Python Matrix sorting via one column

Question

I have a n x 2 matrix of integers. The first column is a series 0,1,-1,2,-2, however these are in the order that they were compiled in from their constituent matrices. The second column is a list of indices from another list.

I would like to sort the matrix via this second column. This would be equivalent to selecting two columns of data in Excel, and sorting via Column B (where the data is in columns A and B). Keep in mind, the adjacent data in the first column of each row should be kept with its respective second column counterpart. I have looked at solutions using the following:

data[np.argsort(data[:, 0])]

But this does not seem to work. The matrix in question looks like this:

matrix([[1, 1],
        [1, 3],
        [1, 7],
        ..., 
        [2, 1021],
        [2, 1040],
        [2, 1052]])

Solution 1

You could use np.lexsort:

numpy.lexsort(keys, axis=-1)

Perform an indirect sort using a sequence of keys.

Given multiple sorting keys, which can be interpreted as columns in a spreadsheet, lexsort returns an array of integer indices that describes the sort order by multiple columns.

---

In [13]: data = np.matrix(np.arange(10)[::-1].reshape(-1,2))

In [14]: data
Out[14]: 
matrix([[9, 8],
        [7, 6],
        [5, 4],
        [3, 2],
        [1, 0]])

In [15]: temp = data.view(np.ndarray)

In [16]: np.lexsort((temp[:, 1], ))
Out[16]: array([4, 3, 2, 1, 0])

In [17]: temp[np.lexsort((temp[:, 1], ))]
Out[17]: 
array([[1, 0],
       [3, 2],
       [5, 4],
       [7, 6],
       [9, 8]])

Note if you pass more than one key to np.lexsort, the last key is the primary key. The next to last key is the second key, and so on.

---

Using np.lexsort as I show above requires the use of a temporary array because np.lexsort does not work on numpy matrices. Since temp = data.view(np.ndarray) creates a view, rather than a copy of data, it does not require much extra memory. However,

temp[np.lexsort((temp[:, 1], ))]

is a new array, which does require more memory.

There is also a way to sort by columns in-place. The idea is to view the array as a structured array with two columns. Unlike plain ndarrays, structured arrays have a sort method which allows you to specify columns as keys:

In [65]: data.dtype
Out[65]: dtype('int32')

In [66]: temp2 = data.ravel().view('int32, int32')

In [67]: temp2.sort(order = ['f1', 'f0'])

Notice that since temp2 is a view of data, it does not require allocating new memory and copying the array. Also, sorting temp2 modifies data at the same time:

In [69]: data
Out[69]: 
matrix([[1, 0],
        [3, 2],
        [5, 4],
        [7, 6],
        [9, 8]])

Solution 2

You had the right idea, just off by a few characters:

>>> import numpy as np
>>> data = np.matrix([[9, 8],
...                   [7, 6],
...                   [5, 4],
...                   [3, 2],
...                   [1, 0]])
>>> data[np.argsort(data.A[:, 1])]
matrix([[1, 0],
        [3, 2],
        [5, 4],
        [7, 6],
        [9, 8]])