Nested for loops problem

Hi all,

Current problem: I have a 6 x 2 matrix, which looks like this:

5    0
3    0
8    2
1    2
3    2
2    2

I want to use a nested for loop with if statements to say "if 0 extract the number in the row beside" and "if 2 extract the number beside", so that I have two sets of data extracted from row 1 based on row 2's values.

Any solutions and in depth explanations especially would be great as need to get my head around these mock questions before an exam in two weeks. Thanks,

Jason.

[Walter Roberson]

What would be the expected output for that?

Could you confirm that the 2 at the bottom of column 1 is to be treated with the same meaning as the 2s in column 2 ?

[Jason Melody]

Both columns have different meanings. They are two separate outcomes from different measures.

I want to be able to separate out column one based on column two's scores.

[Geoff Hayes]

Jason - why do you want to use a nested for loop? Is this part of the assignment requirement? If you are going to use a loop (and I don't think that you have to) just iterate over each element in the second column. If it is 0, then update an array for the "zero value" data. If it is 2, then update a different array for the "two value" data.

[Guillaume]

I agree with Geoff. From the vague description of the desired result, I don't see the need for even one for loop and certainly can't see why you'd need nested loops. What would the inner loop iterate over?

If it's some homework problem that explicitly require the use of 2 loops, then we don't solve homework for you (that's called cheating and can get you excluded). We do provide help if you show that you made decent attempt at solving the problem and ask specific questions.

[Walter Roberson]

I am not clear as to what "beside" here means, especially for ""if 0 extract the number in the row beside" as any given row may have up to two rows that are immediately beside the row.

Answer 1

[Guillaume]

Using no loop at all:

M = [5    0
     3    0
     8    2
     1    2
     3    2
     2    2];
[~, ~, subs] = unique(M(:, 2))
result = accumarray(subs, M(:, 1), [], @(v) {v});
celldisp(result)