# Random Sampling of Repeated Numbers in an Array

조회 수: 20(최근 30일)
Masato Koizumi 2018년 4월 21일
댓글: Jan 2018년 4월 22일
Dear MATLAB Experts,
Hello. I would like to solve this problem. I have a following column vector Q that contains numbers which some are repeating and some do not. I would like to obtain a new column QQ which consists of only the unique values contained in my original Q array. I understand that I could simply use unique(Q).
However, I would like to take one step further.
I actually would like to select randomly from a collection of repeated number. That is, suppose in the following example, I have three 7's in the first, second and third row. I would like to randomly select one 7 from these three rows. Likewise, I have three 8's which I would like to randomly select from row 4,5 and 6.
I have displayed my code. However, I am obtaining a dimensional inconsistency due to random selection of rows. I would greatly appreciate if you could provide me with an advice which I could code this problem efficiently without using excessive number of loops.
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;
while i < length(Q)
QQ(i) = Q(i);
if Ncount(i) > 1
[row col] = find(Q == B(i));
row_select = randsample(row,1);
QQ(i) = QQ(row_select);
end
i = i + 1;
end
Thank you.
Sincerely,
Masato
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
dpb 2018년 4월 21일
편집: dpb 2018년 4월 21일

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.

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### 채택된 답변

John BG 2018년 4월 21일
편집: John BG 2018년 4월 21일

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。

thanks in advance for time and attention

John BG

jgb2012@sky.com

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.

.

##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Jan 2018년 4월 22일
• 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.

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### 추가 답변(3개)

Jan 2018년 4월 21일
편집: Jan 2018년 4월 21일
```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.

##### 댓글 수: 4표시숨기기 이전 댓글 수: 3
Stephen23 2018년 4월 21일
편집: Stephen23 2018년 4월 21일

+1 very thorough!

Note that Jan's submission RunLength also includes the mfile RunLength_M (which does not require compilation). So you can choose between a very efficient mex file or an almost-as-efficient Mfile!

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Bruno Luong 2018년 4월 22일
편집: Bruno Luong 2018년 4월 22일
```p=randperm(length(Q));
[~,I]=unique(Q(p));
RandomSelect=p(I), % Q(RandomSelect) is equal to unique(Q)
```
##### 댓글 수: 1표시숨기기 없음
Masato Koizumi 2018년 4월 22일
~Masato

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dpb 2018년 4월 21일

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
>> ```
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
dpb 2018년 4월 21일
Agreed. Just shows one way to deal with a group; left generalization to OP as "exercise for student" :).

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