發現在MATLAB

Question

交替順序我有一個像向量：

x = [0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 2 1 2 1 2 1 2 1 2 1 2 1 2 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1] 

我怎麼能搶交替序列的開始和結束索引？ （即1和2）

Answer 1

我認爲這是一個很酷的方法，但數值/數組方法會更快：P您可以使用正則表達式！

% convert into a space-separated string 
% (surely there's a better way than converting to cell array first?) 
str = strcat({num2str(x)}) 
% for some reason all elements are separated by more than one space, convert 
% so they're separated by a single space 
str = regexprep(str,' +',' ') 

% find start & end indices (into str) alternating sequences 
[s e]=regexp(str,'\<([^ ]+) ((?!\1)[^ ]+)(?: \1 \2)+(?: \1)?\>'),'start','end') 
% convert these indices into indices into `x` 
% (at the moment s & e include spaces) 
s = (cell2mat(s)+1)/2 
e = (cell2mat(e)+1)/2 

% run i is at x(s(i):e(i)) 

Answer 2

如果你知道你只有一個序列，它總是[1 2 ... 1 2]，你可以簡單地使用strfind

x = [0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 2 1 2 1 2 1 2 1 2 1 2 1 2 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1]; 

idx = strfind(x, [1 2]); 

start = idx(1); 
end = idx(end)+1; 

如果有可能多次出現，或者如果它並不總是1-2，或者如果序列是不完整的（例如，1 2 1，而不是1 2 1 2），你可以使用diff代替：

dx = diff(x); 
alt = dx(2:end)==-dx(1:end-1) & dx(1:end-1)~=0; 

starts = find(diff(alt)>0) + 1; 
ends = find(diff(alt)<0) + 2;