Try this:

def index_2d(myList, v):
    for i, x in enumerate(myList):
        if v in x:
            return (i, x.index(v))

Usage:

>>> index_2d(myList, 3)
(1, 0)

u may use this answer

c="yam"
index=[(i, fruits.index(c)) for i, fruits in enumerate(array) if c in fruits]

You don't need to define no_classes yourself. Use enumerate():

def in_list(c, classes):
    for i, sublist in enumerate(classes):
        if c in sublist:
            return i
    return -1

You can create a temporary mapping from the list2 and use it afterwards to create your output:

list1 = [[0, 1], [2, 3], [4, 5], [6, 7, 8], [9, 10, 11], [12, 13, 14]]
list2 = [[8, 2], [1, 9], [6, 4], [13, 5, 0], [14, 7, 10], [12, 11, 3]]

m = {v: i for i, t in enumerate(list2) for v in t}
out = [[m[list1[i][j]] for j in range(len(t))] for i, t in enumerate(list2)]

print(out)

Prints:

[[3, 1], [0, 5], [2, 3], [2, 4, 0], [1, 4, 5], [5, 3, 4]]

You can use np.where to return a tuple of arrays of x and y indices where a given condition holds in an array.

If a is the name of your array:

>>> np.where(a == 1)
(array([0, 0, 1, 1]), array([0, 1, 2, 3]))

If you want a list of (x, y) pairs, you could zip the two arrays:

>>> list(zip(*np.where(a == 1)))
[(0, 0), (0, 1), (1, 2), (1, 3)]

Or, even better, @jme points out that np.asarray(x).T can be a more efficient way to generate the pairs.

Assuming you don't want to reinvent the wheel I believe this question has the answer you're looking for: Is there a Numpy function to return the first index of something in an array?

import numpy as np
a = np.array([[1,2,4],[4,5,6]])
item = 4
row_indices, col_indices = np.where(a == item)

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)

You don't want a 2D list, you want a dictionary, and luckily, it's super simple to go from a 2D list (where each sublist has only two elements) to a dictionary:

prices = [['Bread',5],['Loaf',100],['Meat_Chicken',2.4],['Meat_Cow',450]]
d = dict(prices)
# {'Bread': 5, 'Loaf': 100, 'Meat_Chicken': 2.4, 'Meat_Cow': 450}

Now all you have to do is query the dictionary (O(1) lookup):

>>> d['Bread']
5

If you want to enable error checking:

>>> d.get('Bread', 'Item not found')
5
>>> d.get('Toast', 'Item not found')
'Item not found'

items_to_find_the_index_of = [9, 8, 8, 0, 3]
index_container = []
for item in items_to_find_the_index_of:
    indexes = []
    for index, sub_lst in enumerate(my2D_list):
        try:
            indexes.append((index, sub_lst.index(item)))
        except ValueError:
            pass
    index_container.append(indexes)

print(index_container)

Output:

[[(2, 0), (3, 0)], [(0, 2), (2, 1), (3, 3)], [(0, 2), (2, 1), (3, 3)], [(1, 2), (2, 2), (4, 2)], [(0, 3), (1, 0), (4, 4)]]

Loop through your list and search each sublist for the string.

Testlist = [
               ["Romeo and Juliet","Shakespeare"],
               ["Othello","Play"],
               ["Macbeth","Tragedy"]
               ]

Value = "Tragedy"

for index, lst in enumerate(Testlist):
  if Value in lst:
    print( index, lst.index(Value) )