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Simplifying a four-dimensional rule table in Matlab: addressing rows and columns of each dimension

Answer 1

+1 A:

I'm not sure if this is what you are asking, but here is how to check to see if the values in two rows are either the same as each other, or if not the same, that at least one is 0

table(a,:,:,:) == table(a+1,:,:,:) | table(a,:,:,:) == 0 | table(a+1,:,:,:) = 0;

if you want to be really clever, but somewhat cryptic,

(diff(table, [], 1) == 0) | table(1:(end-1),:,:,:) == 0 | table(2:end, :,:,:) == 0

for the first dimension

(diff(table, [], 2) == 0) | table(:,1:(end-1),:,:) == 0 | table(:,2:end, :,:) == 0

for the second, and so on

Marc 2010-06-02 00:59:23

That is rather cryptic, but by the look of it it may work for this thank you. What may be a better (and simpler) question is how would I access the first element of the third and fourth dimensions, given that the first element of the first is (a,1,:,:), the second (1,b,:,:) etc?

Cate 2010-06-02 10:44:41

Actually, that checks the entire row/column for it being "empty" as a whole, doesn't it? My issue is checking individual cells in the adjacent rows/columns to see if they match or if either is empty; if for the entire row/column cells either match or at least one is empty then the two rows/columns may be merged.

Cate 2010-06-02 10:56:30

== does an element by element comparison, so [1 2 3 4 5] == [5 4 3 2 1] yields [0 0 1 0 0]I'm not sure what you're asking in your first comment. To access an individual element of a 4 d array, arr(i,j,k,m) to get all the elements that have the first 3 indices == i,j,k, arr(i,j,k,:). etc.

Marc 2010-06-03 13:22:32

I was just saying in this matrix an empty cell holds a 0. Checking element by element == 0 will return 1 if all hold a 0, and return 0 if not. My other question needs clarification though:To access the first element of the 1stD in a 2D matrix = (1,1), the second = (1,2).To access the first element of the 2ndD in a 2D matrix = (1,1), the second = (2,1).Now if I had 3D, the first element of the 3rdD would be (1, :, 1) therefore giving all values of the 2ndD corresponding to the first row. So how do you access the first of a 4D? (1,:,:,1)? (:,1,:,1)? (:,:,1,1)? something else?Thanks

Cate 2010-06-03 14:27:34

Checking element by element == 0 will return 1 if all hold a 0: no, x == 0 will return a matrix of the same size as x, with a 1 wherever x == 0 and a 0 otherwise. Try it.Call the 1st dimension the row, the 2nd dimension the column, and the 3rd dimension the plane.arr(1,:,:) returns a 2D array of elements from the first rowarr(:,1,:) returns a 2D array of elements from the first columnarr(:,:,1) returns a 2D array of elements from the first planearr(1,:,1) is a 1D vector of elements from the first row and the first plane, and so on

Marc 2010-06-03 18:23:48

I have and I think I didn't make myself clear enough. if(x == 0) will check if each element of x is equal to 0 and if indeed it is, it will return a 1 and proceed with the consequence of the if statement. If not, the condition isn't met and nothing happens. This is how I check if the entire plane for that dimension is empty i.e. if(1,:,:,:)==0 etc tells me whether I can remove that row. And I can understand 3 dimensional matrices but the 4 dimensional matrix is what I am unsure of. I think maybe my question is a little too perplexing to answer succinctly!

Cate 2010-06-03 22:05:10

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Simplifying a four-dimensional rule table in Matlab: addressing rows and columns of each dimension

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