CODEWITHSUNDEEP
pythonarraysnumpymultidimensional-array
Kernelf
Kernelf

Reputation: 103

Access one dimensional array with array of array indices in python

x = array([0. , 0.5, 1. , 1.5, 2. , 2.5, 3. , 3.5, 4. , 4.5, 5. ])

nbd = array([array([1], dtype=int64), array([0, 2], dtype=int64), array([1, 3], dtype=int64),
 array([2, 4], dtype=int64), array([3, 5], dtype=int64), array([4, 6], dtype=int64),
 array([5, 7], dtype=int64), array([6, 8], dtype=int64), array([7, 9], dtype=int64),
 array([ 8, 10], dtype=int64), array([9], dtype=int64)], dtype=object)

nbd is array of arrays with indices for x.

I am looking for array of arrays 'x_nbd' such that it is holding x values at nbd indices.

i.e.,

x_nbd = array([array([0.5]), array([0., 1.]), array([0.5, 1.5]), array([1., 2.]),
       array([1.5, 2.5]), array([2., 3.]), array([2.5, 3.5]),
       array([3., 4.]), array([3.5, 4.5]), array([4., 5.]), array([4.5])],
      dtype=object)

I have tried with x_nbd = x[nbd] but no luck. I found a way but it is very slow for large data, as it is running each element through a loop. That is

 x_nbd = np.array([np.array(x[e]) for e in nbd]).

Moreover, it is showing "VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray."

Is there any faster way to get this in python? or How to actually achieve this using numpy in python ?

Upvotes: 0

Views: 440

Answers (2)

hpaulj
hpaulj

Reputation: 231615

Forget about the unnecessary array wrappers. The simple list comprehension is the right way - to go from a list of indices to a list arrays. There's nothing "multidimensional" about this.

In [84]: array=np.array
    ...: int64=np.int64
    ...: x = array([0. , 0.5, 1. , 1.5, 2. , 2.5, 3. , 3.5, 4. , 4.5, 5. ])
    ...: 
    ...: nbd = [array([1], dtype=int64), array([0, 2], dtype=int64), array([1, 3], dtype=int64)
    ...: ,
    ...:  array([2, 4], dtype=int64), array([3, 5], dtype=int64), array([4, 6], dtype=int64),
    ...:  array([5, 7], dtype=int64), array([6, 8], dtype=int64), array([7, 9], dtype=int64),
    ...:  array([ 8, 10], dtype=int64), array([9], dtype=int64)]
In [85]: [x[i] for i in nbd]
Out[85]: 
[array([0.5]),
 array([0., 1.]),
 array([0.5, 1.5]),
 array([1., 2.]),
 array([1.5, 2.5]),
 array([2., 3.]),
 array([2.5, 3.5]),
 array([3., 4.]),
 array([3.5, 4.5]),
 array([4., 5.]),
 array([4.5])]

Now if indexing arrays were all the same size, you could make a 2d array:

In [86]: np.stack(nbd[1:-1])
Out[86]: 
array([[ 0,  2],
       [ 1,  3],
       [ 2,  4],
       [ 3,  5],
       [ 4,  6],
       [ 5,  7],
       [ 6,  8],
       [ 7,  9],
       [ 8, 10]])
In [87]: x[np.stack(nbd[1:-1])]
Out[87]: 
array([[0. , 1. ],
       [0.5, 1.5],
       [1. , 2. ],
       [1.5, 2.5],
       [2. , 3. ],
       [2.5, 3.5],
       [3. , 4. ],
       [3.5, 4.5],
       [4. , 5. ]])

Upvotes: 1

QWERTYL
QWERTYL

Reputation: 1383

You could try

x_nbd = array([array([x[i], nbd[i]]) for i in range(len(x))])

I noticed you have an array([0.5]) at the start of your expected output which I assume to be a typo, but you could easily prepend this element without changing the time complexity.

Upvotes: 0

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