:,2),1) 
对于每个S(i,j),我想计算i,j一起出现的次数。 例如:

X = [1 11 ; 
     2 11; 
     3 11; 
     5 23; 
     6 23;
     1 23; 
     9 24;
     9 25;
     3 23;
    10 23]
unique(X(:,2)) 
11
23
24
25  
S应为:

 0     2     0     0
 0     0     0     0
 0     0     0     1
 0     0     0     0
(我不关心对角线,它可能有也可能没有,而且S可能是对称的)

因为11和23(位于位置1,2)同时出现了两次(即在X(:,1)中具有相同的值)


谢谢

这是一种方法:

[~, ~, n1] = unique(X(:,1));
[~, ~, n2] = unique(X(:,2));
B = accumarray([n2 n1],1);
S = B*B';
这给出了完整的矩阵:

>> S
S =
     3     2     0     0
     2     5     0     0
     0     0     1     1
     0     0     1     1
要删除对角线和下三角形,可以使用

S = triu(B*B',1);
产生

>> S
S =
     0     2     0     0
     0     0     0     0
     0     0     0     1
     0     0     0     0

这是一种方法:

[~, ~, n1] = unique(X(:,1));
[~, ~, n2] = unique(X(:,2));
B = accumarray([n2 n1],1);
S = B*B';
这给出了完整的矩阵:

>> S
S =
     3     2     0     0
     2     5     0     0
     0     0     1     1
     0     0     1     1
要删除对角线和下三角形,可以使用

S = triu(B*B',1);
产生

>> S
S =
     0     2     0     0
     0     0     0     0
     0     0     0     1
     0     0     0     0
请尝试以下操作:

% convert each columns to indices starting from 1
[a,~,aa] = unique(X(:,1));
[b,~,bb] = unique(X(:,2));

% group occurences of col2 according to values of col1
C = accumarray(aa, bb, [], @(x){x});

% keep only occurences of two or more values
C = C(cellfun(@numel,C) > 1);

% in case of three or more values co-occured, generate all pairs
C = cellfun(@(v) nchoosek(v,2), C, 'UniformOutput',false);

% concatenate all pairs
C = cell2mat(C);

% build count matrix
C = sparse(C(:,[1 2]), C(:,[2 1]), 1);
C = full(C);
这种情况下的结果(显然是对称矩阵):

或使用行/列标题打印:

>> [{[]} num2cell(b'); num2cell(b) num2cell(C)]
ans = 
      []    [11]    [23]    [24]    [25]
    [11]    [ 0]    [ 2]    [ 0]    [ 0]
    [23]    [ 2]    [ 0]    [ 0]    [ 0]
    [24]    [ 0]    [ 0]    [ 0]    [ 1]
    [25]    [ 0]    [ 0]    [ 1]    [ 0]
请尝试以下操作:

% convert each columns to indices starting from 1
[a,~,aa] = unique(X(:,1));
[b,~,bb] = unique(X(:,2));

% group occurences of col2 according to values of col1
C = accumarray(aa, bb, [], @(x){x});

% keep only occurences of two or more values
C = C(cellfun(@numel,C) > 1);

% in case of three or more values co-occured, generate all pairs
C = cellfun(@(v) nchoosek(v,2), C, 'UniformOutput',false);

% concatenate all pairs
C = cell2mat(C);

% build count matrix
C = sparse(C(:,[1 2]), C(:,[2 1]), 1);
C = full(C);
这种情况下的结果(显然是对称矩阵):

或使用行/列标题打印:

>> [{[]} num2cell(b'); num2cell(b) num2cell(C)]
ans = 
      []    [11]    [23]    [24]    [25]
    [11]    [ 0]    [ 2]    [ 0]    [ 0]
    [23]    [ 2]    [ 0]    [ 0]    [ 0]
    [24]    [ 0]    [ 0]    [ 0]    [ 1]
    [25]    [ 0]    [ 0]    [ 1]    [ 0]

+1个不错的矩阵乘法技巧,不是很明显它是如何工作的:)+1个不错的矩阵乘法技巧,不是很明显它是如何工作的:)非常好,但之后有点过度。非常好,但之后有点过度。