13

For clever usage of linear indexing or accumarray, I've sometimes felt the need to generate sequences based on run-length encoding. As there is no built-in function for this, I am asking for the most efficient way to decode a sequence encoded in RLE.

Specification:

As to make this a fair comparison I would like to set up some specifications for the function:

  • If optional second argument values of same length is specified, the output should be according to those values, otherwise just the values 1:length(runLengths).
  • Gracefully handle:
    • zeros in runLengths
    • values being a cell array.
  • Output vector should have same column/row format as runLengths

In short: The function should be equivalent to the following code:

function V = runLengthDecode(runLengths, values)
[~,V] = histc(1:sum(runLengths), cumsum([1,runLengths(:).']));
if nargin>1
    V = reshape(values(V), 1, []);
end
V = shiftdim(V, ~isrow(runLengths));
end

Examples:

Here are a few test cases 

runLengthDecode([0,1,0,2])
runLengthDecode([0,1,0,4], [1,2,4,5].')
runLengthDecode([0,1,0,2].', [10,20,30,40])
runLengthDecode([0,3,1,0], {'a','b',1,2})

and their output:

>> runLengthDecode([0,1,0,2])
ans =
     2     4     4

>> runLengthDecode([0,1,0,4], [1,2,4,5].')
ans =    
     2     5     5     5     5

>> runLengthDecode([0,1,0,2].', [10,20,30,40])
ans =
    20
    40
    40

>> runLengthDecode([0,3,1,0],{'a','b',1,2})
ans = 
    'b'    'b'    'b'    [1]
4

4 回答 4

6

为了找出最有效的解决方案，我们提供了一个评估性能的测试脚本。第一个图描述了 vector 长度增长的运行时间runLengths，其中条目均匀分布，最大长度为200。gnovice解决方案的修改是最快的，而Divakar的解决方案位居第二。 速度比较

第二个图使用几乎相同的测试数据，除了它包含一个初始运行长度2000。这主要影响两种bsxfun解决方案，而其他解决方案的表现非常相似。

速度比较

测试表明，修改gnovice的原始答案将是最有效的。

如果您想自己进行速度比较，这里是用于生成上述图的代码。

function theLastRunLengthDecodingComputationComparisonYoullEverNeed()
Funcs =  {@knedlsepp0, ...
          @LuisMendo1bsxfun, ...
          @LuisMendo2cumsum, ...
          @gnovice3cumsum, ...
          @Divakar4replicate_bsxfunmask, ...
          @knedlsepp5cumsumaccumarray
          };    
%% Growing number of runs, low maximum sizes in runLengths
ns = 2.^(1:25);
paramGenerators{1} = arrayfun(@(n) @(){randi(200,n,1)}, ns,'uni',0);
paramGenerators{2} = arrayfun(@(n) @(){[2000;randi(200,n,1)]}, ns,'uni',0);
for i = 1:2
    times = compareFunctions(Funcs, paramGenerators{i}, 0.5);
    finishedComputations = any(~isnan(times),2);
    h = figure('Visible', 'off');
    loglog(ns(finishedComputations), times(finishedComputations,:));
    legend(cellfun(@func2str,Funcs,'uni',0),'Location','NorthWest','Interpreter','none');
    title('Runtime comparison for run length decoding - Growing number of runs');
    xlabel('length(runLengths)'); ylabel('seconds');
    print(['-f',num2str(h)],'-dpng','-r100',['RunLengthComparsion',num2str(i)]);
end
end

function times = compareFunctions(Funcs, paramGenerators, timeLimitInSeconds)
if nargin<3
    timeLimitInSeconds = Inf;
end
times = zeros(numel(paramGenerators),numel(Funcs));
for i = 1:numel(paramGenerators)
    Params = feval(paramGenerators{i});
    for j = 1:numel(Funcs)
        if max(times(:,j))<timeLimitInSeconds
            times(i,j) = timeit(@()feval(Funcs{j},Params{:}));
        else
            times(i,j) = NaN;
        end
    end
end
end
%% // #################################
%% // HERE COME ALL THE FANCY FUNCTIONS
%% // #################################
function V = knedlsepp0(runLengths, values)
[~,V] = histc(1:sum(runLengths), cumsum([1,runLengths(:).']));%'
if nargin>1
    V = reshape(values(V), 1, []);
end
V = shiftdim(V, ~isrow(runLengths));
end

%% // #################################
function V = LuisMendo1bsxfun(runLengths, values)
nn = 1:numel(runLengths);
if nargin==1 %// handle one-input case
    values = nn;
end
V = values(nonzeros(bsxfun(@times, nn,...
    bsxfun(@le, (1:max(runLengths)).', runLengths(:).'))));
if size(runLengths,1)~=size(values,1) %// adjust orientation of output vector
    V = V.'; %'
end
end

%% // #################################
function V = LuisMendo2cumsum(runLengths, values)
if nargin==1 %// handle one-input case
    values = 1:numel(runLengths);
end
[ii, ~, jj] = find(runLengths(:));
V(cumsum(jj(end:-1:1))) = 1;
V = values(ii(cumsum(V(end:-1:1))));
if size(runLengths,1)~=size(values,1) %// adjust orientation of output vector
    V = V.'; %'
end
end

%% // #################################
function V = gnovice3cumsum(runLengths, values)
isColumnVector =  size(runLengths,1)>1;
if nargin==1 %// handle one-input case
    values = 1:numel(runLengths);
end
values = reshape(values(runLengths~=0),1,[]);
if isempty(values) %// If there are no runs
    V = []; return;
end
runLengths = nonzeros(runLengths(:));
index = zeros(1,sum(runLengths));
index(cumsum([1;runLengths(1:end-1)])) = 1;
V = values(cumsum(index));
if isColumnVector %// adjust orientation of output vector
    V = V.'; %'
end
end
%% // #################################
function V = Divakar4replicate_bsxfunmask(runLengths, values)
if nargin==1   %// Handle one-input case
    values = 1:numel(runLengths);
end

%// Do size checking to make sure that both values and runlengths are row vectors.
if size(values,1) > 1
    values = values.'; %//'
end
if size(runLengths,1) > 1
    yes_transpose_output = false;
    runLengths = runLengths.'; %//'
else
    yes_transpose_output = true;
end

maxlen = max(runLengths);

all_values = repmat(values,maxlen,1);
%// OR all_values = values(ones(1,maxlen),:);

V = all_values(bsxfun(@le,(1:maxlen)',runLengths)); %//'

%// Bring the shape of V back to the shape of runlengths
if yes_transpose_output
    V = V.'; %//'
end
end
%% // #################################
function V = knedlsepp5cumsumaccumarray(runLengths, values)
isRowVector = size(runLengths,2)>1;
%// Actual computation using column vectors
V = cumsum(accumarray(cumsum([1; runLengths(:)]), 1));
V = V(1:end-1);
%// In case of second argument
if nargin>1
    V = reshape(values(V),[],1);
end
%// If original was a row vector, transpose
if isRowVector
    V = V.'; %'
end
end
于 2015-02-13T15:24:55.697 回答
5

方法一

这应该相当快。它用于 bsxfun创建大小为numel(runLengths)x的矩阵numel(runLengths)，因此它可能不适合巨大的输入大小。

function V = runLengthDecode(runLengths, values)
nn = 1:numel(runLengths);
if nargin==1 %// handle one-input case
    values = nn;
end
V = values(nonzeros(bsxfun(@times, nn,...
    bsxfun(@le, (1:max(runLengths)).', runLengths(:).'))));
if size(runLengths,1)~=size(values,1) %// adjust orientation of output vector
    V = V.';
end

方法二

这种基于 的方法是对其他答案cumsum中使用的方法的改编。它比方法 1 使用更少的内存。

function V = runLengthDecode2(runLengths, values)
if nargin==1 %// handle one-input case
    values = 1:numel(runLengths);
end
[ii, ~, jj] = find(runLengths(:));
V(cumsum(jj(end:-1:1))) = 1;
V = values(ii(cumsum(V(end:-1:1))));
if size(runLengths,1)~=size(values,1) %// adjust orientation of output vector
    V = V.';
end
于 2015-02-13T15:01:47.720 回答
5

从 R2015a 开始，该函数repelem是执行此操作的最佳选择：

function V = runLengthDecode(runLengths, values)
if nargin<2
    values = 1:numel(runLengths);
end
V = repelem(values, runLengths);
end

对于 R2015a 之前的版本，这是一个快速的替代方案：

替代repelem：

我觉得可以通过使用比预处理掩码更好的零运行长度修复来改进gnovice的方法。所以我试accumarray了一下。似乎这为解决方案提供了另一个推动力：

function V = runLengthDecode(runLengths, values)
%// Actual computation using column vectors
V = cumsum(accumarray(cumsum([1; runLengths(:)]), 1));
V = V(1:end-1);
%// In case of second argument
if nargin>1
    V = reshape(values(V),[],1);
end
%// If original was a row vector, transpose
if size(runLengths,2)>1
    V = V.'; %'
end
end
于 2015-02-19T20:01:55.357 回答
4

这里介绍的解决方案基本上run-length decoding分为两个步骤 -

  1. 复制所有values的最大数量runLengths。
  2. 使用bsxfun's masking 功能从每一列中选择相应的runlengths.

函数代码中的其余内容是处理输入和输出大小以满足问题中设置的要求。

接下来列出的功能代码将是我之前对类似问题的答案之一的“清理”版本。这是代码 -

function V = replicate_bsxfunmask(runLengths, values)

if nargin==1   %// Handle one-input case
    values = 1:numel(runLengths);
end

%// Do size checking to make sure that both values and runlengths are row vectors.
if size(values,1) > 1
    values = values.'; %//'
end
if size(runLengths,1) > 1
    yes_transpose_output = false;
    runLengths = runLengths.'; %//'
else
    yes_transpose_output = true;
end

maxlen = max(runLengths);

all_values = repmat(values,maxlen,1);
%// OR all_values = values(ones(1,maxlen),:);

V = all_values(bsxfun(@le,(1:maxlen)',runLengths)); %//'

%// Bring the shape of V back to the shape of runlengths
if yes_transpose_output
    V = V.'; %//'
end

return;

接下来列出的代码将是混合代码 ( cumsum+ replicate_bsxfunmask)，适用于有大量异常值或非常大的异常值时。同样为了简单起见，目前这仅适用于数字数组。这是实现 -

function out = replicate_bsxfunmask_v2(runLengths, values)

if nargin==1                       %// Handle one-input case
    values = 1:numel(runLengths);
end

if size(values,1) > 1
    values = values.';  %//'
end

if size(runLengths,1) > 1
    yes_transpose_output = true;
    runLengths = runLengths.';  %//'
else
    yes_transpose_output = false;
end

%// Regularize inputs
values = values(runLengths>0);
runLengths = runLengths(runLengths>0);

%// Main portion of code
thresh = 200; %// runlengths threshold that are to be processed with cumsum

crunLengths = cumsum(runLengths); %%// cumsums of runlengths
mask = runLengths >= thresh; %// mask of runlengths above threshold
starts = [1 crunLengths(1:end-1)+1]; %// starts of each group of runlengths

mask_ind = find(mask); %// indices of mask

post_mark = starts(mask);
negt_mark = crunLengths(mask)+1;

if  ~isempty(negt_mark) && negt_mark(end) > crunLengths(end)
    negt_mark(end) = [];
end

%// Create array & set starts markers for starts of runlengths above thresh
marked_out = zeros(1,crunLengths(end));
marked_out(post_mark) = mask_ind;
marked_out(negt_mark) = marked_out(negt_mark) -1*mask_ind(1:numel(negt_mark));

%// Setup output array with the cumsumed version of marked array
out = cumsum(marked_out);

%// Mask for final ouput to decide between large and small runlengths
thresh_mask = out~=0;

%// Fill output array with cumsum and then rep-bsxfun based approaches
out(thresh_mask) = values(out(thresh_mask));

values = values(~mask);
runLengths = runLengths(~mask);

maxlen = max(runLengths);
all_values = repmat(values,maxlen,1);
out(~thresh_mask) = all_values(bsxfun(@le,(1:maxlen)',runLengths)); %//'

if yes_transpose_output
    out = out.';  %//'
end

return;
于 2015-02-15T19:09:44.883 回答