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" bsxfun(@rdivide..." Explain Use in Code

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Katarina Vuckovic
Katarina Vuckovic 4 Feb 2020
편집: Adam 4 Feb 2020

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darova
darova 4 Feb 2020
Please provide sample code

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Adam
Adam 4 Feb 2020
편집: Adam 4 Feb 2020
rgbh_e = bsxfun(@rdivide, rgbh_e, rgbh_e(4,:));
divides rgbh_e by by the 4th row of rgbh_e, on a per row basis - i.e. each element of each row is divided by the element in the same column of the 4th row.
The best way to understand this is to create a small example and test it on command line to see what it is doing. I don't know what rgbh_e is, but for example:
>> a = rand( 4, 5 )
a =
Columns 1 through 4
0.814723686393179 0.63235924622541 0.957506835434298 0.957166948242946
0.905791937075619 0.0975404049994095 0.964888535199277 0.485375648722841
0.126986816293506 0.278498218867048 0.157613081677548 0.8002804688888
0.913375856139019 0.546881519204984 0.970592781760616 0.141886338627215
Column 5
0.421761282626275
0.915735525189067
0.792207329559554
0.959492426392903
>>
>>
>>
>> bsxfun( @rdivide, a, a(4,:) )
ans =
Columns 1 through 4
0.891991704091174 1.15630026617957 0.986517573000511 6.74601203684419
0.99169682555935 0.178357471543757 0.994122925011876 3.42087655104056
0.139030187233434 0.50924781527068 0.162388475001478 5.64029262176829
1 1 1 1
Column 5
0.439567078410234
0.954395782603168
0.825652509356184
1
Here you can see that each column is divided by the 4th row value of that column. This is implemented in recent versions by implicit expansion that does not need bsxfun. It is effectively dividing a (4,5) matrix by a (1,5) array in this case and could be done as
a ./ a(4,:)
equivalently.
As per many cases, investigating syntax on command line with simpler examples can help illuminate what is happening.
Without having looked at the code in the submission you mention I would guess this is a normalisation step that forces values to be in the correct range ( 0 - 1 ) and not get clipped off above 1.

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