Trying to normalize a matrix across all element values.

I have the triu of an 8x8 adjacency matrix A shown below. I would LIKE to normallize all the non-zero elements using normalize(A,'range'), for instance with output values between 0 and 1, but NOT column or row-wise - I'd like to normalize across ALL non-zero values. I haven't been able to find options for this. Any help appreciated!

[ 0 54 70 67 18 13 100 18

0 0 89 67 20 37 47 99

0 0 0 38 20 43 49 13

0 0 0 0 26 30 62 27

0 0 0 0 0 11 91 88

0 0 0 0 0 0 17 18

0 0 0 0 0 0 0 4

0 0 0 0 0 0 0 0 ]

Answer 1

[dpb]

M=[ 0    54    70    67    18    13   100    18
    0     0    89    67    20    37    47    99
    0     0     0    38    20    43    49    13
    0     0     0     0    26    30    62    27
    0     0     0     0     0    11    91    88
    0     0     0     0     0     0    17    18
    0     0     0     0     0     0     0     4
    0     0     0     0     0     0     0     0    ];
M(M~=0)=rescale(M(M~=0),0,1)

M = 8×8

0 0.5208 0.6875 0.6562 0.1458 0.0938 1.0000 0.1458 0 0 0.8854 0.6562 0.1667 0.3438 0.4479 0.9896 0 0 0 0.3542 0.1667 0.4062 0.4688 0.0938 0 0 0 0 0.2292 0.2708 0.6042 0.2396 0 0 0 0 0 0.0729 0.9062 0.8750 0 0 0 0 0 0 0.1354 0.1458 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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[jeet-o]

Ah! Excellent and elegant. Thank you!

Follow up Question - if I wanted to do something similar, but scaled to standard deviation (such as normallize(A, 'scale'), is there a similar approach?

[jeet-o]

Belay that...I think you've already answered it - namely:

A(A~=0)= normalize(A(A~=0),'scale')

Thanks again.

Answer 2

[Matt J]

Normalize across only non-zero elements, or across elements only in the upper triangle? If the latter, then,

A=[ 0    54    0    67    18    13   100    18
    0     0    89    67    0    37    47    99
    0     0     0    38    -1    43    49    13
    0     0     0     0    26    30    62    27
    0     0     0     0     0    11    91    88
    0     0     0     0     0     0    17    18
    0     0     0     0     0     0     0     4
    0     0     0     0     0     0     0     0    ];

idx=triu(true(size(A)),1);

A(idx) =  normalize(A(idx),'range')

A = 8×8

0 0.5446 0.0099 0.6733 0.1881 0.1386 1.0000 0.1881 0 0 0.8911 0.6733 0.0099 0.3762 0.4752 0.9901 0 0 0 0.3861 0 0.4356 0.4950 0.1386 0 0 0 0 0.2673 0.3069 0.6238 0.2772 0 0 0 0 0 0.1188 0.9109 0.8812 0 0 0 0 0 0 0.1782 0.1881 0 0 0 0 0 0 0 0.0495 0 0 0 0 0 0 0 0

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[jeet-o]

Ah, this is very useful, too. I was about to ask a related question: In DB's solution, the zero scale then ended up zeroing out some "weights" in the adjacency matrix (which is not good). This solution keeps zero in the source matrix, so the low one (.0495) is still around (good).

My question is, do the 8 zeros change the scaling in any way compared to, say just a single zero? (I'm guessing not, but I'm a stats novice)

[Matt J]

I don't know what is meant by the "8 zeros". The scaling is determined by the min and max of the matrix subset A(idx), indepndently of how many zeros that contains.

[dpb]

"the zero scale then ended up zeroing out some "weights" in the adjacency matrix"

That's the solution you asked for -- "I would LIKE to normallize all the non-zero elements ...with output values between 0 and 1,"

What is supposed to be the minimum value then, if not zero? Specify it as the lower bound.