我试图看看我是否可以提高for循环和if条件语句的速度。基本上它会将非重复键值查找到数组中并从另一列中获取值。
如果我运行 100000值需要大约13秒，请参阅下面的代码。有没有办法让这个更有效率？ Ps i使用octave 3.8.1，它与matlab一起使用
%test if lookup statment
clear all, clc, tic, clf;
num_to_test=100000 %amount of numbers to test
a1=(1:1:num_to_test)';
a2=(a1.*num_to_test);
array=[a1,a2]; %array where values are stored
lookupval=(randperm(num_to_test,num_to_test/2)/4)'; %lookup these random values of non repeating integers and floats and get another value
amp=[];
freq=[];
found_array=[];
notfound_array=[];
for ii=1:1:rows(lookupval)
if (find(lookupval(ii)==array(:,1))) %if you find a lookup value in array
%disp('found');
[row,col] = find(lookupval(ii) == array(:,1));
amp=[amp;array(row,2)];
freq=[freq;array(row,1)];
found_array=[freq,amp];
else %add lookup value to another array and make amp value zero
notfound_arraytmp=[lookupval(ii),0];
notfound_array=[notfound_array;notfound_arraytmp];
endif
end
comb_array=[found_array;notfound_array];
sort_comb_array=sortrows(comb_array,1); %sort array by first col incrementing
fprintf('\nfinally Done-elapsed time -%4.4fsec- or -%4.4fmins- or -%4.4fhours-\n',toc,toc/60,toc/3600);
答案 0 :(得分：2)
有几个问题，但主要问题可能是你没有预先分配 - 这样追加：amp=[amp;array(row,2)];在MATLAB中通常很慢。不过，你不需要循环。
让我们从一个简单的数组开始，A：
1 500
2 700
3 900
7 1000
9 800
我们的查找值为[2 6 3 9 7];我们希望输出显示这些查找值，在第一列中排序，第二列是A的第二列中的值（它们存在的位置）或零。
lookup = sort(lookup);
output = zeros(length(lookup),2);
output(:,1) = lookup;
[c a b ] = intersect(A(:,1),lookup);
output(b,2) = A(a,2);
输出结果为：
2 700
3 900
6 0
7 1000
9 800
答案 1 :(得分：2)
方法＃1
使用 ismember -
lookupval = sort(lookupval); %// Do sorting at the start
sort_comb_array = [lookupval zeros(size(lookupval))]; %// Setup output array
[idA,idB] = ismember(array(:,1),lookupval); %// Get matching IDs
sort_comb_array(idB(idA),2) = array(idA,2); %// Index into second column
%// of array and get corresponding values
方法＃2
我也会抛弃我最喜欢的 bsxfun ，但对于100,000这样巨大的数据，其内存效率低下可能会让它变慢 -
lookupval = sort(lookupval);
sort_comb_array = [lookupval zeros(size(lookupval))];
[idA,idB] = find(bsxfun(@eq,array(:,1),lookupval(:).')); %//'# Get matching IDs
sort_comb_array(idB,2) = array(idA,2);
答案 2 :(得分：0)
纯粹从效率的角度来看，我会按如下方式重写for循环：
m = 0; % number of omitted values
n = 0; % number of found values
for ii=1:1:rows(lookupval)
[row,col] = find(lookupval(ii) == array(:,1));
if ~isempty(row) %if you find a lookup value in array
%disp('found');
n=n+1;
amp(n)=array(row,2);
freq(n)=;array(row,1);
found_array=[freq,amp];
else %add lookup value to another array and make amp value zero
m=m+1;
notfound_array(2*m-1:2*m)=[lookupval(ii);0];
endif
end
这可以直接使用其输出来保存find调用，而不是在find返回位置时重新计算它，并以更有效的方式增长数组（如this question所示） ）。
答案 3 :(得分：0)
这是一个测试Divakar建议我看到运行它需要八度音程3.8.1的速度。结果如下以及代码。
1）使用2,000,000的ismember更快但使用更多内存
- 经过时间-0.2306sec-或-0.0038mins-
总计为15000001个元素，使用106000008个字节
2）使用与2,000,000的交叉较慢但使用较少的内存
- 经过时间-0.3057sec-或-0.0051mins-
总计为11749047个元素，使用93992376字节
3）使用带有100,000的bskfun会产生错误：内存不足或尺寸对于八度音程的索引类型而言太大
首次测试结果：
clear all, clc, tic, clf;
num_to_test=2000000 %amount of numbers to test
a1=(1:1:num_to_test)';
a2=(a1.*num_to_test);
array=[a1,a2]; %array where values are stored
lookupval=(randperm(num_to_test,num_to_test/2)/4)'; %lookup these random vaules of intergers and floats and get another value
lookupval = sort(lookupval);
sort_comb_array = [lookupval zeros(size(lookupval))];
[idA1,idB1] = ismember(array(:,1),lookupval);
sort_comb_array(idB1(idA1),2) = array(idA1,2);
fprintf('\nfinally Done-elapsed time -%4.4fsec- or -%4.4fmins- or -%4.4fhours-\n',toc,toc/60,toc/3600);
whos
>>>num_to_test = 2000000
>>>
finally Done-elapsed time -0.2306sec- or -0.0038mins- or -0.0001hours-
>>>Variables in the current scope:
Attr Name Size Bytes Class
==== ==== ==== ===== =====
a1 2000000x1 16000000 double
a2 2000000x1 16000000 double
array 2000000x2 32000000 double
idA1 2000000x1 2000000 logical
idB1 2000000x1 16000000 double
lookupval 1000000x1 8000000 double
num_to_test 1x1 8 double
sort_comb_array 1000000x2 16000000 double
Total is 15000001 elements using 106000008 bytes
========================================================================
第二次测试结果：
clear all, clc, tic, clf;
num_to_test=2000000 %amount of numbers to test
a1=(1:1:num_to_test)';
a2=(a1.*num_to_test);
array=[a1,a2]; %array where values are stored
lookupval=(randperm(num_to_test,num_to_test/2)/4)'; %lookup these random vaules of intergers and floats and get another value
lookupval = sort(lookupval);
output = zeros(length(lookupval),2);
output(:,1) = lookupval;
[c a b ] = intersect(array(:,1),lookupval);
output(b,2) =array(a,2);
fprintf('\nfinally Done-elapsed time -%4.4fsec- or -%4.4fmins- or -%4.4fhours-\n',toc,toc/60,toc/3600);
whos
>>>num_to_test = 2000000
>>>
finally Done-elapsed time -0.3057sec- or -0.0051mins- or -0.0001hours-
>>>Variables in the current scope:
Attr Name Size Bytes Class
==== ==== ==== ===== =====
a 250005x1 2000040 double
a1 2000000x1 16000000 double
a2 2000000x1 16000000 double
array 2000000x2 32000000 double
b 250005x1 2000040 double
c 250005x1 2000040 double
lookupval 1000000x1 8000000 double
num_to_test 1x1 8 double
output 1000000x2 16000000 double
Total is 11750016 elements using 94000128 bytes
=======================================================================