Vector (positionx,positiony,value) of a matrix python

Question

I've got a matrix of values, and I want a vector with the positions of the matrix and the value.

First idea: double-bucle and store index and value

Second idea: Create a ndarray whit a sequence

Any other idea most efficient?

Sorry for my english.

Input (image 1C):

[[0 1 2 3 4 5]
 [6 7 8 9 10 11]]

Output:

[[0 0 1][0 1 2] [0 2 3] [0 3 4] [0 4 5] [1 0 6] [1 1 7] [1 2 8] [1 3 9] [1 4 10] [1 5 11] ... ]

Thank you all for your answers, you have helped to solve the problem and understand numpy.

please include sample input+output so it's easier to understand what you're asking. — shx2
– shx2, Commented Nov 16, 2013 at 21:38

Bi Rico · Accepted Answer · 2013-11-16 23:07:50Z

2

Here's a pretty clean way to do it:

import numpy as np
arr = np.random.random((2, 6))
x, y = np.indices(arr.shape)
output = np.column_stack([x.ravel(), y.ravel(), arr.ravel()])

And here's a overly cute way to do it that minimizes memory usage:

import numpy as np
arr = np.random.random((2, 6))
output = np.indices(arr.shape + (1,), dtype=arr.dtype)
output[-1, ..., 0] = arr
output = output.reshape(arr.ndim + 1, -1).T

answered Nov 16, 2013 at 23:07

Bi Rico

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1 Comment

Fran Navarro Over a year ago

I test the indices, but for large arrays indices don't work well. Sorry, I don't say anything about the size of the matrix (and the example was poor). But your approach (stack columns) has been useful.

shx2 · Accepted Answer · 2013-11-16 22:01:34Z

1

Use numpy.ndenumerate.

list(np.ndenumerate(a))
=>
[((0, 0), 0),
 ((0, 1), 1),
 ((0, 2), 2),
 ((0, 3), 3),
 ((0, 4), 4),
 ((0, 5), 5),
 ((1, 0), 6),
 ((1, 1), 7),
 ((1, 2), 8),
 ((1, 3), 9),
 ((1, 4), 10),
 ((1, 5), 11)]

To get the output you specified in the question, i.e. to flatten the indices and the values, you can do:

[ list(idx) + [value] for idx, value in np.ndenumerate(a) ]

answered Nov 16, 2013 at 22:01

shx2

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Comments

Community · Accepted Answer · 2017-05-23 12:09:23Z

1

import numpy as np

mat = np.arange(12).reshape(2,6)

result = np.c_[(np.array(list(np.ndindex(mat.shape))), mat.ravel())]
print(result)

yields

[[ 0  0  0]
 [ 0  1  1]
 [ 0  2  2]
 [ 0  3  3]
 [ 0  4  4]
 [ 0  5  5]
 [ 1  0  6]
 [ 1  1  7]
 [ 1  2  8]
 [ 1  3  9]
 [ 1  4 10]
 [ 1  5 11]]

Caution: Using list as in np.array(list(np.ndindex(mat.shape))) forms a temporary Python list. It's convenient, but not very memory-efficient since Python lists are much bigger than "equivalent" NumPy arrays. You could avoid the temporary Python list by using np.fromiter; for example, (in conjunction with shx2's ndenumerate idea):

import numpy as np
arr = np.arange(12).reshape(2,6)

def ndindex_values(arr):
    for idx, value in np.ndenumerate(arr):
        for coord in idx:
            yield coord
        yield value

result = np.fromiter(
    ndindex_values(arr),
    count=arr.size*(len(arr.shape)+1),
    dtype=arr.dtype).reshape(arr.size, -1)

edited May 23, 2017 at 12:09

CommunityBot

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answered Nov 16, 2013 at 22:01

unutbu

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2 Comments

Fran Navarro Over a year ago

just now i was reading about np.indices to create the array from this approach. Thanks!

unutbu Over a year ago

I can't think of a good reason why you'd want to form this array. That might just be a lack of imagination on my part, but it might be because forming this array is typically not necessary. You might want to post a new question explaining your intended purpose for this array; people might be able to suggest a different approach which avoids it altogether.

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