Cycle counting from 0

2 ビュー (過去 30 日間)
stefan
stefan 2019 年 12 月 11 日
コメント済み: stefan 2019 年 12 月 14 日
I have a matrix [ 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0] for exemple
I'd like to have an outing:
2 x 4 "0" in succession
2 x 2 "0" in succession
1 x 3 "0" in succession
How to program it
Thank you.
  2 件のコメント
Image Analyst
Image Analyst 2019 年 12 月 11 日
編集済み: Image Analyst 2019 年 12 月 11 日
I have no idea what you want. What is an "outing"? Do you mean output? If so, what is the output array you want? I can't tell from your word descriptions. Type it out in numbers like you did for the input matrix.
Stephen23
Stephen23 2019 年 12 月 12 日
stefan's "Answer" moved here:
With this matrix : [ 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 0 ]
I would have liked to have this matrix :
|5 2|
|6 1|
|4 1|
With the 1st column the number of successive 0
With the 2nd column the number of the sequence
thanks

サインインしてコメントする。

採用された回答

Stephen23
Stephen23 2019 年 12 月 12 日
編集済み: Stephen23 2019 年 12 月 12 日
>> M = [0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,1,1,1,1,1,0,0,0,0,0];
>> D = diff([false;M(:)==0;false]);
>> L = find(D<0) - find(D>0);
>> [U,~,X] = unique(L,'stable');
>> V = histc(X,1:max(X));
>> Z = [U(:),V(:)]
Z =
5 2
6 1
4 1
  7 件のコメント
Stephen23
Stephen23 2019 年 12 月 14 日
編集済み: Stephen23 2019 年 12 月 14 日
"I have a vector that can have values from 1 to 15. I would like to know the number of times for all the possible changes: for example:"
from 1 to 2
from 2 to 11
from 14 to 3
....
None of your provided data includes the values 11 or 14, so you are asking me to help you on an unknown algorithm with unknown data. Sorry, but my magic crystal ball is not working today.
"from 2 to 3 and from 3 to 4, I don't have the couples (1,2) and (2,3)."
Well, that makes two of us, because I don't have them either! Nor do you explain what they are, or how they relate to 1, 2, 11, 14, and 3 from your previous comment, or the binary data of your original question.
One clear explanation of the method/algorithm (complete with input and output example data) would actually mean that you get the help that you are looking.
stefan
stefan 2019 年 12 月 14 日
it's a new question.
For example with this matrix:[1 1 1 1 12 12 2 3 2 3 4 5 2 1 10 13 13 14 14 14 14 ]
in output, I would have liked to have as an output vector:
1 12 1
12 2 1
2 3 2
3 1 1
3 4 1
4 5 1
5 2 1
2 1 1
1 10 1
10 13 1
13 14 1
with :
1st column: 1st number
2nd column: 2nd number
2nd column: number of times we have the couple (1st number, 2nd number)

サインインしてコメントする。

その他の回答 (1 件)

Akira Agata
Akira Agata 2019 年 12 月 12 日
How about the following?
x = [0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 0];
str = num2str(x')';
c = regexp(str,'0+','match');
len = cellfun(@numel,c);
output = [(1:max(len))', accumarray(len',1)];
The result becomes like this:
>> output
output =
1 0
2 0
3 0
4 1
5 2
6 1
  2 件のコメント
Stephen23
Stephen23 2019 年 12 月 12 日
stefan's "Answer" moved here:
that's not what I expect:
if I look at the matrix:
there are 2 sequences with 5 consecutive zeros
there is 1 sequence with 6 consecutive zeros
and there are 1 sequences with 4 consecutive zeros
the output matrix is:
5 2
6 1
4 1
Akira Agata
Akira Agata 2019 年 12 月 12 日
You mean, the order is also important?
My output is same as what you expected, except the order of rows and including 0 sequence cases (such as 1~3 consecutive zeros).

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeCreating and Concatenating Matrices についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by