Counting how many times a number occured after a specific number

1 次查看(过去 30 天)
Lets say I have a sequence of numbers ranging from 1 to 4, S = [ 3 2 2 4 3 1 ]. I want a 4x4 matrix which tells me how many times I went from say 3 to 2 or 4 to 1. It would look like this: M = [ 0 0 0 0; 0 1 0 1; 1 1 0 0; 0 0 1 0]. Sounds simple, but I'm out of thoughts. Thanks.

采纳的回答

Kelly Kearney
Kelly Kearney 2020-4-10
This is a good use case for accumarray:
S = [ 3 2 2 4 3 1];
[seq, ~, g] = unique([S(1:end-1)' S(2:end)'], 'rows');
n = accumarray(g, ones(size(g)));
M = zeros(4);
idx = sub2ind(size(M), seq(:,1), seq(:,2));
M(idx) = n;
  3 个评论
Kelly Kearney
Kelly Kearney 2020-4-10
That only works if there is only one instance of each sequence, though:
S = [ 3 2 2 4 3 1 3 2];
[seq, ~, g] = unique([S(1:end-1)' S(2:end)'], 'rows');
% Method 1
n = accumarray(g, 1);
M = zeros(4);
idx = sub2ind(size(M), seq(:,1), seq(:,2));
M(idx) = n
% Method 2
M2 = accumarray(seq, 1, [4 4])
% Check...
assert(isequal(M,M2))
Result:
M =
0 0 1 0
0 1 0 1
1 2 0 0
0 0 1 0
M2 =
0 0 1 0
0 1 0 1
1 1 0 0
0 0 1 0
Error using testsnippets (line 2638)
Assertion failed.
Kelly Kearney
Kelly Kearney 2020-4-10
Though apparently the call to unique is unnecessary... possibly what you were suggesting?
S = [ 3 2 2 4 3 1 3 2];
accumarray([S(1:end-1)' S(2:end)'], 1)
ans =
0 0 1 0
0 1 0 1
1 2 0 0
0 0 1 0
I learned something new!

请先登录,再进行评论。

更多回答(2 个)

Andrei Bobrov
Andrei Bobrov 2020-4-10
accumarray(hankel(S(1:end-1),S(end-1:end)),1)

AB WAHEED LONE
AB WAHEED LONE 2021-12-14
What about m*n matrix
for example , S= [1 2 2 3 1;
1 3 1 3 4;
1 3 4 1 3;
1 3 4 1 3;
1 3 4 1 3];
When i tried to count the same ((1,1),(1,2),(1,3),(1,4),(2,2),(2,3), etc), it just creates the square matrix of max element size.
for example in case of first row X=[1 2 2 3 1], the size of output matrix is 3*3 ,which should have been 4*4.
could you comment on this.

类别

Help CenterFile Exchange 中查找有关 Data Type Conversion 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by