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:
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.
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]]