Pairwise subtraction of rows in the same matrix

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Adedeji
Adedeji 2019 年 5 月 11 日
編集済み: David Goodmanson 2019 年 5 月 12 日
Hello guys...
Please I have this code
How can I subtract each row from all other rows(R) ... like subtracting R1-R1; R1-R2, R1-R3... R1-Rn+1 AND
then R2-R1, R2-R3.... R2-Rn+1 for all the rows
A1 = [...
0 0 1 1 1
0 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
0 0 1 1 1
1 0 1 0 1
0 0 0 0 1
1 1 1 1 1
1 1 1 1 1
0 1 1 1 1
1 0 1 0 1
0 0 1 0 1
1 1 1 1 1
1 1 1 1 1
1 0 1 1 1
0 0 1 1 1
0 1 1 1 1
1 1 1 1 1];
I need this urgently....
Thanks
  6 件のコメント
Adedeji
Adedeji 2019 年 5 月 12 日
It's not a homework but part of an ongoing project of mine
as I explained while replying to David's answer.
I wish to subtract each customers 5 attributes (The Columns) from the other customers'.
The rows are the customers
The columns are their 5 attributes.
Adedeji
Adedeji 2019 年 5 月 12 日
I have tried Diff()
but it's not giving me a pairwise subtraction across all possible combinations of rows there could be.

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madhan ravi
madhan ravi 2019 年 5 月 12 日
編集済み: madhan ravi 2019 年 5 月 12 日
Aa= reshape(a',1,size(a,2),[])- a; % a your matrix , each page of Aa represents each row’s subtraction
A=reshape(permute(Aa,[2,1,3]),size(a,2),[]).'
  7 件のコメント
madhan ravi
madhan ravi 2019 年 5 月 12 日
編集済み: madhan ravi 2019 年 5 月 12 日
Wanted = bsxfun(@minus,reshape(D.',1,size(D,2),[]), D) % the reason was your using version prior to 2016b, that's why when you ask a question you should fill up the field "Release" thereby these kinds of misunderstandings can be avoided
Adedeji
Adedeji 2019 年 5 月 12 日
Exactly!!!!!!
This gives me what I want...
But I also want to sum all the elements for each row and divide it by 5 for all the possible combinations.
So I end up with a 3 by 3 matrix again.
Please how do you advice I do this.

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