while loop matrix problem
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hi,
I want to use a while loop on matrices, to define a new matrix by calculating one row each time.
I'm trying to do it without creating another loop that will go over the columns. Is that possible?
for example:
z5 and k5 are known matrices. This loop defines c5.
i=1;
while i<imax
if z5(i,:) < z5(i-1,:)
c5(i,:) = k5(i,:);
elseif z5(i,:) > z5(i-1,:)
c5(i,:) = z5(i,:);
else
c5(i,:) = c5(i-1,:);
end
i=i+1;
end
It's not doing what I want it to do and I'm not sure what exactly its doing.
Would appreciate any help and clarification on this matter.
10 个评论
Torsten
2019-7-26
result = z5(i,:) < z5(i-1,:)
is not the same as
for j=1:size(z5,2)
result(j) = z5(i,j) < z5(i-1,j)
end
Test it and see the difference.
Galgool
2019-7-26
Bob Thompson
2019-7-26
编辑：Bob Thompson
2019-7-26
if z5(i,:) < z5(i-1,:)
How is this evaluated for each row?
A = [1 2 3;
4 2 1];
It could be argued that the second row of A is greater than the first, because the sum total of the elements is greater than the first row. It could also be argued that the second row is not greater because each individual element is not greater than the corresponding element in the first row.
An inequality comparison statement is really only useful for comparing one piece of information with another. If you would like to complete all of these comparisons at once, and don't want to use a loop, you can use & to have multiple comparisons for one if statement, but this is very tedious for a large z5 matrix.
if z5(i,1) < z5(i-1,1) & z5(i,2) < z5(i-1,2)
Torsten
2019-7-26
As you can see from our answers, it's not possible without looping because comparisons must be done elementwise to get the results you expect to get.
Galgool
2019-7-26
If you test
if z5(i,:) < z5(i-1,:)
the result will be true only if the vector v = z5(i,:) < z5(i-1,:) is a vector of ones at all positions.
So if there is only one case for which z5(i,j) >= z5(i-1,j), all other cases for which z5(i,:) < z5(i-1,:) are ignored, positions for which you intend to set c5(i,:) = k5(i,:), I guess.
Maybe this is what you want to do:
jdx1 = z5(i,:) < z5(i-1,:);
jdx2 = z5(i,:) > z5(i-1,:);
jdx3 = z5(i,:) == z5(i-1,:);
c5(i,jdx1) = k5(i,jdx1);
c5(i,jdx2) = z5(i,jdx2);
c5(i,jdx3) = c5(i-1,jdx3);
Galgool
2019-7-26
Actually you're worng. ab1(i,:) = z5(i,:) < z5(i-1,:); works properly (as you can see in (37,8)):
z5:
ab1:
the cyclist
2019-7-26
编辑：the cyclist
2019-7-26
To reiterate what Torsten is saying ...
To enter an if statement on a vector condition, all elements of that vector have to evaluate to true. MATLAB doesn't discern the if condition element-by-element.
That is the fatal flaw in your algorithm
采纳的回答
Guillaume
2019-7-26
@galgool, before calling someone wrong please learn the language properly, in particular how if works when given a vector as an expression. From the doc:
An expression is true when its result is nonempty and contains only nonzero elements (logical or real numeric). Otherwise, the expression is false.
So, let's look at
if z5(i,:) < z5(i-1,:)
The expression is
z5(i,:) < z5(i-1,:)
The result of that expression is a logical vector (array of 0s and 1s). As per the rule above, the if is true only if all the elements of the vector are 1. I.e. the only time the expression is true is when z5(i, j) < z5(i-1, j) for all j. In general, you should avoid passing a vector to if because people don't know the rule above. The expression you've written is exactly equivalent to:
if all(z(5(i,;) < z5(i-1,:))
which is clearer.
As others have said, you will also have to loop over the columns.
Note that:
i = 2; %your code would error if you actually started at 1!
while i < constant
%some code that doesn't change i or the constant
i = i+1;
end
is more simply written as a for loop:
for i = 2:constant
%some code that doesn't change i or the constant
end
Note that if you didn't have the dependence on the previous c5 row, when z5(i,j) == z5(i-1, j), the whole thing could written without any loop at all.
4 个评论
Galgool
2019-7-28
Guillaume
2019-7-29
That's fine as long as you write a comment explaining exactly what this line does.
My preference, rather than multiplying by logical vectors to select the actual working part of the expression would be:
c5(i, z5(i, :) < z5(i-1, :)) = k5(i, z5(i, :) < z5(i-1, :));
c5(i, z5(i, :) > z5(i-1, :)) = z5(i, z5(i, :) > z5(i-1, :));
c5(i, z5(i, :) == z5(i-1, :)) = c5(i-1, z5(i, :) == z5(i-1, :));
More lines but a lot easier to understand
Galgool
2019-7-29
Guillaume
2019-7-29
Maybe, this is clearer this way:
mask = z5(i, :) < z5(i-1, :); c5(i, mask) = k5(i, mask); %set c5 values for which mask is true to matching k5 values
mask = z5(i, :) > z5(i-1, :); c5(i, mask) = z5(i, mask); %set c5 values for which mask is true to matching z5 values
mask = z5(i, :) == z5(i-1, :); c5(i, mask) = c5(i-1, mask); %set c5 values for which mask is true to matching value on previous row
If there was the dependence on the previous row of c5, this could be done trivially without a loop, but unfortunately with that dependence it does need the loop... unless you are guaranteed not to have two consecutive identical value in any z5 column.
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