I want to run my script more speed. It spent two hours

ar = ['a','a','a','a','a','a'] % this matrix have 3920 rows
r = [1          2                 3                 4                 6                 7                 8                 11               14               17               19               22                    23               24               25] %this matrix have 162 rows
w = [1  2       3       4       5       6       7       10      11      14      17      18        22      23      24      1       2       3       4       5       7       8       9        10      11      15      16      17      23      24      2       3       4       9        10      11      12      13      16      17      18      19      20      23      25        2       6       7       8       9       10      12      14      16      17      19        20      22      23      25] %this matrix have 10000 rows
x1 = 1;
arlin1 = 1;
arcol1 = 1;
rlin1 = 1;
wlin1 = 1;
nlin1 = 1;
ncol1 = 1;
ncol2 = 3921;
v = 0;
v1 = 1;
ctr1 = 1;

for i3 = 1:10000 
    for i2= 1:3920  
        for i1 = 1:27 
            while x1 < 7
                if ar(arlin1, arcol1) == "a"
                    d = ismember(r(rlin1,:), w(wlin1,1:15));
                    sum(d);
                    if sum(d) > 10
                        v = v + 1;
                        break
                    end
                elseif ar(arlin1, arcol1) == "b"
                    d = ismember(r(rlin1,:), w(wlin1,16:30));
                    sum(d);
                    if sum(d) > 10
                        v = v + 1;
                        break
                    end
                elseif ar(arlin1, arcol1) == "c"
                    d = ismember(r(rlin1,:), w(wlin1,31:45));
                    sum(d);
                    if sum(d) > 10
                        v = v + 1;
                        break
                    end
                elseif ar(arlin1, arcol1) == "d"
                    d = ismember(r(rlin1,:), w(wlin1,46:60));
                    sum(d);
                    if sum(d) > 10
                        v = v + 1;
                        break
                    end                
                end  
                arcol1 = arcol1 + 1;
                rlin1 = rlin1 + 1;
                x1 = x1 + 1;
            end 
            x1 = 1;
            arcol1 = 1;
            ctr1 = ctr1 + 6;
            rlin1 = ctr1;
        end 
        n(nlin1,ncol1) = v;
        v = 0;
        ncol1 = ncol1 + 1;
        arlin1 = arlin1 + 1;
        ctr1 = 1;
        rlin1 = ctr1;
    end 
    arlin1 = 1;
    wlin1 = wlin1 + 1;
    nlin1 = nlin1 + 1;
    ncol1 = 1;
end 
nlin1 = 1;
while v1 < 10001
    y = max(n(nlin1,:));  
    n(nlin1, ncol2) = y;   
    nlin1 = nlin1 + 1;
    v1 = v1 + 1;
end

[Walter Roberson]

                    sum(d);

Those calls are useless. They calculate the sum of d and store the result in the special variable "ans". You do not do anything with "ans", so you might as well not call sum(d)

You call sum(d) in the very next if statement, so the seperate line calling sum(d) is just wasted.

[Walter Roberson]

ar = ['a','a','a','a','a','a'] 

That is a char array

if ar(arlin1, arcol1) == "a"

That indexes the char array to get a single char out, and then compares the char to "a" which is a string() object. That involves converting the char to a string() object and then doing the comparison. That is a bit of a waste of time, when you could instead be doing

    if ar(arlin1, arcol1) == 'a'

which is faster for comparing char.

[Walter Roberson]

            while x1 < 7

I do not understand why you implemented that as a while instead of a for x1 = 1:7

It is completely valid to have a break inside of a for

[AIRTON]

Hi Walter

Thanks for your return.

about ar = ['a','a','a','a','a','a'] it changes between letters a, b, c, d; concern for i2= 1:3920; see it below

ar = ['a','a','a','a','a','a'

'a','a','a','a','a','b'

'a','a','a','a','a','c'

'a','a','a','a','a','d'

'a','a','a','a','b','a'

'a','a','a','a','b','b'

'a','a','a','a','b','c'

'a','a','a','a','b','d'

.

.

.

As to the while x1 < 7

it circulates the 162 rows of the variable r which changes too. See it below

r = [1 2 3 4 6 7 8 11 14 17 19 22 23 24 25

1 2 3 4 7 12 13 15 17 18 19 21 22 23 25

1 3 5 8 9 13 14 15 16 17 21 22 23 24 25

1 3 5 9 12 13 14 15 16 17 19 20 21 22 25

3 5 6 9 10 12 13 14 16 17 18 19 20 21 24

.

.

.

If you can help improve my code.

thanks

[Steven Lord]

Can you explain in words not code what you intend this code to do? Assume that we have no knowledge of the problem you're trying to solve (since we don't.) Knowing what you want to do may help us recommend other functions to help perform that task more efficiently.

[Stephen23]

"it changes between letters a, b, c, d"

Yet is a rather indirect way of defining character vectors. Simpler:

ar = ['aaaaaa';'aaaaab';'aaaaac';'aaaaad';'aaaaba';'aaaabb';'aaaabc';'aaaabd';...]

Your code seems to boil down to summing based on some conditions: please describe those conditions in words (not in code).

[dpb]

@Walter Roberson ..."I do not understand why you implemented that as a while instead of a for x1 = 1:7"

Probably because the only example in the documentation for break illustrates its use in a while loop, I'd guess...

Other than the loop would be for x1=1:6, not 1:7, the result would be the same, indeed.

I'd probably structure the multiple if...else if as a switch, case, otherwise block for legibility and try to generalize the code rather than having the magic constants built into the code for the indexing expressions.

But, to actually try to reduce run time significantly, as others have said, having a verbal problem description to work from rather than trying to guess from code would be the better route...

[dpb]

...
d = ismember(r(rlin1,:), w(wlin1,1:15));
if sum(d) > 10
  ...

Another minor efficiency would be to replace sum(d) with nnz(d) since the sum isn't actually of interest -- and the number of nonzero entries equals the sum, anyways...

The final piece

nlin1 = 1;
while v1 < 10001
    y = max(n(nlin1,:));  
    n(nlin1, ncol2) = y;   
    nlin1 = nlin1 + 1;
    v1 = v1 + 1;
end

could be vectorized to

y=max(n,[],2);  % max of each row in n  
n(:,ncol2)=y;   % save

None of these is likely to make more than trivial changes in run time; a nested loop of

for i3 = 1:10000 
  for i2= 1:3920  
    for i1 = 1:27 
      while x1 < 7
        ...

is

>> 10000*3920*27*6
ans =
   6.3504e+09
>> 

iterations which is going to take a while...

[AIRTON]

Hi Walter,

Greets

As I said, my code have slowed two hours, but

with yours notes I have perfected it and now it runs at one hour and 17 minutes.

I thank you Very much!

Please, see like it is.

ar = 3920 arrangements
r = 162 combinations
w = 10000 combinations

n = zeros(10000,3921);

arlin1 = 1;
arcol1 = 1;
rlin1 = 1;
wlin1 = 1;
nlin1 = 1;
ncol1 = 1;
ncol2 = 3921;
v = 0;
ctr1 = 1;

for i3 = 1:10000
    for i2= 1:3920  
        for i1 = 1:27 
            for i0 = 1:6
                if ar(arlin1, arcol1) == 1
                    d = ismembc(r(rlin1,:), w(wlin1,1:15));                                                         
                elseif ar(arlin1, arcol1) == 2
                    d = ismembc(r(rlin1,:), w(wlin1,16:30));
                elseif ar(arlin1, arcol1) == 3
                    d = ismembc(r(rlin1,:), w(wlin1,31:45));
                elseif ar(arlin1, arcol1) == 4
                    d = ismembc(r(rlin1,:), w(wlin1,46:60));     
                end  
                sum(d);
                if sum(d) > 10
                    v = v + 1;
                    break
                end
                arcol1 = arcol1 + 1;
                rlin1 = rlin1 + 1;
            end   
            arcol1 = 1;
            ctr1 = ctr1 + 6;
            rlin1 = ctr1;
        end 
        n(nlin1,ncol1) = v;
        v = 0;
        ncol1 = ncol1 + 1;
        arlin1 = arlin1 + 1;
        ctr1 = 1;
        rlin1 = ctr1;
    end 
    arlin1 = 1;
    wlin1 = wlin1 + 1;
    nlin1 = nlin1 + 1;
    ncol1 = 1;
end  
y=max(n,[],2);

Let me tell him my two principal goals:

• To come to the values of “y” (last row of the code – “y=max(n,[],2);”)
• To run it in the Lesser time possible.

I attach three zip files.

If can you help me?

I thank you Very much.

[dpb]

Well, we understand the end result is the y array; the problem is we don't know what is the underlying problem definition/description other than trying to reverse engineer your code. That is a nonproductive way to approach the problem.

What is needed is the problem to be solved verbally described as basic functional requirements. From such a description one can devise an algorithm to solve the problem as did your brute-force iterative solution above; the issue is that as you've seen, trying to optimize code may produce somewhat faster results but real improvements in speed (like order of magnitude) will have to come from better algorithms. Better algorithms will only come from having a statement of the problem.

That said, your code above still has the totally superfuous sum(d); line which is doing nothing but wasting time, particularly since you're doing it twice...

  ...
  d = ismembc(r(rlin1,:), w(wlin1,46:60));
end
sum(d);
if sum(d) > 10
  v = v + 1;
  break
end

should be

  ...
  d = ismembc(r(rlin1,:), w(wlin1,46:60));
end
%sum(d);  % wasted operation; perhaps useful for debugging, but a total waste in production code
if nnz(d)>10  % just count logicals, no need to add...
  v = v + 1;
  break
end
...