`numpy:lexsort`

Signature

def lexsort ( keys , axis = -1 )

Summary

Perform an indirect stable sort using a sequence of keys.

Extended Summary

Given multiple sorting keys, lexsort returns an array of integer indices that describes the sort order by multiple keys. The last key in the sequence is used for the primary sort order, ties are broken by the second-to-last key, and so on.

Parameters

keys : (k, m, n, ...) array-like: The k keys to be sorted. The last key (e.g, the last row if keys is a 2D array) is the primary sort key. Each element of keys along the zeroth axis must be an array-like object of the same shape.
axis : int, optional: Axis to be indirectly sorted. By default, sort over the last axis of each sequence. Separate slices along axis sorted over independently; see last example.

Returns

indices : (m, n, ...) ndarray of ints: Array of indices that sort the keys along the specified axis.

Examples

Sort names: first by surname, then by name.

import numpy as np
surnames =    ('Hertz',    'Galilei', 'Hertz')
first_names = ('Heinrich', 'Galileo', 'Gustav')
ind = np.lexsort((first_names, surnames))
ind

✓

[surnames[i] + ", " + first_names[i] for i in ind]

✓

Sort according to two numerical keys, first by elements of ``a``, then breaking ties according to elements of ``b``:

a = [1, 5, 1, 4, 3, 4, 4]  # First sequence
b = [9, 4, 0, 4, 0, 2, 1]  # Second sequence
ind = np.lexsort((b, a))  # Sort by `a`, then by `b`
ind
[(a[i], b[i]) for i in ind]

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Compare against `argsort`, which would sort each key independently.

np.argsort((b, a), kind='stable')

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To sort lexicographically with `argsort`, we would need to provide a structured array.

x = np.array([(ai, bi) for ai, bi in zip(a, b)],
             dtype = np.dtype([('x', int), ('y', int)]))
np.argsort(x)  # or np.argsort(x, order=('x', 'y'))

✓

The zeroth axis of `keys` always corresponds with the sequence of keys, so 2D arrays are treated just like other sequences of keys.

arr = np.asarray([b, a])
ind2 = np.lexsort(arr)
np.testing.assert_equal(ind2, ind)

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Accordingly, the `axis` parameter refers to an axis of *each* key, not of the `keys` argument itself. For instance, the array ``arr`` is treated as a sequence of two 1-D keys, so specifying ``axis=0`` is equivalent to using the default axis, ``axis=-1``.

np.testing.assert_equal(np.lexsort(arr, axis=0),
                        np.lexsort(arr, axis=-1))

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For higher-dimensional arrays, the axis parameter begins to matter. The resulting array has the same shape as each key, and the values are what we would expect if `lexsort` were performed on corresponding slices of the keys independently. For instance,

x = [[1, 2, 3, 4],
     [4, 3, 2, 1],
     [2, 1, 4, 3]]
y = [[2, 2, 1, 1],
     [1, 2, 1, 2],
     [1, 1, 2, 1]]
np.lexsort((x, y), axis=1)

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Each row of the result is what we would expect if we were to perform `lexsort` on the corresponding row of the keys:

for i in range(3):
    print(np.lexsort((x[i], y[i])))

✓

Aliases

numpy.lexsort