how to arrange in order index by number of position

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nadia nadi
nadia nadi 2015-10-6
评论： Guillaume 2015-10-7
Dear,
I have this
row =[4 5 2 2 2 1];
its index as Known is [1 2 3 4 5 6]; I want to arrange this index according to the number of the row from right to left, elements of row is the unfilled position in the final row I want to put the index in it, so I wrote this but I stuck with the filling. I will explain it more. Starting from right to left, for
idx=[1 2 3 4 5 6];
idx=1 take position 1 as in row=1 , idx=2 take position 2, then idx=3 take position 2 again instead number idx 2,then idx=4 take position 2 again instead idx 3, then idx=5 take position 5, then idx=6 take position 4, I should get [5 2 6 3 4 1]. I need to arrange the index for any row in this method. What I wrote is very wrong.
I appreciate any help.
row =[4 5 2 2 2 1];
fliprow=flip(row)
% Mapping Row
idx=find(row);
S=[];
for s=1:g
d= fliprow(:,s)
S=[S d]
end
S
Nadia,
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Guillaume
Guillaume 2015-10-7
Your diagram clarifies things a lot. You've never mentioned shifting/sliding numbers before, so that step was not apparent.
nadia nadi
nadia nadi 2015-10-7
I am so sorry,
I thought what I explained is clear. Its confusing and difficult to me to program it.
regards,

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采纳的回答

Guillaume
Guillaume 2015-10-7
Now that we have clear explanations, it's easy to compute the result of your algorithm.
As mentioned in my comments, I'm going to use the natural left-to-right indexing. You're free to permute the input and output with fliplr.
The following may not be the most efficient implementation in term of memory (it does a fair number of reallocation). It's a straightforward implementation of your algorithm:
permutationrule = [1 2 2 2 5 4] %your input
permutated = []; %the output
for idx = 1:numel(permutationrule)
insertionpoint = permutationrule(idx);
permutated = [permutated(1:insertionpoint-1), idx, permutated(insertionpoint:end)];
end
It's simple, permutationrule(idx) tells you where idx needs to be inserted in permutated. So you jut have to concatenate, the portion of permutated before the insertion point (i.e. permutated(1:insertionpoint-1)), the insertion (i.e. idx), and the portion of permutated from the insertion point to the end (i.e. permutated(insertionpoint:end)).
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nadia nadi
nadia nadi 2015-10-7
Thanks Guillaume ,
that's great. I will try to use it for any size for permutationrule .
Great thanks,
Nadia
  
Guillaume
Guillaume 2015-10-7
The code already works for any length of permutationrule.

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更多回答（2 个）

arich82
arich82 2015-10-6
I wonder if, perhaps, there is an error in the expected solution you posted, and if it should instead be:
S = [5, 2, 4, 6, 3, 1].';
If so, this looks suspiciously like the permutation swaps used in Numerical Recipes for e.g. bringing the permuted L*U back into A order:
row =[4 5 2 2 2 1];
n = numel(row);
S = (1:n).';
for k = n:-1:1
swap = S(k, :);
S(k, :) = S(row(k), :);
S(row(k), :) = swap;
end
If not, you'll need to more clearly describe the intermediate result after each step, as I can't seem to get anything close to your expected result.
Please accept this answer if it helps, or clarify in the comments if I'm misinterpreting something.
  1 个评论
nadia nadi
nadia nadi 2015-10-7
Dear,
Thank you very much for helping me. That's really will help me. Its exactly as you expect, is a permutation. But a little but different. I know what I wrote in my code is totally wrong, but to write general formula for this permutation is a bit tricky.
This is an explanation for what I want to do.
The string (452221) is mapped into a g-th order permutation as follows: For the string position index k ranging from g down to 1, place k at position i of the permutation (i.e., Pi). That is Pi = k where i = f ( j ). Here f is a counting function that determines i as the j-th unfilled position in the permutation counting from right-to-left. A simple example illustrates the above. If g = 6, a stringencoding obeying the range bounds for each position is: (452221) and its corresponding permutation is:(526341). I will try work on it and what I can do.
Many thanks for both of you.
Nadia,

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Image Analyst
Image Analyst 2015-10-6
I've read this many times and still can't understand what you're trying to say. First of all "row" is only one row, so "number of the row from right to left" is just going to be 1 always. Unless row means the vector and not the row. Why don't you rename your vector "vec" instead of row? I don't even know why you're talking about row(s) when everything you've mentioned is just a single row - there is no column vector or 2D matrix.
If you just want to reorder some vector, vec2, according to some sorting order given in vec1 from right to left (reverse order), then just do
vec1 =[4 5 2 2 2 1];
vec2 = vec2(fliplr(vec1));
  1 个评论
nadia nadi
nadia nadi 2015-10-7
Dear,
thanks for pay attention for my question, I explain it in my comment above. Its a permutation of [1 2 3 4 5 6] depending on [4 5 2 2 2 1].
Many thanks

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