Searching matrices in Mathematica 8 - Trying to find other elements on the same row as X

Question

The text in italics describes my general goal, if anyone is interested. Question is underneath.

I am trying to graph the energy levels of simple molecules using Mathematica 8. My method is crude, and goes as this:

1. Find eigenvalues of simple Hückel matrix.
2. Delete duplicates and determine size of list.
3. Evaluate the number of degeneracies by comparing duplicate list with no-duplicate list.
4. Create a n x 2 zero matrix where n is the number of unique energy levels.

5. Fill first column with unique energy levels, second column with degeneracies.

The matrix generated in step 5 can look like this:

(1  2)
(3  1)   ==    M
(-1 1)

I wish to evaluate the maximum of column 2, and then find the value of the element in the same row, but in column 1. In this case, the answer I am looking for is 1.

These commands both evaluate to -1:

Extract[M[[All, 1]], M[[Max[M[[All, 2]]], 1]]]
M[[Max[M[[All, 1]]], 1]]

which is not the answer I want.

Any tips?

EDIT: This

Part[Part[Position[M, Max[M[[All, 2]]]], 1], 1]

works, but I don't understand why I have to use Part[] twice.

Answer 1

The inner Part gives you the first occurance of the maximum. Position returns a list of positions, even if there is only one element that has the maximum value, like this:

 M = {{2, 2}, {2, 3}, {2, 2}, {1, 1}}

 {{2, 2}, {2, 3}, {2, 2}, {1, 1}}

Position[M, Max[M[[All, 2]]]]

 {{2, 2}}

So you want the first element in the first element of this output. You could condense your code like this:

Position[M, Max[M[[All, 2]]]][[1, 1]]

However, one thing that I think your code needs to handle better is this case:

 M = {{3, 2}, {2, 3}, {2, 2}, {1, 1}}

3, 2}, {2, 3}, {2, 2}, {1, 1}}

Position[M, Max[M[[All, 2]]]]

{{1, 1}, {2, 2}}

You will get the wrong answer with your code in this case.

Better would be:

M[[All, 1]][[Position[M[[All, 2]], Max[M[[All, 2]]]][[1, 1]] ]]

Or alternatively

M[[Position[M[[All, 2]], Max[M[[All, 2]]]][[1, 1]], 1]]

Answer 2

m = {{1, 2}, {3, 1}, {-1, 1}}
max = Max[m[[All, 2]]]

So find the position of the max and replace the second column with the first:

pos=Position[m, max] /. {x_,_}:>{x,1}
{{1,1}}

Then take the first element from pos, i.e. {1,1} and sub use it in Part

m[[Sequence @@ First[pos]]]
1

But having said that I prefer something like this:

Cases[m, {x_, max} :> x]
{1}

The result is a list. You could either use First@Cases[...] or you might want to keep a list of results to cover cases where the maximum value occurs more than once in a column.

Answer 3

If you only want a single column one value in the case of duplicate maximum values in column two I suggest that you make use of Ordering:

m = {{1, 3}, {1, 8}, {5, 7}, {2, 2}, {1, 9}, {4, 9}, {5, 6}};

m[[ Ordering[m[[All, 2]], -1], 1 ]]

{4}