Dynamic Access of Multi dimensional Python Array

Question

I am a python newbie. I was confused on how to access array element dynamically.

I have a list b= [1,2,5,8] that I dynamically obtain so its length can vary. With help of this list I need to update multi-dimensional array as mArr[1] [2] [5] [8] . The length of the list and array dimension matches as given in the example

Basically, I am looking a technique to access a multi dimensional array with respect to the list "b" as in the form of : marr[b]. This m-array is also dynamically created.

I tried looking on to tutorials of numpy but was not figure out the solution.Am I missing something?

Thanks in advance.

JBernardo · Accepted Answer · 2011-10-17 03:54:43Z

2

if the dimensions are [1,2,5,8] you can use numbers 0, 0..1, 0..4, 0..7 for each dimension.

Numpy lets you access positions with tuples:

shape = [1, 2, 5, 8]
pos = [0, 1, 1, 3]

my_array = np.ones(shape)
my_array[tuple(pos)] # will return 1

edited Oct 17, 2011 at 3:54

answered Oct 17, 2011 at 3:47

JBernardo

33.6k13 gold badges92 silver badges120 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Winston Ewert Over a year ago

no need to do that reshape thing. Just pass the shape as parameter to np.ones

iinception Over a year ago

tx for the quick reply..can you elaborate it more..i am confused in third line...what if, if I have b=[3,4,7,8,10,20] next time, how to access my array as myarr[3][4][7][8][10][20]....

Winston Ewert Over a year ago

@iinception, the third line creates an array, only the fourth line is concerned with accessing.

iinception Over a year ago

tx all...tuple was the keyword I was looking forr.. :)

JBernardo Over a year ago

@iinception, now I understood. [1,2,5,8] is not the dimension, so you just use b = (1,2,5,8) and myarr[b] or myarr[tuple(b)] if b is a list

|

imm · Accepted Answer · 2011-10-17 03:52:18Z

1

You could create a function like:

def array_update(b, marr, value):
  if len(b) > 1:
    return array_update(b[1:], marr[b[0]], value)
  marr[b[0]] = value

Given b=[1,2,5,8], to set the value of mArr[1][2][5][8] to foo, you would call:

array_update(b, mArr, 'foo')

answered Oct 17, 2011 at 3:52

imm

5,9391 gold badge29 silver badges32 bronze badges

Collectives™ on Stack Overflow

Dynamic Access of Multi dimensional Python Array

2 Answers 2

8 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

8 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related