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Substitute numbers in array

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majed majed
majed majed on 22 Mar 2016
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]

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Answers (4)

Azzi Abdelmalek
Azzi Abdelmalek on 22 Mar 2016
Edited: Azzi Abdelmalek on 22 Mar 2016
X=[1 2 3 4 5 6 7 8]
XX=[fliplr(X(1:4)) fliplr(X(5:end))]

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Azzi Abdelmalek
Azzi Abdelmalek on 22 Mar 2016
This case is different, there are 11 element (11 is odd)
majed majed
majed majed on 22 Mar 2016
The numbers or values is just from 1 to 8
Azzi Abdelmalek
Azzi Abdelmalek on 22 Mar 2016
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)]

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Stephen Cobeldick
Stephen Cobeldick on 22 Mar 2016
Edited: Stephen Cobeldick on 22 Mar 2016
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

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Jan
Jan on 22 Mar 2016
Edited: Jan on 22 Mar 2016
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))

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Stephen Cobeldick
Stephen Cobeldick on 22 Mar 2016
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
Steven Lord on 22 Mar 2016
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)))

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Suraj Sudheer Menon
Suraj Sudheer Menon on 22 Jun 2020
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.

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