计数在矩阵的列中运行

Question

我有一个矩阵1s,并-1s用随机穿插0s:

%// create matrix of 1s and -1s

hypwayt = randn(10,5);
hypwayt(hypwayt > 0) =  1;
hypwayt(hypwayt < 0) = -1;

%// create numz random indices at which to insert 0s (pairs of indices may  
%// repeat, so final number of inserted zeros may be < numz)

numz = 15;
a = 1;
b = 10;
r = round((b-a).*rand(numz,1) + a);
s = round((5-1).*rand(numz,1) + a);

for nx = 1:numz
    hypwayt(r(nx),s(nx)) = 0
end

输入:

hypwayt =

-1     1     1     1     1
 1    -1     1     1     1
 1    -1     1     0     0
-1     1     0    -1     1
 1    -1     0     0     0
-1     1    -1    -1    -1
 1     1     0     1    -1
 0     1    -1     1    -1
-1     0     1     1     0
 1    -1     0    -1    -1

我想计算nonzero一列中重复元素的次数,以产生这样的东西:

基本思路是(由@rayryeng提供)对于每一列,每次你点击一个唯一的数字,你开始递增一个累积的运行计数器,并且每次你达到与前一个相同的数字时它会递增.一旦你点击一个新的数字,它就会被重置为1,除了你点击0的情况,所以它是0

预期产出:

hypwayt_runs =

 1     1     1     1     1
 1     1     2     2     2
 2     2     3     0     0
 1     1     0     1     1
 1     1     0     0     0
 1     1     1     1     1
 1     2     0     1     2
 0     3     1     2     3
 1     0     1     3     0
 1     1     0     1     1

实现这一目标的最简洁方法是什么？

Answer 1

我想应该有更好的方法，但这应该有效

使用cumsum，diff，accumarray＆bsxfun

%// doing the 'diff' along default dim to get the adjacent equality
out = [ones(1,size(A,2));diff(A)];

%// Putting all other elements other than zero to 1 
out(find(out)) = 1;

%// getting all the indexes of 0 elements
ind = find(out == 0);

%// doing 'diff' on indices to find adjacent indices
out1 = [0;diff(ind)];

%// Putting all those elements which are 1 to zero and rest to 1
out1 = 0.*(out1 == 1) + out1 ~= 1;

%// counting each unique group's number of elements
out1 = accumarray(cumsum(out1),1);

%// Creating a mask for next operation
mask = bsxfun(@le, (1:max(out1)).',out1.');

%// Doing colon operation from 2 to maxsize
out1 = bsxfun(@times,mask,(2:size(mask,1)+1).');    %'

%// Assign the values from the out1 to corresponding indices of out
out(ind) = out1(mask);

%// finally replace all elements of A which were zero to zero
out(A==0) = 0

结果：

输入：

>> A

A =

-1     1     1     1     1
 1    -1     1     1     1
 1    -1     1     0     0
-1     1     0    -1     1
 1    -1     0     0     0
-1     1    -1    -1    -1
 1     1     0     1    -1
 0     1    -1     1    -1
-1     0     1     1     0
 1    -1     0    -1    -1

输出：

>> out

out =

 1     1     1     1     1
 1     1     2     2     2
 2     2     3     0     0
 1     1     0     1     1
 1     1     0     0     0
 1     1     1     1     1
 1     2     0     1     2
 0     3     1     2     3
 1     0     1     3     0
 1     1     0     1     1