MATLAB中的行程解码

Question

为了巧妙地使用线性索引accumarray,我有时觉得需要根据游程编码生成序列.由于没有内置函数,我要求最有效的方法来解码在RLE中编码的序列.

规格:

为了使这个公平比较,我想为该功能设置一些规范:

• 如果values指定了相同长度的可选第二个参数,则输出应该根据这些值,否则只是值1:length(runLengths).
• 优雅处理:
  • 零 runLengths
  • values 是一个单元阵列.
• 输出向量应具有相同的列/行格式 runLengths

简而言之:该函数应该等效于以下代码:

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

例子:

以下是一些测试用例

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})

和他们的输出:

>> 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]

Answer 1

为了找出哪个是最有效的解决方案,我们提供了一个评估性能的测试脚本.第一曲线示出了运行时的矢量的长度生长runLengths,其中所述条目被均匀地最大长度200的分布式的修改gnovice的溶液是最快的,与Divakar的溶液是第二位.

第二个图使用几乎相同的测试数据,除了它包括初始长度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.

测试表明,一个修改的gnovice的原来的答复将是最高效的.

如果你想自己进行速度比较,这里是用于生成上图的代码.

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

Answer 2

从R2015a开始,该功能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,这是一个快速的选择:

替代repelem:

我感觉gnovice的方法可以通过使用比预处理掩码更好的零runLength修复来改进.所以我试accumarray了一下.这似乎为解决方案提供了另一个提升:

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

Answer 3

方法1

这应该相当快.它用于 bsxfun创建大小为numel(runLengths)x 的矩阵numel(runLengths),因此可能不适合大的输入大小.

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2

这种方法基于cumsum,是对其他答案中使用的方法的改编.它使用的内存少于方法1.

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