counting column bits with the same weight in binary array

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Henry Buck
Henry Buck el 7 de Sept. de 2015
Comentada: Henry Buck el 9 de Sept. de 2015
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 comentarios
Henry Buck
Henry Buck el 8 de Sept. de 2015
Editada: dpb el 8 de Sept. de 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 el 8 de Sept. de 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...

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dpb
dpb el 7 de Sept. de 2015
Editada: dpb el 8 de Sept. de 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 comentarios
dpb
dpb el 8 de Sept. de 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 el 9 de Sept. de 2015
Hi dpb, It works perfectly. Thanks for your help, Henry

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