将参数传递给CellFun Matlab [英] Passing argument to cellfun matlab

问题描述

您好，我有一个char单元格数组(由下划线分隔)，我想将其转换为double. 我在for循环中执行此操作，但是由于尺寸很大，因此需要花费很多时间. 我想使用cellfun，但是我不知道如何传递定界符.

Hello I have a cell array of char (separated by underscore) that I would like to convert to double. I do it in a for loop, but since the dimensions are very big, it takes a lot of time. I would like to use cellfun, but I don't know how to pass the delimiter.

你能帮我吗?

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

As requested by Divakar

>> 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%

推荐答案

您可以使用这样的匿名函数-

You can use anonymous function like this -

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

使用regexp和单线的另一种方法-

Another approach with regexp and a one-liner -

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

---

面向性能的解决方案

方法1

---

Performance oriented solutions

Approach #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表明，strcat可能成为瓶颈.要消除这一点，您可以对该部分使用基于cumsum的方法.接下来列出-

Benchmarking the above solution suggests strcat could prove to be the bottleneck. To get rid of that you can use a cumsum based approach for that part. This is listed next -

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，这是使用sscanf的，并且看起来非常快.这是代码-

Now, this one uses sscanf and appears to be really fast. Here's the code -

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

---

基准化

按照@ CST-Link的要求，这是将他的克拉肯" eval与approach #3进行比较的基准.基准测试代码看起来像这样-

---

Benchmarking

As requested by @CST-Link, here's the benchmark comparing his "Kraken" eval against approach #3. The benchmarking code would look something like this -

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) '%'])

这是代码运行几次后的基准测试结果-

And here are the benchmark results when the code is run for a few number of times -

>> 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%

查看结果并观察到一些积极的提速中的一些负面提速(表明在这些情况下，sscanf会更好)，我的观点是坚持使用sscanf.

Looking at the results and observing some negative speedups (indicating sscanf to be better in those cases) among some positive speedups, my opinion would be to stick with sscanf.

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