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counting column bits with the same weight in binary array

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Henry Buck
Henry Buck 2015년 9월 7일
댓글: Henry Buck 2015년 9월 9일
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
  댓글 수: 3
Henry Buck
Henry Buck 2015년 9월 8일
편집: dpb 2015년 9월 8일
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 2015년 9월 8일
>> [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...

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dpb
dpb 2015년 9월 7일
편집: dpb 2015년 9월 8일
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.
  댓글 수: 3
dpb
dpb 2015년 9월 8일
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 2015년 9월 9일
Hi dpb, It works perfectly. Thanks for your help, Henry

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