list's index() method for nested structures OR how to match against particular value of composed element stored in list and get it's index

Question

It is possible to get index of particular element in list using index() method, following way (borrowed from here):

>>> ["foo", "bar", "baz"].index('bar')
1 

Is it possible to apply the same principle to nested structures (if not what is the closest most pythonic way)? The result should looks like this:

In [20]: list
Out[20]: [(0, 1, 2), (3, 4, 5)]
In [21]: list.someMagicFunctionHere(,4,)
Out[20]: 1

Answer 1

Search for the element in a container in your list:

def get_nested_index(list_, element):
    for index, container in enumerate(list_):
        if element in container:
             return index

    # If we made it here, it wasn't in any of the containers
    raise ValueError("{element} not in any element in list".format(element=element))

-

>>> get_nested_index([(0, 1, 2), (3, 4, 5)], 4)
1

You can read about enumerate here if you haven't seen it before.

Answer 2

adding to Navith's solution index...

def get_nested_index(list_, element):
    lt = []
    for index, container in enumerate(list_):
        if element in container:
             t = index, container.index(element)
        lt.append(t)
        t = ()
    return lt

print(get_nested_index([(0, 1, 4, 2), (3, 4, 5)], 4))

output [(0, 2), (1, 1)]