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How to replace zeros in one matrix with another matrix

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J
J 2016 年 12 月 19 日
コメント済み: Guillaume 2016 年 12 月 19 日
Hello, how to replace zeros in one matrix with another matrix using if statement? Dimensions are the same; first matrix has 10 zeros and second matrix has 10 numbers. Thanks for answers J
  3 件のコメント
Geoff Hayes
Geoff Hayes 2016 年 12 月 19 日
J's answer moved here
Matrix B=[1 0 0 0 0;2 3 0 0 0;4 5 6 0 0;7 8 9 10 0;11 12 13 14 15] You can see that "triu" of matrix B is made of zeros. I need to replace zeros in matrix B using IF statement with matrix A=[1 2 3 4 5;6 7 8 9 10]. Orientation is set and number of zeros is matching matrix A.
Guillaume
Guillaume 2016 年 12 月 19 日
The first option in my answer matches this scenario. if is not needed.

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回答 (3 件)

Andrei Bobrov
Andrei Bobrov 2016 年 12 月 19 日
A = [1:3, 0,0 7,12,0]';% your first array
B = randi(20,8,1); % your second array
t = A == 0;
A(t) = B(t);

Guillaume
Guillaume 2016 年 12 月 19 日
An example would clarify what you want. Possibly:
To replace all zero values in A by consecutive values in B, B must have as many elements as there are zeros in A:
assert(numel(B) == numel(A == 0)); %B must have as many elements as there are zeros in A
A(A == 0) = B %replace zeros values in A by values in B
or to replace zeros values in A by values in B in the same position, B must be the same size as A:
assert(isequal(size(B), size(A)));
A(A == 0) = B(A == 0)
An if is most likely not needed.

arwa
arwa 2016 年 12 月 19 日
x = [0 0 0 0 0 0 0 0 0 0]; z = [1 2 3 4 5 6 7 8 9 10]; for i=1:1:10 x(1:i)=z(1:i); end
then you can get x = [1 2 3 4 5 6 7 8 9 10] like z instead of zeros..
Do you mean this??

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