if condition in for loop
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Hello ,
I have a problem in finding the same rows between two matrices. I wrote the code as follows
function neigh_no = neigh_search(hex_center,neigh)
[~, n] =size(hex_center);
[~, n1]= size(neigh);
if(n~=n1)
neigh=neigh';
end
neigh_no=[];
for i=1:length(hex_center)
for j=1:length(neigh)
if hex_center(i,:)== neigh(j,:)
neigh_no=[neigh_no i];
end
end
end
end
Suppose if the hex_center array values are
0 0
0 1.7320
0 3.4641
0 5.1961
3 0
3 1.7320
3 3.4641
3 5.1961
6 0
6 1.7320
6 3.4641
6 5.1961
9 0
9 1.7320
9 3.4641
9 5.1961
1.5 0.8660
1.5 2.5980
1.5 4.3301
4.5 0.8660
4.5 2.5980
4.5 4.3301
7.5 0.8660
7.5 2.5980
7.5 4.33012
and the neigh values are
9 5.1961
7.5 6.0621
6 5.1961
6 3.4641
7.5 2.5980
9 3.4641
if you observe that the rows 11,12,15,16 and 24 in hex_center matrix has same rows in the second matrix. But in the output it displays only the rows 11 and 15.Can you please help me? its so important> I know its simple. But I dont know where I did mistake.
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回答 (2 件)
Image Analyst
2015 年 9 月 21 日
That usually doesn't work for floating point numbers, just integers and inverse powers of 2. You'll have to check for a tolerance like the code examples in the FAQ show you how to do: http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F
2 件のコメント
Star Strider
2015 年 9 月 21 日
編集済み: Star Strider
2015 年 9 月 21 日
Try this one line of code instead of the entire loop structure:
[neigh, neigh_no] = ismember(neigh, hex_center, 'rows');
Also, since ‘neigh_no’ returns one value that is 0, the documentation for ismember explains ‘The output array, Locb, contains 0 wherever A is not a member of B.’ (The Locb variable in the documentation is ‘neigh_no’ here.)
EDIT — Added link to ismember documentation.
3 件のコメント
Image Analyst
2015 年 9 月 22 日
Compute the distances. I think you can use pdist() if you have the stats toolbox. Then find distances less than about 1.75 times the separation to find the immediate neighbors.
Star Strider
2015 年 9 月 22 日
You can introduce a tolerance with the ismember function by using ismembertol (introduced in R2015a).
If you have an earlier version, consider using the newest version of round (with two arguments) to round to a specific number of decimal places. If you don’t have that version of round, you can simulate it with this function to reduce argument precision for ismember:
roundn = @(x,n) round(x*10.^n).*10.^(-n);
It rounds ‘x’ to ‘n’ places to the right of the decimal (n>0) or to the left of the decimal (n<0). For n=0 it’s the usual round function.
Either of these should solve the tolerance problem. (Your data produced the desired result with my original code, so I didn’t initially suggest ismembertol or rounding.)
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