将参数传递给cellfun matlab

Question

你好我有一个char的单元格数组(由下划线分隔),我想转换为double.我是在for循环中完成的,但由于尺寸很大,需要花费很多时间.我想用cellfun,但我不知道如何通过分隔符.

你能帮助我吗？

listofwords = {'02_04_04_52';'02_24_34_02'};
for i = 1 : size(listofwords,1)
    listofwords_double(i,:) = str2double(strsplit(listofwords{i},'_'))./1000;
end

listofwords_double2= cellfun(@strsplit , listofwords);

基准

按照Divakar的要求

>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -46.3398%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -46.4068%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -47.1129%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -46.2882%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -46.2325%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -46.0161%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -46.9728%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -46.4267%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -46.2867%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -46.3031%

Answer 1

你可以使用这样的匿名函数 -

listofwords_double2= cellfun(@(x) strsplit(x,'_') , listofwords,'uni',0)

另一种方法regexp和单线 -

cell2mat(cellfun(@(x) str2double(regexp(x,'_','Split'))./1000 , listofwords,'uni',0))

性能导向解决方案

方法#1

N = 4; %// Edit this to 10 in your actual case
cat_cell = strcat(listofwords,'_');
one_str = [cat_cell{:}];
one_str(end)=[];
sep_cells = regexp(one_str,'_','Split');
out = reshape(str2double(sep_cells),N,[]).'./1000; %//'# desired output

方法#2

对上述解决方案进行基准测试strcat可能会成为瓶颈.要摆脱这种情况,您可以使用cumsum基于该方法的方法.这是下一个 -

N = 4; %// Edit this to 10 in your actual case

lens = cellfun(@numel,listofwords);
tlens = sum(lens);
idx = zeros(1,tlens); %// Edit this to "idx(1,tlens)=0;" for more performance
idx(cumsum(lens(1:end-1))+1)=1;
idx2 = (1:tlens) + cumsum(idx);

one_str(1:max(idx2))='_';
one_str(idx2) = [listofwords{:}];

sep_cells = regexp(one_str,'_','Split');
out = reshape(str2double(sep_cells),N,[]).'./1000; %//'# desired output

方法#3

现在,这个使用sscanf并且看起来非常快.这是代码 -

N = 4; %// Edit this to 10 in your actual case
lens = cellfun(@numel,listofwords);
tlens = sum(lens);
idx(1,tlens)=0;
idx(cumsum(lens(1:end-1))+1)=1;
idx2 = (1:tlens) + cumsum(idx);

one_str(1:max(idx2)+1)='_';
one_str(idx2) = [listofwords{:}];
delim = repmat('%d_',1,N*numel(lens));
out = reshape(sscanf(one_str, delim),N,[])'./1000; %//'# desired output

标杆

按照要求通过@ CST-Link的,这里是比较他的"海怪"的标杆eval反对approach #3.基准测试代码看起来像这样 -

clear all

listofwords = repmat({'02_04_04_52_23_14_54_672_0'},100000,1);
for k = 1:50000
    tic(); elapsed = toc(); %// Warm up tic/toc
end

tic
N = 9; %// Edit this to 10 in your actual case
lens = cellfun(@numel,listofwords);
tlens = sum(lens);
idx(1,tlens)=0;
idx(cumsum(lens(1:end-1))+1)=1;
idx2 = (1:tlens) + cumsum(idx);

one_str(1:max(idx2)+1)='_';
one_str(idx2) = [listofwords{:}];
delim = repmat('%d_',1,N*numel(lens));
out = reshape(sscanf(one_str, delim),N,[])'./1000; %//'# desired output
time1 = toc;
clear out delim one_str idx2 idx tlens lens N

tic
n_numbers = 1+sum(listofwords{1}=='_');
n_words   = numel(listofwords);
listofwords_double = zeros(n_numbers, n_words);
for i = 1:numel(listofwords)
        temp = ['[', listofwords{i}, ']'];
        temp(temp=='_') = ';';
        listofwords_double(:,i) = eval(temp);
end;
listofwords_double = (listofwords_double / 1000).';
time2 = toc;
speedup = ((time1-time2)/time2)*100;
disp(['Speedup with EVAL over NO-LOOP-SSCANF = ' num2str(speedup) '%'])

以下是代码运行几次时的基准测试结果 -

>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = 0.30609%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = 0.012241%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -2.3146%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = 0.33678%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -1.8189%
>> benchmark1
Speedup with EVAL over NO-LOOP-SSCANF = -0.12254%

sscanf在一些积极的加速中,观察结果并观察一些负面加速(在这些情况下表明更好),我的意见是坚持sscanf.