How to do xor operation?

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Darsana P M
Darsana P M 2017-10-22
评论： Jan 2020-2-2
How to do xor operation in cell array. Suppose i have bits x={'1' '1' '1' '0' '1' '1' '0' '1'}; and v={'1' '1' '0' '0' '1' '0' '1' '1'}; Should i use cell function? Can somebody help me with the correct code.
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Stephen23
Stephen23 2017-10-22
@Darsana P M: why are you storing this data in cell arrays? Your code would be a lot simpler and more efficient if you stored character data in a char array, or used logical/numeric arrays. Cell arrays do not seem to serve any useful purpose here.

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采纳的回答

Birdman
Birdman 2017-10-22
x={'1' '1' '1' '0' '1' '1' '0' '1'};
v={'1' '1' '0' '0' '1' '0' '1' '1'};
for i=1:1:length(x)
xx(i)=str2double(cell2mat(x(i)));
vv(i)=str2double(cell2mat(v(i)));
if(xx(i)==vv(i))
XOR(i)=0;
else
XOR(i)=1;
end
end
Hope this helps.
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Darsana P M
Darsana P M 2017-10-22
thanks a lot sir
Jan
Jan 2017-10-22
@Çevikalp Sütunç: In the line:
xx(i)=str2double(cell2mat(x(i)))
the cell2mat can be replaced by a simpler cell indexing with curly braces:
xx(i) = str2double(x{i})
But str2double works on cell strings directly also:
xx(i) = str2double(x(i))
or better outside the loop:
xx = str2double(x)

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更多回答（4 个）

Stephen23
Stephen23 2017-10-22
编辑：Stephen23 2017-10-22
Ugly loops are not required, this is all you need:
>> x = {'1' '1' '1' '0' '1' '1' '0' '1'};
>> v = {'1' '1' '0' '0' '1' '0' '1' '1'};
>> xor([x{:}]-'0',[v{:}]-'0')
ans =
0 0 1 0 0 1 1 0
If you had stored your data in a simpler character array, then all you would need is this:
>> x = '11101101'; v = '11001011';
>> xor(x-'0',v-'0')
ans =
0 0 1 0 0 1 1 0
Compare with Çevikalp Sütunç's answer: which one is going to be more efficient, is easier to understand the purpose of, and will be easier to debug?
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Darsana P M
Darsana P M 2017-10-22
Thanks a lot sir. I think this is much more simpler. I took them as cell array because i was dealing with hexa decimal values. I was trying out this with small values.
So this gives xor of x and v. If suppose i have few vectors as x1={'1' 1' 0 1 1 0 1 0} x2={ 1 1 0 0 1 1 1 0} upto xn same for v1,v2 and so on. So while applying loops how can i access the entire set of elements and find the xor of each. ie xor(x1,v1), xor(x2,v2) etc

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KL
KL 2017-10-22
Another method maybe,
x = {'1' '1' '1' '0' '1' '1' '0' '1'};
v = {'1' '1' '0' '0' '1' '0' '1' '1'};
res = logical(zeros(size(x)));
res(str2double(x)~=str2double(v)) = logical(1);
As Stephen says, having a cell array here is unnecessary.
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Darsana P M
Darsana P M 2017-10-23
编辑：Stephen23 2017-10-23
If suppose i have few vectors as x1={'1' 1' 0 1 1 0 1 0} x2={ 1 1 0 0 1 1 1 0} upto xn same for v1,v2 and so on. So while applying loops how can i access the entire set of elements and find the xor of each. ie xor(x1,v1), xor(x2,v2) etc
Stephen23
Stephen23 2017-10-23
"If suppose i have few vectors as x1={'1' 1' 0 1 1 0 1 0} x2={ 1 1 0 0 1 1 1 0} upto xn..."
Do not do this. Magically accessing variables names is complex, slow, buggy, and hard to debug. Read this to know why:
https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval
Much neater, faster, and more efficient would be if you simply stored all of your data in one array (which could be an ND numeric array, or a cell array). Using indexing is simple, fast, easy to debug, and very efficient.

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Jan
Jan 2017-10-22
编辑：Jan 2017-10-22
It is strange, that the inputs are char's, but there is no need to convert them:
x = {'1' '1' '1' '0' '1' '1' '0' '1'};
v = {'1' '1' '0' '0' '1' '0' '1' '1'};
L = {'0', '1'};
Result = L(~strcmp(x, v) + 1)
Christina Christofidi
Hey, I have an employment about the network coding. I need to run a file with some numbers and do XOR between them. You can help me????
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Jan
Jan 2020-2-2
Please do not attach a new question in the section for answers of another question. Create your won question instead. By the way, this seems to be a job for a simple xor comand.

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