Asked by adam
on 6 Mar 2014

Hi guys, have tried searching but can't find anything to help, maybe my problem is too simple! lol Anyway, I'm running a nested FOR loop, but the array I save my results to only keeps the last "run" of results. Can someone please help me to store/concatenate the results in a single array? EG the resultant array should have 12 rows, not 4. with column one going 1,2,3,4,1,2,3,4,1,2,3,4 and column two going 10,10,10,10,20,20,20,20,30,30,30,30.

Cheers in advance! (first line currently commented out as the last 8 rows stay zero at the minute)

%tableA = zeros(12,2);

for i=1:4

for j=1:3

answerA=i*1

answerB=j*10

tableA(i,:)=[answerA answerB]

end

end

P.S. this isn't the real code I'm using, but a dumbed down version just so I can get the correct syntax to use to apply to the bigger problem! Cheers.

Answer by Thomas
on 6 Mar 2014

Edited by MathWorks Support Team
on 28 Nov 2018

Accepted Answer

for i=1:4

for j=1:3

answerA(i,j)=i*1;

answerB(i,j)=j*10;

% tableA(i,:)=[answerA answerB]

end

end

table=[reshape(answerA,[],1) reshape(answerB,[],1)]

adam
on 6 Mar 2014

thanks mate, works a treat!

Adithya Prabhu
on 22 Dec 2017

can you please explain how this works?

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Answer by dpb
on 6 Mar 2014

As this is clearly early learning, let's go with a hint-based approach rather than just throwing out an answer -- might make it more beneficial to discover the answer sorta' on your own...

%tableA = zeros(12,2);

for i=1:4

for j=1:3

answerA=i*1

answerB=j*10

tableA(i,:)=[answerA answerB]

...

1) preallocating is a_good_thing (tm) so bring that back out of retirement but -- first, for ease use a variable for the upper limit on the two loops so you can modify them and easily compute the size of the array needed--

nr=4; % rows

nc=3; % columns

table=zeros(nr,nc); % preallocate

2) NB: that answerA is invariant with j so move it outside the inner loop--no sense doing stuff over and over that is the same answer every time--

for i=1:nr

answerA=i*1; % of course the '*1' isn't needed but it's tutorial, so ok...

3) Now to the nub of your ? --

for j=1:nc

AnswerB=j*10; % NB: this doesn't give what you say you want, either...

table(?,?) = ???

Walk thru the steps and follow what i is each pass -- it should be apparent what the next row index should be; what computation would generate that?

After that, then look at how to get the indices for the two values in the proper columns -- it's a similar idea. Or, of course, instead of building a vector to store two at a time, store the individual i,j elements in their correct location when generate them is more direct.

4) Rethink the whole process and see if you cannot actually compute the whole thing in a vectorized form and eschew the loops entirely. That's "the Matlab way".

Chew on that a while and then come back... :)

Answer by PANKAJ PANDYA
on 3 Oct 2016

Edited by PANKAJ PANDYA
on 3 Oct 2016

if true

% i think you would like it

% i liked the question. it really checked my aptitude

table=zeros(12,2);

for k=0:4:8

for i=1:4

for j=1:2

if mod(j,2)==0

table(i+k,j)=i;

else

table(i+k,j)=10*(k/4+1);

end

end

end

end

end

it works fine

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Answer by adam
on 6 Mar 2014

Hi guys,

the first answer worked great, however I can't use a step size less than 1 (as indices can't be less than 1 right?)

Do I need a different method in order to achieve what the below code is trying to do?

(I just get "??? Attempted to access answerA(1.5,1); index must be a positive integer or logical." error)

for i=1:0.5:4

for j=1:3

answerA(i,j)=i*1;

answerB(i,j)=j*10;

end

end

table=[reshape(answerA,[],1) reshape(answerB,[],1)]

Moving my answerA outside the j loop didn't work either, but that's a minor thing.

Cheers

adam
on 6 Mar 2014

sorry guys, think I've sorted it. Just added an index value as below:

index=0;

for i=1:0.5:4

for j=1:3

index=index+1;

answerA(index)=i*1;

answerB(index)=j*10;

end

end

table=[reshape(answerA,[],1) reshape(answerB,[],1)]

Any thoughts or comments are still greatly appreciated though! Cheers Adam

dpb
on 6 Mar 2014

BINGO!!! on that lesson... :)

Note however you still have answerA computed 24 times when it's only needed to be done 8 by not moving the j-invariant portion of the code outside the loop on j

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Answer by Sawan Kumar Jindal
on 3 Oct 2016

Edited by Sawan Kumar Jindal
on 3 Oct 2016

The first thing to do is assign zero value to each location of the matrix and then, use for loops to store the value. You have not initialised the memory for all the locations.

Code ex:

for i=1:4

for j=1:3

answerA(i,j)=i*1;

answerB(i,j)=j*10;

tableA(i,:)=[answerA answerB]

end

end

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## Kyle (view profile)

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## Aroosh Amjad (view profile)

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