# Random Sampling of Repeated Numbers in an Array

##### 2 Comments

Agree with Jan if you only save the value there's no difference; appears that what your problem is that it matters which one is selected as there's something unique about position, not just magnitude. If that isn't so, then there is no difference and may as well just use the results of `unique` as one seven itself is as good as another; only if **which** seven makes a difference.

### Accepted Answer

Hi Masato

this is John BG jgb2012@sky.com

I have done the following corrections to your code

1.- the mismatch error was cause by the i=i+1 in the while loop attempting to index +1 beyond the length of the vector.

a for loops suffices.

2.- there's no need for the variable row_select

3.- there's no need for the find producing row_select

4.- return 0, not 1, in QQ, for all those single values of Q that do not imply random selection.

Setting those to 1 may be misleading, because 1 is a possible index for a sub-selection within the partial ranges.

Q = [7 7 7 8 8 8 10 18 27 42 65 49 54 65 78 78 78 82 87 98 98]; B = unique(Q) Ncount = histc(Q, B) i = 1; for i=1:1:length(B) % < length(Q) QQ(i) = 0 % Q(i); % or QQ(i) = 1, but 1 may be index of random selection, thus potentially confusing if Ncount(i) > 1 % [row col] = find(Q == B(i)); % no need for variable row_select QQ(i) = randsample(Ncount(i),1); end % i = i + 1; end

この回答が役に立つと判断した場合は、回答として回答をマークするようにしてください。

他の読者には、この答えは、親指をクリックして検討してください便利な投票リンクを見つける。

時間と注意を事前に感謝

If you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance for time and attention

John BG

additional comments:

1.- as usual, and as a compliment, Jan Simon is on the top of his game always providing extremely useful insight and solutions to all questions he contributes to,

but it my opinion for this particular answer there's no need for any additional functions like runlength.m

https://uk.mathworks.com/matlabcentral/fileexchange/241-runlength-m?s_tid=srchtitle

or RunLength.m

https://uk.mathworks.com/matlabcentral/fileexchange/41813-runlength?s_tid=srchtitle

many people do not even have a complier installed, which fills up the screen with all the checks and the suggestion to install one, or to download something from a website. Again, I find RunLength a powerful function, but in the context of generating random selection of sub-sections, there's no need for such advanced function.

.

##### 2 Comments

- Does this output match the question?

QQ = [2, 1, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 2]

- There is no need for
`i = 1`before the loop`for i=1:1:length(B)`. - A proper pre-allocation of QQ would accelerate the code.
- The FileExchange submission RunLength contains the M-file RunLength_M, so you do not need a C-compiler.

### More Answers (3)

B = unique(Q); while i < length(Q) [row col] = find(Q == B(i)); i = i + 1; end

This cannot work, because `B` is shorter than `Q`, when `Q` is not unique. `i` must be `<= length(B)`.

I do not understand the purpose of the code. All elements with the same value are identical, so it does not matter which e.g. 7 you select.

[EDITED] Another solution using FEX: RunLength:

Q = [7 7 7 8 8 8 10 18 27 42 65 49 54 65 78 78 78 82 87 98 98 7 7]'; [B, N, Index] = RunLength(Q); Select = floor(Index + rand(size(N)) .* N) Value = Q(Select)

This considers the 2 different blocks of 7s separately. But if Q is sorted, this might be efficient also.

##### 4 Comments

p=randperm(length(Q)); [~,I]=unique(Q(p)); RandomSelect=p(I), % Q(RandomSelect) is equal to unique(Q)

On the assumption made in above comment...

>> [B,ia]=unique(Q); >> isMult3=(NCount==3); >> arrayfun(@(x) randperm(x,1),Ncount(isMult3)) % random index into matching groups ans = 1 3 2 >>

Now, fix up to get the index to the original location --

>> ix3=arrayfun(@(x) randperm(x,1),Ncount(isMult))+ia(find(Ncount==3))-1 ix3 = 1 5 15 >> Q(ix3) Q = 7 8 78 >>

### See Also

### Categories

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!