Question

我有一个像：

这样的矢量

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]

如何获取交替序列的开始和结束索引？（即1和2）

Answer 1

我认为这是一个很酷的方法，虽然数值/数组方法会更快：P你可以使用正则表达式！

% 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个序列，并且它总是[1 2 ... 1 2]，你只需使用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;

如果可能有多次出现，或者它不总是1-2，或者序列不完整（例如1 2 1，而不是1 2 1 2），则可以使用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;

在MATLAB中查找交替序列

2 个答案: