清空筛选条件
清空筛选条件

using multiple if statement to replace values from another matrix

1 次查看（过去 30 天）
Phat Pumchawsaun
Phat Pumchawsaun 2017-3-12
评论： Walter Roberson 2017-3-12
Hi,
I have two matrix (A and B in attached excel file) which have the same dimension (853x13). I would like to use multiple "if" function as following code;
C = zeros(size(B));
for k = 1:length(B);
l = 1:length(A);
if B(k,2) > B(k,3);
if B(k,4:13) > A(l,4:13);
C(k,4:13) == A(l,4:13);
else
C(k,4:13) = B(k,4:13);
end
else
C(k,4:13) = B(k,4:13);
end
end
I want matrix "C" to have both values from A and B if it follows the fist line of if condition. However, the new matrix "C" has values only from matrix B. How should I correct it? Since the length of matrix A and B is so huge, I attached excel file of both matrix.
Thanks

请先登录，再回答此问题。

回答（1 个）

Walter Roberson
Walter Roberson 2017-3-12
Your line
C(k,4:13) == A(l,4:13);
is a comparison, not an assignment.
  2 个评论
Walter Roberson
Walter Roberson 2017-3-12
You have
l = 1:length(A);
so l is a vector.
You have
if B(k,4:13) > A(l,4:13);
l is a vector, as indicated above, so A(l,4:13) is indexing A with subscripts that are both vectors. Since l is 1:length(A) then as long as A has more rows than columns, A(l,4:13) means the same as A(:,4:13) does -- a rectangular portion extracted from an array.
Meanwhile, B(k,4:13) is a vector. So your > operator is comparing a vector to a rectangular array. In all versions before R2016b, that would be an error. In R2016b and later, the result is the same as if you had used
if bsxfun(@gt, B(k,4:13), A(l,4:13))
which in turn is the same as
if repmat(B(k,4:13), length(A), 1) > A(l,4:13)
which repeats the row of B into equivalent size of A's rows. You end up with a rectangular matrix ">" a rectangular matrix. The result of that ">" operator is going to be a rectangular matrix of true and false values.
When "if" is asked to process a vector or matrix of true and false values, then it considers the condition to hold only if all of the values are non-zero (true). In your 2D matrix case, that would be equivalent to
if all(all( repmat(B(k,4:13), length(A), 1) > A(l,4:13) ))
And in your case that does not happen to be the case.
My guess at what you want:
nrow = size(B,1);
for k = 1:nrow
if B(k,2) > B(k,3)
C(k,4:13) = min(A(k,4:13), B(k,4:13));
else
C(k,4:13) = B(k,4:13);
end
end
Or in vectorized form with no loop:
C = zeros(size(B));
C(:,4:13) = B(:,4:13);
mask = B(:,2) > B(:,3);
C(mask, 4:13) = min( A(mask,4:13), B(mask,4:13) );

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 MATLAB Report Generator 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by