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loop through a column and display the number of elements until the same value occcurs

Asked by Locks on 17 Mar 2013

Hi,

how am I able to loop through a coulmn and get as the output the numbers of rowselement between the first and the next element which is equal to the first element. Here is an example, if this vector:

1

2

3

4

5

7

9

2

1

5

6

8

9

2

4

5

6

now I would like to loop through the column and display the number of elements between the two numbers which are equal.

I tried to use something like this: for i:numel(matrix_call)

   for n:
      if matrix(i,2)==matrix(i+n,2)

but I am not sure how to do that.

The output for the example I mentioned above should be a new vector which displays in which row the next element, which is equal to the one we are looking at, is following, such as:

vector:

9 (row in which the one is occuring for the second time)

8 (in row 8 there is the number "2" for the second time)

NaN (there is no second number "3")

15 (number "4" occcurs in line 15 for the second time)

and so on

1 Comment

Azzi Abdelmalek on 17 Mar 2013

Is it possible that an element can be repeated more then twice?

Locks

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2 Answers

Answer by Azzi Abdelmalek on 17 Mar 2013
Edited by Azzi Abdelmalek on 17 Mar 2013
Accepted answer

This gives the number of element between two consecutive same numbers.

EDIT

A=[1 2 3 4 5 7 9 2 1 1 5 6 8 9 2 4 5 6]'
m=numel(A);
out=nan(1,m);
numb=out;
for k=1:m
  B=A(k:end);
  idx=find(B==B(1),2);
  n=numel(idx);
  if n>1
      out(k)=idx(2)-idx(1)+k;
      numb(k)=numel(unique(B(idx(1)+1:idx(2)-1)));
  end
end
[A out' numb']

18 Comments

Azzi Abdelmalek on 17 Mar 2013

You can post a link where you will upload your data file. Posting an image showing your data is not useful because we can't copy it.

Locks on 17 Mar 2013

how can I post a link/ how am I able to upload the data anywhere so I am able to post a link? I have posted another question including the data, I hope it is clearer what I am up to

Azzi Abdelmalek
Answer by Azzi Abdelmalek on 17 Mar 2013
A=[1 2 3 4 5 7 9 2 1 5 6 8 9 2 4 5 6]'
for k=1:numel(A)
  B=A(k:end)
  idx=find(B==B(1))
  n=numel(idx)
  if n==1
      out(k)=nan
  else
      out(k)=idx(2)-idx(1)+k
  end
end

1 Comment

Locks on 17 Mar 2013

yes it is possible, in fact the same number occcurs more than 100 time and I need a vector to see where the next number which is equal to the one I am looking for is occuring. as a alternative, it woudl also be possible, I think, to create a matrix which shows the numbers of elements between the first time the numbers are equal, between 2nd and 3rd occurence etc.

Azzi Abdelmalek

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