Generalise slicing operation in a NumPy array

Question

This question is based on this older question:

Given an array:

 In [122]: arr = np.array([[1, 3, 7], [4, 9, 8]]); arr Out[122]: array([[1, 3, 7], [4, 9, 8]]) 

And given its indices:

 In [127]: np.indices(arr.shape) Out[127]: array([[[0, 0, 0], [1, 1, 1]], [[0, 1, 2], [0, 1, 2]]]) 

How would I be able to stack them neatly one against the other to form a new 2D array? This is what I'd like:

 array([[0, 0, 1], [0, 1, 3], [0, 2, 7], [1, 0, 4], [1, 1, 9], [1, 2, 8]]) 

This solution by Divakar is what I currently use for 2D arrays:

def indices_merged_arr(arr):
    m,n = arr.shape
    I,J = np.ogrid[:m,:n]
    out = np.empty((m,n,3), dtype=arr.dtype)
    out[...,0] = I
    out[...,1] = J
    out[...,2] = arr
    out.shape = (-1,3)
    return out

Now, if I wanted to pass a 3D array, I need to modify this function:

def indices_merged_arr(arr):
    m,n,k = arr.shape   # here
    I,J,K = np.ogrid[:m,:n,:k]   # here
    out = np.empty((m,n,k,4), dtype=arr.dtype)   # here
    out[...,0] = I
    out[...,1] = J
    out[...,2] = K     # here
    out[...,3] = arr
    out.shape = (-1,4)   # here
    return out

But this function now works for 3D arrays only - I can't pass a 2D array to it.

Is there some sort of way I can generalise this to work for any dimension? Here's my attempt:

def indices_merged_arr_general(arr):
    tup = arr.shape   
    idx = np.ogrid[????]   # not sure what to do here....
    out = np.empty(tup + (len(tup) + 1, ), dtype=arr.dtype) 
    for i, j in enumerate(idx):
        out[...,i] = j
    out[...,len(tup) - 1] = arr
    out.shape = (-1, len(tup)
    return out

I'm having trouble with this line:

idx = np.ogrid[????]   

How can I get this working?

Answer 1

Here's the extension to handle generic ndarrays -

def indices_merged_arr_generic(arr, arr_pos="last"):
    n = arr.ndim
    grid = np.ogrid[tuple(map(slice, arr.shape))]
    out = np.empty(arr.shape + (n+1,), dtype=np.result_type(arr.dtype, int))

    if arr_pos=="first":
        offset = 1
    elif arr_pos=="last":
        offset = 0
    else:
        raise Exception("Invalid arr_pos")        

    for i in range(n):
        out[...,i+offset] = grid[i]
    out[...,-1+offset] = arr
    out.shape = (-1,n+1)

    return out

Sample runs

2D case :

In [252]: arr
Out[252]: 
array([[37, 32, 73],
       [95, 80, 97]])

In [253]: indices_merged_arr_generic(arr)
Out[253]: 
array([[ 0,  0, 37],
       [ 0,  1, 32],
       [ 0,  2, 73],
       [ 1,  0, 95],
       [ 1,  1, 80],
       [ 1,  2, 97]])

In [254]: indices_merged_arr_generic(arr, arr_pos='first')
Out[254]: 
array([[37,  0,  0],
       [32,  0,  1],
       [73,  0,  2],
       [95,  1,  0],
       [80,  1,  1],
       [97,  1,  2]])

3D case :

In [226]: arr
Out[226]: 
array([[[35, 45, 33],
        [48, 38, 20],
        [69, 31, 90]],

       [[73, 65, 73],
        [27, 51, 45],
        [89, 50, 74]]])

In [227]: indices_merged_arr_generic(arr)
Out[227]: 
array([[ 0,  0,  0, 35],
       [ 0,  0,  1, 45],
       [ 0,  0,  2, 33],
       [ 0,  1,  0, 48],
       [ 0,  1,  1, 38],
       [ 0,  1,  2, 20],
       [ 0,  2,  0, 69],
       [ 0,  2,  1, 31],
       [ 0,  2,  2, 90],
       [ 1,  0,  0, 73],
       [ 1,  0,  1, 65],
       [ 1,  0,  2, 73],
       [ 1,  1,  0, 27],
       [ 1,  1,  1, 51],
       [ 1,  1,  2, 45],
       [ 1,  2,  0, 89],
       [ 1,  2,  1, 50],
       [ 1,  2,  2, 74]])

Answer 2

For large arrays, AFAIK, senderle's cartesian_product is the fastest way ¹ to generate cartesian products using NumPy :

---

In [372]: A = np.random.random((100,100,100))

In [373]: %timeit indices_merged_arr_generic_using_cp(A)
100 loops, best of 3: 16.8 ms per loop

In [374]: %timeit indices_merged_arr_generic(A)
10 loops, best of 3: 28.9 ms per loop

---

Here is the setup I used to benchmark. Below, indices_merged_arr_generic_using_cp is a modification of senderle's cartesian_product to include the flattened array beside with the cartesian product:

import numpy as np
import functools

def indices_merged_arr_generic_using_cp(arr):
    """
    Based on cartesian_product
    http://stackoverflow.com/a/11146645/190597 (senderle)
    """
    shape = arr.shape
    arrays = [np.arange(s, dtype='int') for s in shape]
    broadcastable = np.ix_(*arrays)
    broadcasted = np.broadcast_arrays(*broadcastable)
    rows, cols = functools.reduce(np.multiply, broadcasted[0].shape), len(broadcasted)+1
    out = np.empty(rows * cols, dtype=arr.dtype)
    start, end = 0, rows
    for a in broadcasted:
        out[start:end] = a.reshape(-1)
        start, end = end, end + rows
    out[start:] = arr.flatten()
    return out.reshape(cols, rows).T

def indices_merged_arr_generic(arr):
    """
    https://stackoverflow.com/a/46135084/190597 (Divakar)
    """
    n = arr.ndim
    grid = np.ogrid[tuple(map(slice, arr.shape))]
    out = np.empty(arr.shape + (n+1,), dtype=arr.dtype)
    for i in range(n):
        out[...,i] = grid[i]
    out[...,-1] = arr
    out.shape = (-1,n+1)
    return out

---

¹ Note that above I actually used senderle's cartesian_product_transpose . For me, this is the fastest version. For others, including senderle, cartesian_product is faster.

Answer 3

ndenumerate iterates on the elements, as opposed to the dimensions in the other solutions. So I don't expect it to win the speed tests. But here's a way of using it

In [588]:  arr = np.array([[1, 3, 7], [4, 9, 8]])
In [589]: arr
Out[589]: 
array([[1, 3, 7],
       [4, 9, 8]])
In [590]: list(np.ndenumerate(arr))
Out[590]: [((0, 0), 1), ((0, 1), 3), ((0, 2), 7), ((1, 0), 4), ((1, 1), 9), ((1, 2), 8)]

In py3 * unpacking can be used in a tuple, so the nested tuples can be flattened:

In [591]: [(*ij,v) for ij,v in np.ndenumerate(arr)]
Out[591]: [(0, 0, 1), (0, 1, 3), (0, 2, 7), (1, 0, 4), (1, 1, 9), (1, 2, 8)]
In [592]: np.array(_)
Out[592]: 
array([[0, 0, 1],
       [0, 1, 3],
       [0, 2, 7],
       [1, 0, 4],
       [1, 1, 9],
       [1, 2, 8]])

And it generalizes nicely to more dimensions:

In [593]: arr3 = np.arange(24).reshape(2,3,4)
In [594]: np.array([(*ij,v) for ij,v in np.ndenumerate(arr3)])
Out[594]: 
array([[ 0,  0,  0,  0],
       [ 0,  0,  1,  1],
       [ 0,  0,  2,  2],
       [ 0,  0,  3,  3],
       [ 0,  1,  0,  4],
       [ 0,  1,  1,  5],
       ....
       [ 1,  2,  3, 23]])

---

With these small samples, it's actually faster than @Diakar's function. :)

In [598]: timeit indices_merged_arr_generic(arr)
52.8 µs ± 271 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [599]: timeit indices_merged_arr_generic(arr3)
66.9 µs ± 434 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [600]: timeit np.array([(*ij,v) for ij,v in np.ndenumerate(arr)])
21.2 µs ± 40.5 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [601]: timeit np.array([(*ij,v) for ij,v in np.ndenumerate(arr3)])
59.4 µs ± 1.28 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

But for a large 3d array it is much slower

In [602]: A = np.random.random((100,100,100))
In [603]: timeit indices_merged_arr_generic(A)
50.3 ms ± 141 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [604]: timeit np.array([(*ij,v) for ij,v in np.ndenumerate(A)])
2.39 s ± 11.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

And with `@unutbu's - slower for small, faster for big:

In [609]: timeit indices_merged_arr_generic_using_cp(arr)
104 µs ± 1.78 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [610]: timeit indices_merged_arr_generic_using_cp(arr3)
141 µs ± 1.09 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [611]: timeit indices_merged_arr_generic_using_cp(A)
31.1 ms ± 1.28 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Answer 4

We can use the following oneliner:

from numpy import hstack, array, meshgrid

hstack((
        array(meshgrid(*map(range, t.shape))).T.reshape(-1,t.ndim),
        t.flatten().reshape(-1,1)
       ))

Here we first use map(range, t.shape) to construct an iterable of range s. By using np.meshgrid(..).T.reshape(-1, t.dim) we construct the first part of the table: an n×m matrix with n the number of elements of t , and m the number of dimensions, we then add a flattened version of t at the right.