Access array[][]...[] element with N "[]" automatically

Question

First I apologize if this question has already been asked. I've search for it but found nothing.

Given the following code :

RHO = RHOarray(D,nbInter)
    print(RHO)
    for k in range(N):
        for l in range(D):
            j=0
            Coord=[0 for i in range(D)]
            while ((X[k][l]>mini+(j+1)*step) and (j<nbInter)):
                j+=1
            Coord[l]+=j 
            RHO[ ]
for k in range(len(RHO)):
    RHO[k]/=N
return RHO

RHO is a D-dimensions array, of size nbInter. (RHOarray is a function that creates such an array). The problem is that according to D's value, I'd need to access RHO[n] if D = 1, RHO[n][m] if D = 2 ... RHO[n][m]...[whatever indice] as D increases. Every indices would actually be Coord[0], Coord[1]... and so on, so I know what I have to put into each indices. To make the question "simpler" : how can I do so the number of "[...]" after RHO increases as much as D does ?

I hope I've been clear. I'm french and even in my native language it's still tricky to explain. Thanks in advance

ps The initial code is longer but I don't think there's need to show it entirely.

You could probably illuminate your problem by providing some sample input with the desired ouput. — user2390182, Jul 11 '18 at 15:32
So given something like `Coord = [1,2,3,4,5]`, you want `RHO[1][2][3][4][5]`? — chepner, Jul 11 '18 at 15:32
@chepner yes, exactly. But Coord's size depends directly on D, so does the number of indices after RHO. — Tiranyk, Jul 11 '18 at 15:39

blue_note · Accepted Answer · 2018-07-11T15:58:12.717

1

You could simply do something like

 indices = [1, 2, 3, 4, 5]
 value = X

 for i in indices:
     value = value[i]
 print(value)

If X is a numpy array, you could directly do X[(1,2,3)] instead of X[1, 2, 3] or X[1][2][3]

edited Jul 11 '18 at 15:58

answered Jul 11 '18 at 15:36

blue_note

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I don't understand how that solves the problem... I don't need to print anything, nor to change an indice's value. But I need the amount of indices to change according to an argument. – Tiranyk Jul 11 '18 at 15:41
Printing is not the point. what the for loop does is create the value `X[1][2][3][4][5]` from an array of indices – blue_note Jul 11 '18 at 15:46
The temporary name `value` "iterates" over the nested object. You don't need to generate the syntactic expression in order to get the value. – chepner Jul 11 '18 at 15:50
@chepner: what?? – blue_note Jul 11 '18 at 15:57
@blue_note That was meant to explain to Tiranyk that the loop calculates the expression he wants, whether or not it literally constructs the equivalent "hard-coded" expression. – chepner Jul 11 '18 at 16:24
@chepner: aah, ok – blue_note Jul 11 '18 at 16:43
@blue_note thanks a lot for your editing, I understood. – Tiranyk Jul 11 '18 at 17:00

Access array[][]...[] element with N "[]" automatically

1 Answers1