Reorder a matrix relative to another

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Matthew
Matthew 2015-3-2
编辑： Stephen23 2015-3-2
I have investigated all day the different ways to do this, and I have not been able to come up with a plan to achieve exactly what i need. Below are my sample matrices.
a=[5;20;50;30;10]
index=[1;2;3;4;5]
b=[10;50;5;20;50]
c=[44;22;11;88;55]
"a" and "index" are the proper order of the matrix. Matrix "b" is identical is size and shape, but the values are scrambled in order. I need a way to establish an order to move "b" to look like "a", and then apply that reordering pattern to "c". The only matrix i am interested in is the reordered "c"
In the end i want the new reordered matrix to be the following:
cReordered=[11;88;22;55;44]
I am looking for the most efficient way to do this.
I have used:
[bnew, ia, ib] = intersect(a, b, 'stable')
but it only reorders "b", by generating a "bnew". and this will match the pattern in "a", but that is all.
Any advice would be appreciated. thank you.
  2 个评论
Stephen23
Stephen23 2015-3-2
编辑：Stephen23 2015-3-2
Are the values in a, b, and index unique? It seems that the vector b should actually be
b = [10;50;5;20;30];
(i.e. replacing the second 50 with 30), is this correct?
Matthew
Matthew 2015-3-2
Sorry i typed that wrong. Yes, the values in a and values in b will be identical, just those in b will be out of order.
b=[10;50;5;20,30]

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Stephen23
Stephen23 2015-3-2
编辑：Stephen23 2015-3-2
Assuming that the values in both of a and b are unique, and that they each contain exactly the same values, then the vector b original question needs have the second 50 replaced with a 30, like this:
>> a = [5;20;50;30;10];
>> x = [1;2;3;4;5];
>> b = [10;50;5;20;30];
>> c = [44;22;11;88;55];
Then the solution is very easy using MATLAB's indexing ability:
>> [~,xx] = sort(a);
>> [~,yy] = sort(b);
>> d(xx) = c(yy)
d =
11 88 22 55 44
Which matches the desired output [11;88;22;55;44]. Note that this requires that the values in a and b are unique!

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