How to: Count length of identical consecutive elements in an array

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Jay Alan
Jay Alan 2018-11-28
评论： Jay Alan 2018-11-28
An earlier answer provided to counting identical consecutive elements in an array is as follows (tabulate command added by me)
X = [0 0 0 0 1 1 1 0 1 1 1 1 0 0]; % Sample data
d= [true, diff(X) ~= 0, true]; % TRUE if values change
n = diff(find(d)); % Number of repetitions
tabulate(n)
This code combines the consecutive element lengths of different elements (0s and 1s) in the answer
--> However, am trying to determine length of consecutive elements of "each element type" (not combine consecutive lengths of different element values)
Output content would be something like:
Array entry 0: consecutive length 4, # of occurences=1
Array entry 0: consecutive length 2, # of occurences=1
Array entry 0: consecutive length 1, # of occurences=1
Array entry 1: consecutive length 4, # of occurences=1
Array entry 1, consecutive length 3, # of occurences=1
  1 个评论
Jay Alan
Jay Alan 2018-11-28
编辑：Jay Alan 2018-11-28
The following seems to work; any elegant solutions would be greatly appreciated
rr=[ NaN
NaN
NaN
0.0382
-0.0211
0.0252
0.0165
-0.0417
-0.0128
0.0667
-0.0040
-0.0329
-0.0340
0.0172
-0.0085
0.0128
0.0210
-0.0295
-0.0260
0.0260]
rr1=rr>0;
b1 = diff([0 rr1' 0] ==1);
c1 = find(b1==-1) - find(b1==1);
tabulate(c1)
%--
rr0=rr<0;
b0 = diff([0 rr0' 0] ==1);
c0 = find(b0==-1) - find(b0==1);
tabulate(c0)

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回答（1 个）

KSSV
KSSV 2018-11-28
A = [0 0 0 0 1 1 1 0 1 1 1 1 0 0]; % Sample data
ii = zeros(size(A));
jj = A > 0;
ii(strfind([0,jj(:)'],[0 1])) = 1;
idx = cumsum(ii).*jj;
out = accumarray( idx(jj)',A(jj)',[],@(x){x'});
Reference: https://in.mathworks.com/matlabcentral/answers/104614-grouping-continuous-nonzero-in-a-row-vector
  1 个评论
Jay Alan
Jay Alan 2018-11-28
Thank you very much for the suggestion; am unable to get outputs with your code and will need to analyze further (with my sincere apologies for the delay). Thanks once again.

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