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 looking for a way to find an item and retrieve is "index" (Is it OK to refer to the location of item such as "a" - that is inside a Tuple as index on the same level as array?)

Now my first thought was to use something like a simple helper function such as:

def myindex(nestedlist, item):
    for i in nestedlist:
        if item in i:
         index = []
         index.append(i)
         index.append(i.index(item))
         return index

But I'm sure you can guess that such a function won't do much good, especially since I don't know in advance how many levels the array might have, and what each level might contain (in terms of data type/structure)

Any hint in the right direction would be highly appreciate!

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.

Answer 2

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

def find_index(array, item, index=None):
    if not index:
        index = []
    try:
        i = array.index(item)
    except:
        for new_array in array:
           if hasattr(new_array, '__iter__'):
               i = find_index(new_array, item, index+[array.index(new_array)])
               if i:
                   return i
    else:
        return index + [i]
    return None

This gives:

>>> find_index(array, 1)
>>> find_index(array, "1")
[0]
>>> find_index(array, 2)
[1]
>>> find_index(array, "4")
[2, 0]

Answer 3

I'm a bit late to the party, but I spent several minutes on it so I feel like it ought to be posted anyway :)

def getindex(container, target, chain=None):
    if chain is None: chain = list()
    for idx, item in enumerate(container):
        if item == target:
            return chain + [idx]
        if isinstance(item, collections.Iterable) and not isinstance(item, str):
            # this should be ... not isinstance(item, basestring) in python2.x
            result = getindex(item, target, chain + [idx])
            if result:
                return result
    return None