using index() on multidimensional lists

Question

For a one dimensional list, the index of an item is found as follows:

 a_list = ['a', 'b', 'new', 'mpilgrim', 'new']
 a_list.index('mpilgrim')

What is the equivalent for a 2 or n dimensional list?

Edit: I have added an example to clarify: If I have a 3 dimensional list as follows

b_list = [
          [1,2],
          [3,4],
          [5,6],
          [7,8]
         ],
         [
          [5,2],
          [3,7],
          [6,6],
          [7,9]
         ]

Now lets say I want to identify a certain value in this list. If I know the index of the 1st and 2nd dimesion but don't know the zero-th index for the value I want, how do I go about finding the zero-th index?

Would it be something like:

  target_value = 7
  b_list[0].index(target_value)

With the output being an integer: 0

Answer 1

I don't know of an automatic way to do it, but if

a = [[1,2],[3,4],[5,6]]

and you want to find the location of 3, you can do:

x = [x for x in a if 3 in x][0]

print 'The index is (%d,%d)'%(a.index(x),x.index(3))

The output is:

The index is (1,0)

Answer 2

For two dimensional list; you can iterate over rows and using .index function for looking for item:

def find(l, elem):
    for row, i in enumerate(l):
        try:
            column = i.index(elem)
        except ValueError:
            continue
        return row, column
    return -1

tl = [[1,2,3],[4,5,6],[7,8,9]]

print(find(tl, 6)) # (1,2)
print(find(tl, 1)) # (0,0)
print(find(tl, 9)) # (2,2)
print(find(tl, 12)) # -1