Use list of nested indices to access list element

Question

How can a list of indices (called "indlst"), such as [[1,0], [3,1,2]] which corresponds to elements [1][0] and [3][1][2] of a given list (called "lst"), be used to access their respective elements? For example, given

    indlst = [[1,0], [3,1,2]]
    lst = ["a", ["b","c"], "d", ["e", ["f", "g", "h"]]]
    (required output) = [lst[1][0],lst[3][1][2]]

The output should correspond to ["b","h"]. I have no idea where to start, let alone find an efficient way to do it (as I don't think parsing strings is the most pythonic way to go about it).

EDIT: I should mention that the nested level of the indices is variable, so while [1,0] has two elements in it, [3,1,2] has three, and so forth. (examples changed accordingly).

You have a list and you have indices, just reference elements of your list using these indices. — Iron Fist
– Iron Fist, Commented Dec 3, 2016 at 5:14
@Iron The difficult part comes from not knowing beforehand how many indices there will be to reference a single given element. So instead of being able to reference an element with i and j as would be typical, I don't know how many variables will be necessary. — amateur3057
– amateur3057, Commented Dec 3, 2016 at 5:23

f5r5e5d · Accepted Answer · 2016-12-03 10:40:36Z

Recursion can grab arbitrary/deeply indexed items from nested lists:

indlst = [[1,0], [3,1,2]]
lst = ["a", ["b","c"], "d", ["e", ["f", "g", "h"]]]
#(required output) = [lst[1][0],lst[3][1][2]]


def nested_lookup(nlst, idexs):
    if len(idexs) == 1:
        return nlst[idexs[0]]
    return nested_lookup(nlst[idexs[0]], idexs[1::])

reqout = [nested_lookup(lst, i) for i in indlst]
print(reqout) 

dindx = [[2], [3, 0], [0], [2], [3, 1, 2], [3, 0], [0], [2]]    
reqout = [nested_lookup(lst, i) for i in dindx]
print(reqout)
['b', 'h']
['d', 'e', 'a', 'd', 'h', 'e', 'a', 'd']

I also found that arbitrary extra zero indices are fine:

    lst[1][0][0]
    Out[36]: 'b'

    lst[3][1][2]
    Out[37]: 'h'

    lst[3][1][2][0][0]
    Out[38]: 'h'

So if you actually know the max nesting depth you can fill in the index list values by overwriting your (variable number, shorter) index list values into the max fixed length dictionary primed with zeros using the .update() dictonary method
Then directly hard code the indices of nested list, which ignores any "extra" hard coded zero valued indices

below hard coded 4 depth:

def fix_depth_nested_lookup(nlst, idexs):
    reqout = []
    for i in idexs:
        ind = dict.fromkeys(range(4), 0)
        ind.update(dict(enumerate(i)))
        reqout.append(nlst[ind[0]][ind[1]][ind[2]][ind[3]])
    return reqout

print(fix_depth_nested_lookup(lst, indlst))
['b', 'h']

Aakash Makwana · Accepted Answer · 2016-12-03 05:00:05Z

1

you can try this code block:

required_output = []
for i,j in indlst:
    required_output.append(lst[i][j])

answered Dec 3, 2016 at 5:00

Aakash Makwana

7545 silver badges9 bronze badges

Comments

Ahasanul Haque · Accepted Answer · 2016-12-03 06:31:42Z

0

You can just iterate through and collect the value.

>>> for i,j in indlst:
...     print(lst[i][j])
... 
b
f

Or, you can use a simple list comprehension to form a list from those values.

>>> [lst[i][j] for i,j in indlst]
['b', 'f']

Edit:

For variable length, you can do the following:

>>> for i in indlst:
...     temp = lst
...     for j in i:
...         temp = temp[j]
...     print(temp)
... 
b
h

You can form a list with functions.reduce and list comprehension.

>>> from functools import reduce
>>> [reduce(lambda temp, x: temp[x], i,lst) for i in indlst]
['b', 'h']

N.B. this is a python3 solution. For python2, you can just ignore the import statement.

edited Dec 3, 2016 at 6:31

answered Dec 3, 2016 at 4:59

Ahasanul Haque

11.2k6 gold badges45 silver badges61 bronze badges

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