Attempt to reference field of non-structure array.

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Cat
Cat 2015-2-22
编辑： Cat 2015-2-24
So I'm trying to write a code to count the inversions in an array. So in the array [6,4,3,5,1,2] there are 12 inversions. 6 is greater than 5 of the numbers to the right of it, 4 is greater than three of the numbers to the right of it, and so on. However, I need it so that the function doesn't count zero. So if there was a zero in the place of the 3, 6 would only be greater than 4 of the numbers to the right of it. (zero counts as a space, not as a number. I got some help, and this is the code I have to far
while (a <= array.length)
while (b <= array.length)
if array(a)>array(b);
count = count + 1;
else
count = count + 0;
end
b = b + 1;
end
a = a + 1;
end
When I try to run an example array through it and then try to display the count, it gives me the error "Attempt to reference field of non-structure array." Any help with what I'm doing wrong? Or how to add to the function so it doesn't count being greater as an inversion?
UPDATE: fixing the error of using left in place of right
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Image Analyst
Image Analyst 2015-2-23
What's your definition of inversion? And 6 doesn't have any numbers to the left of it. 4 has 1 number to the left of it, not 3. And this isn't .Net - there is no .length method to the variables. Use the length() function.
Cat
Cat 2015-2-24
编辑：Cat 2015-2-24
So sorry, I meant numbers to the right. And yes, it has to be compared to every number to the right. It's part of a larger function. We are trying to see if a matrix is solvable as a square puzzle. http://www.cs.bham.ac.uk/~mdr/teaching/modules04/java2/TilesSolvability.html here is a link with more information on the inversions and the goal of the function if it helps

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采纳的回答

Andrew Newell
Andrew Newell 2015-2-24
Suppose we start with the example from your link:
array = [12 1 10 2 7 11 4 14 5 0 9 15 8 13 6 3];
To eliminate the zero entries, just do this:
array = array(array~=0);
A pretty good way of counting the inversions expresses them as a logical relation, e.g.,
array(2:end) > array(1)
ans =
0 0 0 0 0 0 1 0 0 1 0 1 0 0
Then you can just add up the 1's to get the number of inversions. You can do this in a loop:
count = 0;
for ii=1:length(array)-1
count = count + sum(array(ii+1:end)<array(ii));
end
There are more compact ways of doing this, but they're pretty cryptic.

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Chris McComb
Chris McComb 2015-2-23
I think that your problem is that array is not a structure. In order to get the length of a, use:
length(array)
instead of:
array.length
  1 个评论
Cat
Cat 2015-2-24
Do you have any suggestions for how to not include the zero? I should have put this in the original question. http://www.cs.bham.ac.uk/~mdr/teaching/modules04/java2/TilesSolvability.html This link contains a really good example of the inversions and how it is supposed to play into a bigger function

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