根据另一个数组的值（未排序，但分组）将 NumPy 数组拆分为子数组

Question

Suppose I have two NumPy arrays

x = [[1, 2, 8],
     [2, 9, 1],
     [3, 8, 9],
     [4, 3, 5],
     [5, 2, 3],
     [6, 4, 7],
     [7, 2, 3],
     [8, 2, 2],
     [9, 5, 3],
     [10, 2, 3],
     [11, 2, 4]]
y = [0, 0, 1, 0, 1, 1, 2, 2, 2, 0, 0] 

Note: (values in x are not sorted in any way. I chose this example to better illustrate the example) (These are just two examples of x and y . values of x and y can be arbitrarily many different numbers and y can have arbitrarily different numbers, but there are always as many values in x as there are in y )

I want to efficiently split the array x into sub-arrays according to the values in y .

My desired outputs would be

z_0 = [[1, 2, 8],
       [2, 9, 1],
       [4, 3, 5],
       [10, 2, 3],
       [11, 2, 4]]
z_1 = [[3, 8, 9],
       [5, 2, 3],
       [6, 4, 7],]
z_2 = [[7, 2, 3],
       [8, 2, 2],
       [9, 5, 3]]

Assuming that y starts with zero and is not sorted but grouped, what is the most efficient way to do this?

Note: This question is the unsorted version of this question: Split a NumPy array into subarrays according to the values (sorted in ascending order) of another array

Answer 1

One way to solve this is to build up a list of filter indexes for each y value and then simply select those elements of x . For example:

z_0 = x[[i for i, v in enumerate(y) if v == 0]]
z_1 = x[[i for i, v in enumerate(y) if v == 1]]
z_2 = x[[i for i, v in enumerate(y) if v == 2]]

Output

array([[ 1,  2,  8],
       [ 2,  9,  1],
       [ 4,  3,  5],
       [10,  2,  3],
       [11,  2,  4]])
array([[3, 8, 9],
       [5, 2, 3],
       [6, 4, 7]])
array([[7, 2, 3],
       [8, 2, 2],
       [9, 5, 3]])

If you want to be more generic and support different sets of numbers in y , you could use a comprehension to produce a list of arrays eg

z = [x[[i for i, v in enumerate(y) if v == m]] for m in set(y)]

Output:

[array([[ 1,  2,  8],
       [ 2,  9,  1],
       [ 4,  3,  5],
       [10,  2,  3],
       [11,  2,  4]]),
 array([[3, 8, 9],
       [5, 2, 3],
       [6, 4, 7]]),
 array([[7, 2, 3],
       [8, 2, 2],
       [9, 5, 3]])]

If y is also an np.array and the same length as x you can simplify this to use boolean indexing:

z = [x[y==m] for m in set(y)]

Output is the same as above.

Answer 2

Just use list comprehension and boolean indexing

x = np.array(x)
y = np.array(y)

z = [x[y == i] for i in range(y.max() + 1)]

z
Out[]: 
[array([[ 1,  2,  8],
        [ 2,  9,  1],
        [ 4,  3,  5],
        [10,  2,  3],
        [11,  2,  4]]),
 array([[3, 8, 9],
        [5, 2, 3],
        [6, 4, 7]]),
 array([[7, 2, 3],
        [8, 2, 2],
        [9, 5, 3]])]

Answer 3

Slight variation.

from operator import itemgetter
label = itemgetter(1)

Associate the implied information with the label... (index,label)

y1 = [thing for thing in enumerate(y)]

Sort on the label

y1.sort(key=label)

Group by label and construct the results

import itertools
d = {}
for key,group in itertools.groupby(y1,label):
    d[f'z{key}'] = [x[i] for i,k in group]

---

Pandas solution:

>>> import pandas as pd
>>> >>> df = pd.DataFrame({'points':[thing for thing in x],'cat':y})
>>> z = df.groupby('cat').agg(list)
>>> z       
                                                points
cat
0    [[1, 2, 8], [2, 9, 1], [4, 3, 5], [10, 2, 3], ...
1                    [[3, 8, 9], [5, 2, 3], [6, 4, 7]]
2                    [[7, 2, 3], [8, 2, 2], [9, 5, 3]]