CODEWITHSUNDEEP
pythonarraysnumpypandas
blaz
blaz

Reputation: 4248

Indexing the unique rows of an array

I would like to get the indices of the unique rows in an array. A unique row should have its own index (starting with zero). Here is an example:

import numpy as np

a = np.array([[ 0.,  1.],
              [ 0.,  2.],
              [ 0.,  3.],
              [ 0.,  1.],
              [ 0.,  2.],
              [ 0.,  3.],
              [ 0.,  1.],
              [ 0.,  2.],
              [ 0.,  3.],
              [ 1.,  1.],
              [ 1.,  2.],
              [ 1.,  3.],
              [ 1.,  1.],
              [ 1.,  2.],
              [ 1.,  3.],
              [ 1.,  1.],
              [ 1.,  2.],
              [ 1.,  3.]])

In the above array there are six unique rows:

import pandas as pd
b = pd.DataFrame(a).drop_duplicates().values

    array([[ 0.,  1.],   
           [ 0.,  2.],
           [ 0.,  3.],
           [ 1.,  1.],
           [ 1.,  2.],
           [ 1.,  3.]])

Each row represents an index (0, 1, 2, 3, 4 ,5). In order to get the indices of unique rows in array a, the result would be:

[0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 4, 5, 3, 4, 5, 3, 4, 5]

How can I get to this result in an efficient way?

Upvotes: 0

Views: 78

Answers (3)

blaz
blaz

Reputation: 4248

This is what I got:

b = pd.DataFrame(a).drop_duplicates()
indexed_rows = np.zeros(a.shape[0], dtype=int)
for index, i in enumerate(a):
    for unique_row, j in enumerate(b.values):
        if np.all(i==j):
            indexed_rows[index] = unique_row

The returned result is:

array([0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 4, 5, 3, 4, 5, 3, 4, 5])

​

Upvotes: 0

B. M.
B. M.

Reputation: 18648

A pure numpy solution :

av = a.view(np.complex)
_,inv = np.unique(av,return_inverse=True)

Then inv is :

array([0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 4, 5, 3, 4, 5, 3, 4, 5], dtype=int64)

np.complexis for packing the two components, preserving order. for other types, other approaches are possible.

Upvotes: 3

Maciej A. Czyzewski
Maciej A. Czyzewski

Reputation: 1529

Solution without numpy and pandas:

a = [[0, 1],
     [0, 2],
     [0, 3],
     [0, 1],
     [0, 2],
     [0, 3],
     [0, 1],
     [0, 2],
     [0, 3],
     [1, 1],
     [1, 2],
     [1, 3],
     [1, 1],
     [1, 2],
     [1, 3],
     [1, 1],
     [1, 2],
     [1, 3]]

b = []

#= ALGORITHM

point = -1                                               # Increment
cache = [[-1 for x in range(1000)] for x in range(1000)] # Change to dynamic

for i in a:
    x = i[0]; y = i[1]

    # Check what's going on here...
    # print("x: {0} y: {1} --> {2} (cache)".format(x, y, cache[x][y]))

    if cache[x][y] == -1:
        point += 1
        cache[x][y] = point
        b.append(point)
    else:
        b.append(cache[x][y])

#= TESTING

print(b) # [0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 4, 5, 3, 4, 5, 3, 4, 5]

Upvotes: 0

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