Storing values from nested FOR loop (array only saves last run of results)
19 views (last 30 days)
Show older comments
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.
2 Comments
Kyle
on 15 Jul 2015
tableA = zeros(12,2); for j = 1:4
for m = 1:4
for i = 1:4
answerA = i;
answerB = 10;
tableA(i,:) = [answerA answerB];
end
answerA = m;
answerB = 20;
tableA(m+4,:) = [answerA answerB];
end
answerA = j;
answerB = 30;
tableA(j+8,:) = [answerA answerB];
end
There is probably an easier way, but this works.
Aroosh Amjad
on 13 Feb 2017
any body can give me exact code for storing values in "ForLoop".? i would be very thankful to you
Accepted Answer
Thomas
on 6 Mar 2014
Edited: MathWorks Support Team
on 28 Nov 2018
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)]
2 Comments
More Answers (4)
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... :)
PANKAJ PANDYA
on 3 Oct 2016
Edited: 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
0 Comments
Sawan Kumar Jindal
on 3 Oct 2016
Edited: 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
0 Comments
See Also
Categories
Find more on Matrix Indexing in Help Center and File Exchange
Products
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!