Edit: I used number 1,2 and 3 just for my example. The integers could be different. There could be different "mode" numbers (the most frequent numbers) in different columns.
Identify the mode in each column and eliminate them
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I am curious to see if the following could be done without loops.
I have a 7-by-N matrix with all integers. In each column of the matrix, there is one integer that appears 4 time (hence the most frequent integer).
This there any way to eliminate the most frequent integers in every column and to output the rest of the integers in a 3-by-N matrix without using loops?
Example: a = [1 1 2 2 2 2 3; 1 2 2 2 2 3 3; 2 2 2 2 3 3 3]';
Output should be [1 1 3; 1 3 3; 3 3 3]'
The order of numbers in each column could be randomized unlike in my example. Thank you very much and no this is not a homework problem.
採用された回答
Andrei Bobrov
2013 年 6 月 6 日
編集済み: Andrei Bobrov
2013 年 6 月 7 日
reshape(a(ismember(a,[1,3])),[],size(a,2));
or
hankel([1 1 3],[3 3 3]);
ADD
s = size(a);
b = ones(s);
idxl = fliplr(triu(b,1) + tril(b,diff(s))) > 0;
out = reshape(a(idxl),[],s(2));
その他の回答 (2 件)
Iain
2013 年 6 月 6 日
a = a';
out = reshape(a(a~=2),[],3)';
That will work for your example, but not necessarily what you're dealing with.
Roger Stafford
2013 年 6 月 7 日
Let 'a' be an array in which one element of each row is repeated the same number of times for all rows with no other element in each row occurring as often as this.
b = a.';
b = reshape(b(bsxfun(@ne,mode(b),b)),[],size(b,2)).';
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