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,第三列是累积和,一旦索引发生变化就会重置:

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

如何让第三列避免循环？

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

因此,如果有一种简单的方法将上面矩阵的第四列转换为,那么问题就会得到解决

O =

 0
 0
 0
 6
 6
15
15
15
15
45
45

由于CS-O提供了所需的输出.

我将不胜感激任何建议.

Answer 1

cumsum和diff基础方法,因此可能是良好的性能 -

%// 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)];

标杆

如果你关心性能,这里有一些针对基于的其他解决方案的基准测试accumarray.

基准测试代码(删除注释以获得紧凑性) -

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.

Answer 2

使用accumarray带有自定义功能:

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

无论索引更改是否为1(如您的示例中所示),都可以进行此操作.

以下方法更快,因为它避免了细胞.在他的回答中查看@Divakar的优秀基准测试(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(如示例所示):

   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. 如果索引值可以改变超过1(即可能有"跳过"值):这需要一个小的修改,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);