If loop using cell arrays
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Hi,
I have 2 matrices:
A with 24 arrays of each different length (i.e. A{1} 2362x1, A{2} 2341x1) defining salaries
B with 24 arrays of each different length (i.e. A{1} 2362x1, A{2} 2341x1) contains either a 0 or a 1
And last I have 2 boundary values C= 2600 and D= 1700.
I want to check the following:
if B(each value) == 0 && A(each value) > C
A(this value)= C
elseif B(each value) == 1 && A(each value) > D
A(this value)= D
end
Can anyone help me out?
1 个评论
Walter Roberson
2015-9-14
There is no such thing as an "if loop". There are "for loop" and there are "while loop" and there are "if statement". A loop would have to execute its body multiple times, but "if" executes the body only 1 time or 0 times.
采纳的回答
Guillaume
2015-9-14
You don't need any if. Just use logical operations. Your first condition is to replace elements when B is false (i.e. B==0 or simply ~B) and A greater than C, so is simply writen as A(~B & A>C). Similarly your second condition is simply A(B==1 & A>D) (which simplifies to A(B & A>D). You just have to apply that to each cell of the cell arrays:
for cellidx = 1:numel(A)
a = A{cellidx};
b = B{cellidx};
a(~b & a>C) = C;
a(b & a>D) = D;
A{celldix} = a;
end
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Titus Edelhofer
2015-9-14
Hi,
I'm not 100% sure I understand what you want to do, but I guess it should be like this
for i=1:length(A)
if all(B{i}(:)==0 & A{i}(:)>C)
A{i}(:) = C;
elseif all(B{i}(:)==1 & A{i}(:)>D)
A{i}(:) = D;
end
end
Titus
4 个评论
Titus Edelhofer
2015-9-14
length(A) is the length of the cell array. The code is (just as Guillaume's code) totally independent of the sizes of each A{i}, B{i} (of course as long as they match) :).
Titus
Both codes have the following preconditions:
assert(isequal(numel(A), numel(B)));
assert(all(cellfun(@(a,b) isequal(size(a), size(b)), A, B)));
Titus Edelhofer
2015-9-14
I wish I could vote for a comment! Nice assert calls.
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