CODEWITHSUNDEEP
pythonlist
Gabriel
Gabriel

Reputation: 42449

Access element in list using indexes stored in another list

I'm looking for a general way to access elements in a list using indexes stored in another list.

For example, I have the list:

b = [[[[[0.2], [3]], [[4.5], [78]], [[1.3], [0.23]], [[6.], [9.15]]],
[[[3.1], [44]], [[1.], [66]], [[0.18], [2.3]], [[10], [7.5]]],
[[[3], [4.]], [[12.3], [12]], [[7.8], [3.7]], [[1.2], [2.1]]]]]

and I need to access the element whose index are stored in:

c = [0, 0, 0, 1, 0]

that is:

3

This won't work:

b[c[0]][c[1]][c[2]][c[3]][c[4]]

because the shape of b changes with every run of my code, which is why I need a general way of using c to access the element in b.

Something like:

b[*c]

that I would've bet would work, but it doesn't.

Upvotes: 2

Views: 108

Answers (2)

Julien Spronck
Julien Spronck

Reputation: 15433

You could use a recursive function. A recursive function is a function that calls itself. In this case, each time I call the function, I decrease the dimension of its two arguments.

b = [[[[[0.2], [3]], [[4.5], [78]], [[1.3], [0.23]], [[6.], [9.15]]],
[[[3.1], [44]], [[1.], [66]], [[0.18], [2.3]], [[10], [7.5]]],
[[[3], [4.]], [[12.3], [12]], [[7.8], [3.7]], [[1.2], [2.1]]]]]

c = [0, 0, 0, 1, 0]

def getitem(arr, indices):
    if isinstance(indices, int):
        return arr[indices]
    if len(indices) == 1:
        return arr[indices[0]]
    item = indices[0]
    new_indices = indices[1:]
    return getitem(arr[item], new_indices)

print getitem(b, c) ## prints 3

Upvotes: 2

Jared
Jared

Reputation: 26427

Use reduce (or functools.reduce for forward-compatible with Python 3)

>>> def getitems(data, keys):
...     return reduce(lambda a, b: a[b], [data]+keys)
... 
>>> getitems(b, c)
3

This assumes that keys is always a list.

Upvotes: 4

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