每次列中缺少索引时将一个矩阵插入另一个矩阵

Question

假设我有矩阵 A：

A = [... %ID  Time   X
    1   0    0.7
    1   1    0.1
    1   2   -0.5
    1   3   -1.6
    3   0    0.3
    3   1    0.2
    3   2    0.0
    3   3   -0.2
    4   0    0.7
    4   1   -0.1
    4   2   -0.3
    4   3    0.2
    5   0   -0.5
    5   1   -0.4
    5   2   -0.9
    5   3   -0.4
    8   0    0.5
    8   1    1.0
    8   2    0.3
    8   3    0.4
    ];

你怎么会得到这样的东西（即矩阵 B）：

ID  Time      X
1     0      0.7
1     1      0.1
1     2     -0.5
1     3     -1.6
2   -999    -999
2   -999    -999
2   -999    -999
2   -999    -999
3     0      0.3
3     1      0.2
3     2      0.0
3     3     -0.2
4     0      0.7
4     1     -0.1
4     2     -0.3
4     3      0.2
5     0     -0.5
5     1     -0.4
5     2     -0.9
5     3     -0.4
6   -999    -999
6   -999    -999
6   -999    -999
6   -999    -999
7   -999    -999
7   -999    -999
7   -999    -999
7   -999    -999
8     0      0.5
8     1      1.0
8     2      0.3
8     3      0.4

任何时候“id”列中存在非连续间隙，我想添加一个带有“-999s”的单独矩阵，并用适当的（即连续的）ID 号标记“id”列。请注意 ID 5 和 ID 8 之间如何存在两个间隙 - 理想情况下，我可以填写两次缺失值并相应地标记它们（例如，ID 6 和 ID 7）。

我尝试了以下代码，但没有成功。请注意，“数据”是一个类似于上面的矩阵 A 的矩阵。 'filler' 是一个 -999s 的 3x4 矩阵：

-999 -999 -999
-999 -999 -999
-999 -999 -999
-999 -999 -999

示例代码：

ii = 1; %Starting counter
kk = ii+4; %So I don't start by indexing a row which doesn't yet exist

for ii = 1:4:length(data);
    if data(kk,1) ~= data(kk-4,1) + 1; %If there's a gap between the ID values greater than 1, i.e. they are non-consecutive
        data = [data(1:kk-1, :); filler; data(kk:end, :)]; %Append the filler column to the part of the matrix where another ID should be
    elseif data(kk,1) == -999; 
        end
end

Answer 1

accumarray 的经典工作：

fillval = -999;
A = [...
    1   0    0.7
    1   1    0.1
    1   2   -0.5
    1   3   -1.6
    3   0    0.3
    3   1    0.2
    3   2    0.0
    3   3   -0.2
    4   0    0.7
    4   1   -0.1
    4   2   -0.3
    4   3    0.2
    5   0   -0.5
    5   1   -0.4
    5   2   -0.9
    5   3   -0.4
    8   0    0.5
    8   1    1.0
    8   2    0.3
    8   3    0.4
    ];

%// Compact solution
X = accumarray(A(:,1),(1:size(A,1)).',[],@(x) {A(x,:)});
X(cellfun(@isempty,X)) = {fillval*ones(size(X{1}))};
out = cell2mat(X);

解释：

% indices of groups
gidx = A(:,1);
% vector of all indices
N = 1:size(A,1);  
% create groups with accumarray
X = accumarray(gidx(:),N(:),[],@(x) {A(x,:)});
% create logical vector for missing indices
mask = cellfun(@isempty,X);
% form fill matrix with same size as single group
fillmat = fillval*ones(size(X{1}));
% expand A to complete matrix with fill values
X(mask) = {fillmat};
% transform cell array to double array
out = cell2mat(X);

Answer 2

这里没有任何循环的解决方案（A是初始矩阵，B是结果）。目前假设 ID 值重复 4 次，并且在 2. 列中您有值 0:3 。不确定它是否比 Tommasos 解决方案更快，因为 ismember 对于非常大的矩阵可能需要很长时间。

% create template B to fill in values
B = [repelem(A(1,1):A(end,1),1,4).' repmat(0:3,1,A(end,1)-A(1,1)+1).' repmat(-999,(A(end,1)-A(1,1)+1)*4,1)];

% copy values from column 3 of A to column 3 of B where the first 2 columns are the same
[~,loc] = ismember(A(:,1:2),B(:,1:2),'rows');
B(loc,3) = A(:,3);

% Where column 3 == -999, also set column 2 to -999
B(B(:,3)==-999,2)=-999;

Answer 3

我向您推荐以下解决方案，在该解决方案中，我尝试尽可能地将流程矢量化：

A = [
    1   0    0.7
    1   1    0.1
    1   2   -0.5
    1   3   -1.6
    3   0    0.3
    3   1    0.2
    3   2    0.0
    3   3   -0.2
    4   0    0.7
    4   1   -0.1
    4   2   -0.3
    4   3    0.2
    5   0   -0.5
    5   1   -0.4
    5   2   -0.9
    5   3   -0.4
    8   0    0.5
    8   1    1.0
    8   2    0.3
    8   3    0.4
 ];

A_min = min(A(:,1));
A_max = max(A(:,1));
A_seq = A_min:A_max;
A_seq_cnt = numel(A_seq) * 4;

B_idx1 = repmat(A_seq,4,1);
B = [B_idx1(:) repmat((ones(A_seq_cnt,1) * -999),1,2)];

for n = unique(A(:,1)).'
    B((B(:,1) == n),2:3) = A((A(:,1) == n),2:3);
end

结果：

B =

 1     0   0.7
 1     1   0.1
 1     2  -0.5
 1     3  -1.6
 2  -999  -999
 2  -999  -999
 2  -999  -999
 2  -999  -999
 3     0   0.3
 3     1   0.2
 3     2     0
 3     3  -0.2
 4     0   0.7
 4     1  -0.1
 4     2  -0.3
 4     3   0.2
 5     0  -0.5
 5     1  -0.4
 5     2  -0.9
 5     3  -0.4
 6  -999  -999
 6  -999  -999
 6  -999  -999
 6  -999  -999
 7  -999  -999
 7  -999  -999
 7  -999  -999
 7  -999  -999
 8     0   0.5
 8     1     1
 8     2   0.3
 8     3   0.4

Answer 4

你可以像下面这样更正它：

result = zeros(4*max(data(:,1)),3); % initialization
last_idx = 1; % counter from 1 to max
idx = 1;
len = length(data(:,1));
while(idx < len)
    range = (4*(last_idx-1)+1):(4*last_idx);
    if data(idx,1) ~= last_idx % find missed index
        result(range,:) = repmat([last_idx -9 -9],4,1);
    else
        result(range,:) = data(idx:idx+3,:);
        idx = idx + 4;
    end
    last_idx = last_idx + 1;
end