Question

我一直试图用一种聪明的方式来做这件事。给定具有以下结构的矩阵（或单元格）：

（请注意，上述矩阵/单元格在两列中都未排序，并且包含负数）。

如何根据特定列的唯一值对其进行分组。例如。第二列分组所需的输出类似于：

B{1} = [-1]
B{2} = [-1]
B{3} = [1,2]
B{4} = [2]
B{5} = [3,4]
B{6} = [5]
B{7} = [2,6]
B{-2} = [7]

提前致谢！

Answer 1

您可以使用accumarray：

[~,~,subs] = unique(A(:,2));
values  = accumarray(subs,A(:,1),[],@(x) {x});
ofGroup = accumarray(subs,A(:,2),[],@(x) {x(1)});

out = [ofGroup values]

out = 

    [-2]    [         7]
    [ 1]    [        -1]
    [ 2]    [        -1]
    [ 3]    [2x1 double]
    [ 4]    [         2]
    [ 5]    [2x1 double]
    [ 6]    [         5]
    [ 7]    [2x1 double]

如果您真的坚持您的建议订单，您可以执行以下操作，但我认为不应该这样做。

% positives
pos = A( A(:,2) >= 0 , :);

[~,~,subs] = unique(pos(:,2));
posvalues  = accumarray(subs,pos(:,1),[],@(x) {x});
posofGroup = accumarray(subs,pos(:,2),[],@(x) {x(1)});

% negatives
neg = A( A(:,2) < 0 , :);

[~,~,subs] = unique(neg(:,2));
negvalues  = flipud( accumarray(subs,neg(:,1),[],@(x) {x}) );
negofGroup = flipud( accumarray(subs,neg(:,2),[],@(x) {x(1)}) );

out = [posofGroup posvalues; negofGroup negvalues ]

out = 

    [ 1]    [        -1]
    [ 2]    [        -1]
    [ 3]    [2x1 double]
    [ 4]    [         2]
    [ 5]    [2x1 double]
    [ 6]    [         5]
    [ 7]    [2x1 double]
    [-2]    [         7]

Answer 2

怎么样：

[group, ~, subs] = unique(A(:,2))
B = accumarray(subs, A(:,1), [], @(x){x'})

结果

B= 

    [  7]
    [ -1]
    [ -1]
    [2,1]
    [  2]
    [4,3]
    [  5]
    [2,6]

和group将B的索引与其代表的组号相匹配

此外，如果您附上订购，那么您可以这样做：

[group, ~, subs] = unique(A(end:-1:1,2), 'stable');
B = flipud(accumarray(subs, A(end:-1:1,1), [], @(x){x'}));
group = flipud(group);

B = 

    [        -1]
    [        -1]
    [1x2 double]
    [         2]
    [1x2 double]
    [         5]
    [1x2 double]
    [         7]

group =

     1
     2
     3
     4
     5
     6
     7
    -2

根据另一列的值将一列矩阵的元素分组到单元阵列中

2 个答案: