Cycle counting from 0
显示 更早的评论
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 个评论
Image Analyst
2019-12-11
编辑：Image Analyst
2019-12-11
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
2019-12-12
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
采纳的回答
>> 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 个评论
Stephen23
2019-12-14
"I have another question:"
What have you tried so far?
Stephen23
2019-12-14
Try using diff.
stefan
2019-12-14
"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
2019-12-14
更多回答（1 个）
Akira Agata
2019-12-12
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 个评论
Akira Agata
2019-12-12
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).
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