MATLAB中索引的累积和

Question

考虑以下矩阵，其中第一列是索引，第二列 - 是值，第三列 - 是索引更改后重置的累积和：

1     1     1     % 1   
1     2     3     % 1+2
1     3     6     % 3+3
2     4     4     % 4
2     5     9     % 4+5
3     6     6     % 6
3     7    13    % 6+7
3     8    21    % 13+8
3     9    30    % 21+9
4    10    10    % 10
4    11    21    % 10+11

如何获得第三列避免循环？

我尝试以下方法：

  A = [1 1;...                 % Input
       1 2;...
       1 3;...
       2 4;...
       2 5;...
       3 6;...
       3 7;...
       3 8;...
       3 9;...
       4 10;...
       4 11];
  CS = cumsum(A(:,2));         % cumulative sum over the second column

  I = [diff(data(:,1));0];     % indicate the row before the index (the first column)  
                               % changes
  offset=CS.*I;                % extract the last value of cumulative sum for a given 
                               % index

  offset(end)=[]; offset=[0; offset] %roll offset 1 step forward

  [A, CS, offset]

结果是：

ans =

 1     1     1     0
 1     2     3     0
 1     3     6     0
 2     4    10     6
 2     5    15     0
 3     6    21    15
 3     7    28     0
 3     8    36     0
 3     9    45     0
 4    10    55    45
 4    11    66     0

如果有一种简单的方法可以将上面矩阵的第四列转换为

O =

 0
 0
 0
 6
 6
15
15
15
15
45
45

因为 CS-O 给出了想要的输出。

如果有任何建议，我将不胜感激。

Answer 1

您的策略实际上可能是我所做的。您的最后一步可以通过这种方式实现：（但请记住，您的方法假设连续索引。您当然可以通过 offset=[0; CS(1:end-1).*(diff(A(:,1))~=0)]; 更改此设置，但仍需要排序索引。）

I = find(offset);
idxLastI = cumsum(offset~=0);
hasLastI = idxLastI~=0; %// For the zeros at the beginning
%// Combine the above to the output
O = zeros(size(offset));
O(hasLastI) = offset(I(idxLastI(hasLastI)));
out = CS-O;

这应该类似于 Divakar 的 cumsum-diff 方法。

Answer 2

将accumarray 与自定义函数一起使用：

result = accumarray(A(:,1), A(:,2), [], @(x) {cumsum(x)});
result = vertcat(result{:});

无论索引更改是否以 1 为步长（如您的示例），这都有效。

---

以下方法更快，因为它避免了单元格。在his answer 中查看@Divakar 的出色基准测试（并查看他的解决方案，这是最快的）：

1. 如果索引更改始终对应于增加 1（如您的示例中所示）：

   last = find(diff(A(:,1)))+1; %// index of last occurrence of each index value
   result = A(:,2); %// this will be cumsum'd, after correcting for partial sums
   correction = accumarray(A(:,1), A(:,2)); %// correction to be applied for cumsum
   result(last) = result(last)-correction(1:end-1); %// apply correction
   result = cumsum(result); %// compute result
   

2. 如果索引值的变化幅度超过 1（即可能存在“跳过”值）：这需要稍作修改以稍微减慢速度。

   last = find(diff(A(:,1)))+1; %// index of last occurrence of each index value
   result = A(:,2); %// this will be cumsum'd, after correcting for partial sums
   correction = accumarray(A(:,1), A(:,2), [], @sum, NaN); %// correction
   correction = correction(~isnan(correction)); %// remove unused values
   result(last) = result(last)-correction(1:end-1); %// apply correction
   result = cumsum(result);
   

Answer 3

基于cumsum 和diff 的方法可能对性能有好处 -

%// cumsum values for the entire column-2
cumsum_vals = cumsum(A(:,2));

%// diff for column-1
diffA1 = diff(A(:,1));

%// Cumsum after each index
cumsum_after_each_idx = cumsum_vals([diffA1 ;0]~=0);

%// Get cumsum for each "group" and place each of its elements at the right place
%// to be subtracted from cumsum_vals for getting the final output
diffA1(diffA1~=0) = [cumsum_after_each_idx(1) ; diff(cumsum_after_each_idx)];

out = cumsum_vals-[0;cumsum(diffA1)];

---

基准测试

如果您关心性能，这里有一些基于accumarray 的其他解决方案的基准测试。

基准代码（为了紧凑而删除了 cmets）-

A = ..  Same as in the question

num_runs = 100000; %// number of runs

disp('---------------------- With cumsum and diff')
tic
for k1=1:num_runs
    cumsum_vals = cumsum(A(:,2));
    diffA1 = diff(A(:,1));
    cumsum_after_each_idx = cumsum_vals([diffA1 ;0]~=0);
    diffA1(diffA1~=0) = [cumsum_after_each_idx(1) ; diff(cumsum_after_each_idx)];
    out = cumsum_vals-[0;cumsum(diffA1)];
end
toc,clear cumsum_vals  diffA1 cumsum_after_each_idx out

disp('---------------------- With accumarray - version 1')
tic
for k1=1:num_runs
    result = accumarray(A(:,1), A(:,2), [], @(x) {cumsum(x)});
    result = vertcat(result{:});
end
toc, clear result

disp('--- With accumarray - version 2 (assuming consecutive indices only)')
tic
for k1=1:num_runs
    last = find(diff(A(:,1)))+1; %// index of last occurrence of each index value
    result = A(:,2); %// this will be cumsum'd, after correcting for partial sums
    correction = accumarray(A(:,1), A(:,2)); %// correction to be applied for cumsum
    result(last) = result(last)-correction(1:end-1); %// apply correction
    result = cumsum(result); %// compute result
end
toc, clear last result correction

disp('--- With accumarray - version 2 ( general case)')
tic
for k1=1:num_runs
    last = find(diff(A(:,1)))+1; %// index of last occurrence of each index value
    result = A(:,2); %// this will be cumsum'd, after correcting for partial sums
    correction = accumarray(A(:,1), A(:,2), [], @sum, NaN); %// correction
    correction = correction(~isnan(correction)); %// remove unused values
    result(last) = result(last)-correction(1:end-1); %// apply correction
    result = cumsum(result);
end
toc

结果 -

---------------------- With cumsum and diff
Elapsed time is 1.688460 seconds.
---------------------- With accumarray - version 1
Elapsed time is 28.630823 seconds.
--- With accumarray - version 2 (assuming consecutive indices only)
Elapsed time is 2.416905 seconds.
--- With accumarray - version 2 ( general case)
Elapsed time is 4.839310 seconds.