Substitute numbers in array

How are you everyone!

I have the array X=[1 2 3 4 5 6 7 8] I want to flip this array and change the numbers as next

1 to 5 and 5 to 1

2 to 6 and 6 to 2

3 to 7 and 7 to 3

4 to 8 and 8 to 4

So the result which I want after flipping and substitution is

XX=[4 3 2 1 8 7 6 5]

Answer 1

[Azzi Abdelmalek]

X=[1 2 3 4 5 6 7 8] 
XX=[fliplr(X(1:4)) fliplr(X(5:end))]

[majed majed]

Thanks for you answer Sometimes maybe the array doesn't has the last form , it cold be like this

X=[1 3 6 5 3 3 2 8 7 6 5]

Or

X=[5 6 4 3 7 8 3 2 1 7 8]

How we can solve it? Thank you again

[Azzi Abdelmalek]

Ok, for this example X=[1 2 3 4 5 6 7 8 9 10 11], what are you expecting as result?

[Azzi Abdelmalek]

Maybe you want this:

X=[1 2 3 4 5 6 7 8 9 10 11]
n=fix(numel(X)/2)
XX=[fliplr(X(1:n)) fliplr(X(n+1:end))]

[majed majed]

The result I want is as I mentioned befor, just substitute the numbers as I've mentioned and flip the array

[Azzi Abdelmalek]

This case is different, there are 11 element (11 is odd)

[majed majed]

The numbers or values is just from 1 to 8

[Azzi Abdelmalek]

Now if you want to flip just 8 element:

X=[1 2 3 4 5 6 7 8 9 10 11]
XX=[fliplr(X(1:4)) fliplr(X(5:8)) X(8+1:end)]

Answer 2

[Stephen23]

Try this function. The matrix M defines any arbitrary values to swap. Note that these values are not used as indices so this is a general solution to the problem.

>> M = [1:4;5:8].'; % each row specifies one pair of values to swap
M =
   1     5
   2     6
   3     7
   4     8
>> fun = @(X)fliplr(reshape(M(:,[2,1]),1,[])*bsxfun(@eq,X,M(:)));

and the examples you gave are:

>> fun([1,2,3,4,5,6,7,8])
ans =
     4     3     2     1     8     7     6     5
>> fun([1,3,6,5,3,3,2,8,7,6,5])
ans =
     1     2     3     4     6     7     7     1     2     7     5
>> fun([5,6,4,3,7,8,3,2,1,7,8])
ans =
     4     3     5     6     7     4     3     7     8     2     1

Answer 3

[Jan]

A job for a lookup table:

X   = [1, 2, 3, 4, 5, 6, 7, 8]
LUT = [5, 6, 7, 8, 1, 2, 3, 4]
Result = LUT(fliplr(X))

[Stephen23]

Nice use of indexing :)

Note that it is not a general solution suitable for all data: if

 X   = [1,1e9]

what will LUT have to be?

[Steven Lord]

In that case I would make the LUT a sparse column vector.

X = [1, 1e9];
LUT = sparse(X, 1, [2, 73]);
Y = full(flip(LUT(X)))

Answer 4

[Suraj Sudheer Menon]

The following could be an approach:-

sub=[5 6 7 8 1 2 3 4];
XX=flip(X);
for i=1:numel(X)
    XX(i)=sub(XX(i));
end
%XX contains neccesary values.