How to rearrange array based upon index array

Question

I\'m looking for a one line solution that would help me do the following.

Suppose I have

array = np.array([10, 20, 30, 40, 50])

I\

Answer 1

for those whose index is 2d array, you can use map function. Here is an example:

a = np.random.randn(3, 3)
print(a)
print(np.argsort(a))

print(np.array(list(map(lambda x, y: y[x], np.argsort(a), a))))

the output is

[[-1.42167035  0.62520498  2.02054623]
 [-0.17966393 -0.01561566  0.24480554]
 [ 1.10568543  0.00298402 -0.71397599]]
[[0 1 2]
 [0 1 2]
 [2 1 0]]
[[-1.42167035  0.62520498  2.02054623]
 [-0.17966393 -0.01561566  0.24480554]
 [-0.71397599  0.00298402  1.10568543]]

Answer 2

If you want to sort it but descending:

a = np.array([1,2,3,4,5])
np.argsort(a)
> array([0, 1, 2, 3, 4])
np.argsort(-a)
> array([4, 3, 2, 1, 0])

Answer 3

You can simply use your "index" list directly, as, well, an index array:

>>> arr = np.array([10, 20, 30, 40, 50])
>>> idx = [1, 0, 3, 4, 2]
>>> arr[idx]
array([20, 10, 40, 50, 30])

It tends to be much faster if idx is already an ndarray and not a list, even though it'll work either way:

>>> %timeit arr[idx]
100000 loops, best of 3: 2.11 µs per loop
>>> ai = np.array(idx)
>>> %timeit arr[ai]
1000000 loops, best of 3: 296 ns per loop

Answer 4

For those who have the same confusion, I am actually looking for a slightly different version of "rearrange array based upon index". In my situation, the index array is indexing the target array instead of the source array. In other words, I am try to rearrange an array based on its position in the new array.

In this case, simply apply an argsort before indexing. E.g.

>>> arr = np.array([10, 20, 30, 40, 50])
>>> idx = [1, 0, 3, 4, 2]
>>> arr[np.argsort(idx)]
array([20, 10, 50, 30, 40])

Note the difference between this result and the desired result by op.

One can verify back and forth

>>> arr[np.argsort(idx)][idx] == arr
array([ True,  True,  True,  True,  True])
>>> arr[idx][np.argsort(idx)] == arr
array([ True,  True,  True,  True,  True])