Reputation: 434
I have a multidimensional list F
, holding elements of some type. So, if for example the rank is 4, then the elements of F
can be accessed by something like F[a][b][c][d]
.
Given a list L=[a,b,c,d]
, I would like to access F[a][b][c][d]
. My problem is that my rank is going to be changing, so I cannot just have F[L[0]][L[1]][L[2]][L[3]]
.
Ideally, I would like to be able to do F[L]
and get the element F[a][b][c][d]
. I think something like this can be done with numpy
, but for the types of arrays that I'm using, numpy
is not suitable, so I want to do it with python lists.
How can I have something like the above?
Edit: For a specific example of what I'm trying to achieve, see the demo in Martijn's answer.
Upvotes: 2
Views: 3744
Reputation: 1121416
You can use the reduce()
function to access consecutive elements:
from functools import reduce # forward compatibility
import operator
reduce(operator.getitem, indices, somelist)
In Python 3 reduce
was moved to the functools
module, but in Python 2.6 and up you can always access it in that location.
The above uses the operator.getitem()
function to apply each index to the previous result (starting at somelist
).
Demo:
>>> import operator
>>> somelist = ['index0', ['index10', 'index11', ['index120', 'index121', ['index1220']]]]
>>> indices = [1, 2, 2, 0]
>>> reduce(operator.getitem, indices, somelist)
'index1220'
Upvotes: 12
Reputation: 31339
Something like this?
def get_element(lst, indices):
if indices:
return get_element(lst[indices[0]], indices[1:])
return lst
Test:
get_element([[["element1"]], [["element2"]], "element3"], [2])
'element3'
get_element([[["element1"]], [["element2"]], "element3"], [0, 0])
['element1']
Or if you want an iterative version:
def get_element(lst, indices):
res = lst
for index in indices:
res = res[index]
return res
Test:
get_element([[["element1"]], [["element2"]], "element3"], [1, 0])
['element2']
Upvotes: 3