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I need to figure out how I can find all the index of a value in a 2d numpy array.

For example, I have the following 2d array:

([[1 1 0 0],
  [0 0 1 1],
  [0 0 0 0]])

I need to find the index of all the 1's and 0's.

1: [(0, 0), (0, 1), (1, 2), (1, 3)]
0: [(0, 2), (0, 3), (1, 0), (1, 1), (the entire all row)]

I tried this but it doesn't give me all the indexes:

t = [(index, row.index(1)) for index, row in enumerate(x) if 1 in row]

Basically, it gives me only one of the index in each row [(0, 0), (1, 2)].

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4
  • is this actually a numpy array? Commented Nov 27, 2014 at 16:48
  • yes, its <type 'numpy.ndarray'>. I actually have a large 2d array and I got that from extracting an image. Commented Nov 27, 2014 at 16:51
  • are there only ones and zeros? Commented Nov 27, 2014 at 17:03
  • 1
    not in my actual 2d array, but in the example yes. Commented Nov 27, 2014 at 17:19

3 Answers 3

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72

You can use np.where to return a tuple of arrays of x and y indices where a given condition holds in an array.

If a is the name of your array:

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

If you want a list of (x, y) pairs, you could zip the two arrays:

>>> list(zip(*np.where(a == 1)))
[(0, 0), (0, 1), (1, 2), (1, 3)]

Or, even better, @jme points out that np.asarray(x).T can be a more efficient way to generate the pairs.

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2 Comments

Just a note: zip(*x) can be slow for large x, where x = np.where(a == 1). If a 2d array is fine, np.asarray(x).T is quite a bit faster, and if you really want a list-of-lists, np.asarray(x).T.tolist() is marginally faster.
Is it possible to do something similar but with the index where the 2d array is nearest to a given value, e.g. np.where_is_nearest_to(0.99)?
23

Using numpy, argwhere may be the best solution:

import numpy as np

array = np.array([[1, 1, 0, 0],
                  [0, 0, 1, 1],
                  [0, 0, 0, 0]])

solutions = np.argwhere(array == 1)
print(solutions)

>>>
[[0 0]
 [0 1]
 [1 2]
 [1 3]]
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15

The problem with the list comprehension you provided is that it only goes one level deep, you need a nested list comprehension:

a = [[1,0,1],[0,0,1], [1,1,0]]

>>> [(ix,iy) for ix, row in enumerate(a) for iy, i in enumerate(row) if i == 0]
[(0, 1), (1, 0), (1, 1), (2, 2)]

That being said, if you are working with a numpy array, it's better to use the built in functions as suggested by ajcr.

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