为了巧妙地使用线性索引或accumarray
,我有时觉得需要根据run-length encoding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。
为了使这个公平比较,我想为该功能设置一些规范:
values
,则输出应该根据这些值,否则只是值1:length(runLengths)
。runLengths
values
是一个单元格数组。runLengths
简而言之:该函数应该等同于以下代码:
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
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})
及其输出:
>> 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]
答案 0 :(得分:6)
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。第一个图描绘了矢量runLengths
长度增长的运行时间,其中条目均匀分布,最大长度为200. modification of gnovice's solution是最快的, Divakar &#39 ; s解决方案位居第二。
第二个图使用几乎相同的测试数据,但它包括长度2000
的初始运行。这主要影响两个bsxfun
解决方案,而其他解决方案的表现非常相似。
测试表明modification gnovice 的original answer将是最高效的。
如果您想自己进行速度比较,请参阅上面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
答案 1 :(得分:6)
从R2015a开始,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,这是一个快速的选择:
repelem
:我感觉 gnovice 的方法可以通过使用比预处理掩码更好的zero-runLength修复来改进。
所以我给了accumarray
一个镜头。这似乎为解决方案带来了又一次提升:
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
答案 2 :(得分:5)
这应该相当快。它用
bsxfun
要创建大小为numel(runLengths)
x numel(runLengths)
的矩阵,因此它可能不适合大量输入。
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
这种方法基于cumsum
,是对this other answer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
答案 3 :(得分:4)
此处介绍的解决方案基本上分两步执行run-length decoding
-
values
复制到最大runLengths
。bsxfun
的屏蔽功能从每列中选择相应的runlengths
。功能代码中的其他内容是处理输入和输出大小以满足问题中设置的要求。
下面列出的功能代码是one of my previous answers to a similar problem的“已清理”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;
下面列出的代码将是混合代码(cumsum
+ replicate_bsxfunmask
),并且当您拥有大量异常值或非常大的异常值时,它们将非常适合。另外,为了简单起见,目前这只适用于数字数组。这是实施 -
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;