CODEWITHSUNDEEP
pythonarraysnumpy
hlin117
hlin117

Reputation: 22260

How to rearrange array based upon index array

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'd like to rearrange it based upon an input ordering. If there were a numpy function called arrange, it would do the following:

newarray = np.arrange(array, [1, 0, 3, 4, 2])
print newarray

    [20, 10, 40, 50, 30]

Formally, if the array to be reordered is m x n, and the "index" array is 1 x n, the ordering would be determined by the array called "index".

Does numpy have a function like this?

Upvotes: 72

Views: 114380

Answers (5)

Mike
Mike

Reputation: 11

Numpy has a feature called Integer Array Indexing that makes this easy:

array = np.array([10,20,30,40,50])
print(array)
ind = np.array([1,0,3,4,2])
print(ind)
newarray = array[ind]
print(newarray)
[10 20 30 40 50]
[1 0 3 4 2]
[20 10 40 50 30]

Use an index array as the starting element.

Also works for multi-dimensional arrays:

array = np.array([[10,20,30,40,50],[60,70,80,90,100]])
print(array)
ind = np.array([1,0,3,4,2])
print(ind)
newarray = array[:, ind]
print(newarray)
[[ 10  20  30  40  50]
 [ 60  70  80  90 100]]
[1 0 3 4 2]
[[ 20  10  40  50  30]
 [ 70  60  90 100  80]]

This reorders by column.

Upvotes: 1

Dong Justin
Dong Justin

Reputation: 137

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])

Upvotes: 9

Pau
Pau

Reputation: 534

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])

Upvotes: 1

Jiaming Huang
Jiaming Huang

Reputation: 101

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]]

Upvotes: 9

DSM
DSM

Reputation: 353039

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

Upvotes: 97

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