counting column bits with the same weight in binary array

1 view (last 30 days)
I have a binary array and I want to count the column with the same weight.
I want to do it with function...m-file
I hope that someone can help me to solve it.
Thanks Henry
  3 Comments
Henry Buck
Henry Buck on 8 Sep 2015
Edited: dpb on 8 Sep 2015
For example:
19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 1 1 0 0
0 1 0 0 0 1 0 0 0 1 1 0 1 0 1 0 1 1 0 1
1 0 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 0 1 1
0 0 1 0 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1
1 1 0 0 1 0 0 0 1 0 1 1 1 1 0 0 0 0 0 1
0 1 1 1 1 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0
0 0 1 1 1 1 1 1 0 1 0 0 0 0 0 1 1 1 1 0
There is 6 col with the same weight (two 1 bits) - col=0,3,6,7,11,15.
There is 8 col with the same weight (three 1 bits) - col=1,9,10,12,13,14,17,18.
There is 5 col with the same weight (four 1 bits) - col=2,4,5,16,19.
dpb
dpb on 8 Sep 2015
>> [0:size(a,2)-1;sum(a)]
ans =
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
2 3 4 2 4 4 2 2 2 3 3 2 3 3 3 2 4 3 3 4
>>
So, turns out not as far off as thought if you count from the left rather than using the header row as the column number. Other than missing one '2' (col 8), seems ok...

Sign in to comment.

Accepted Answer

dpb
dpb on 7 Sep 2015
Edited: dpb on 8 Sep 2015
n=hist(sum(array),0:size(a,1));
N th element of n will contain counts for that number of bits from zero to size(array,1); ie, for all possible totals from no bits set to all.
ADDENDUM
>> [0:size(a,1);hist(sum(a),0:size(a,1))]
ans =
0 1 2 3 4 5 6 7
0 0 7 8 5 0 0 0
>> sum(ans(2,:))==size(a,2) % sanity check...
ans =
1
>>
Total number does equal number of columns in array...
Actually, there are 7,8,5 of wt=2,3,4, respectively.
  3 Comments
dpb
dpb on 8 Sep 2015
If it's a subset of rows, only the array indices to refer to the subset...
r1=3; r2=6;
n=hist(sum(array(r1:r2,:)),0:size(a,1));
Henry Buck
Henry Buck on 9 Sep 2015
Hi dpb, It works perfectly. Thanks for your help, Henry

Sign in to comment.

More Answers (0)

Tags

Community Treasure Hunt

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

Start Hunting!