Replace zero in a matrix with value in previous row
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Hi,
Can you please help me on how can I replace all zeroes in a matrix with the value in previous row?
e.g. if value in row 3 column 4 is 0, it should pick value in row 2 column 4.
I can do it using a for loop but I dont want to use that.
Thanks.
0 件のコメント
採用された回答
Azzi Abdelmalek
2014 年 2 月 1 日
編集済み: Azzi Abdelmalek
2014 年 2 月 1 日
A=[1 2 3 4;4 5 0 0;1 0 0 1 ;0 1 1 1]
while any(A(:)==0)
ii1=A==0;
ii2=circshift(ii1,[-1 0]);
A(ii1)=A(ii2);
end
4 件のコメント
Jan
2016 年 11 月 4 日
@Mido: Please open a new thread for a new question.
match = (A(:, 3)==0);
A(match, 3) = A(match, 2);
その他の回答 (5 件)
Shivaputra Narke
2014 年 2 月 1 日
Now answer to your comment...
while(all(a(:))) a(find(a==0))=a(find(a==0)-1) end
3 件のコメント
Captain Karnage
2022 年 12 月 2 日
That expression doesn't work as written in a single line. Without a ; after the assignment, it gets an error 'Error: Illegal use of reserved keyword "end".` If I add the ; however, it doesn't error but it still doesn't work. I'm not sure why, because logically it seems like it should, but I just get the original a out when I run it.
DGM
2022 年 12 月 3 日
編集済み: DGM
2022 年 12 月 3 日
The loop is never entered at all. You could make some modifications.
a = [0 5 9 13; 2 6 0 0; 3 0 0 15; 0 8 12 16]
na = numel(a);
while ~all(a(:)) % loop runs until there are no zeros
idx = find(a==0);
a(idx) = a(mod(idx-2,na)+1);
end
a
The redundant find() can be removed. Since this is based on decrementing the linear indices, this will fill zeros at the top of a column with content from the bottom of the prior column. Using mod() allows the wrapping behavior to extend across the ends of the array. Note that a(1,1) is filled from a(16,16).
Whether this wrapping behavior is intended or acceptable is a matter for the reader to decide.
Andrei Bobrov
2014 年 2 月 1 日
l = A == 0;
ii = bsxfun(@plus,size(A,1)*(0:size(A,2)-1),cumsum(~l));
out = A;
out(l) = A(ii(l));
0 件のコメント
Shivaputra Narke
2014 年 2 月 1 日
May be this code can help...
% where a is your matrix a(find(a==0))=a(find(a==0)-1)
2 件のコメント
Amit
2014 年 2 月 1 日
Lets say your matrix is A
[m,n] = size(A);
An = A';
valx = find(~An); % This will give you zeros elements linear index
valx = valx(valx-n > 0);
An(valx) = An(valx-n);
A = An';
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