Find index of nested item in python

Question

I\'ve been working with some relatively complex arrays such as:

array = [ \"1\", 2, [\"4\", \"5\", (\"a\", \"b\")], (\"c\", \"d\")]

and I was l

Answer 1

What you want is something like:

def myindex(lst, target):
    for index, item in enumerate(lst):
        if item == target:
            return [index]
        if isinstance(item, (list, tuple)):
            path = myindex(item, target)
            if path:
                return [index] + path
    return []

Being recursive, this will deal with arbitrary depth of nesting (up to the recursion limit).

For your example array, I get:

>>> myindex(array, "a")
[2, 2, 0]

---

As Adam alludes to in the comments, explicitly checking instance types isn't very Pythonic. A duck-typed, "easier to ask for forgiveness than permission" alternative would be:

def myindex(lst, target):
    for index, item in enumerate(lst):
        if item == target:
            return [index]
        if isinstance(item, str): # or 'basestring' in 2.x
            return []
        try:
            path = myindex(item, target)
        except TypeError:
            pass
        else:
            if path:
                return [index] + path
    return []

The specific handling of strings is necessary as even an empty string can be iterated over, causing endless recursion.