CODEWITHSUNDEEP
pythonnumpy
Amelio Vazquez-Reina
Amelio Vazquez-Reina

Reputation: 96264

Get x,y,value 1D arrays from 2D numpy array (linear indexing)

Say I have a 2D array like the one below:

array([3, 6, 7
       1,-1, 3])

I would like to get in 3 separate arrays the x, y and value of the array . In other words:

x      = [0, 1, 0,  1, 0, 1]
y      = [0, 0, 1,  1, 2, 2]
values = [3, 1, 6, -1, 7, 3]

How can I do this?

For reference, this is what MATLAB calls linear indexing.

Upvotes: 4

Views: 3586

Answers (6)

den.run.ai
den.run.ai

Reputation: 5933

You could do 2 loops (or nested list comprehension) with 2 enumerate(iterable) inside.

>>> import numpy as np
>>> ar1=np.ones([3,4])
>>> ar1
array([[ 1.,  1.,  1.,  1.],
       [ 1.,  1.,  1.,  1.],
       [ 1.,  1.,  1.,  1.]])
>>> ar1[::2,::3]=-1
>>> ar1[::3,:21]=-2
>>> ar1
array([[-2., -2., -2., -2.],
       [ 1.,  1.,  1.,  1.],
       [-1.,  1.,  1., -1.]])


>>> x=[];y=[];values=[]
>>> for i,ari in enumerate(ar1):
...   for j,valj in enumerate(ari):
...     print i,j,valj
...     x+=[i];y+=[j];values+=[valj]
...
0 0 -2.0
0 1 -2.0
0 2 -2.0
0 3 -2.0
1 0 1.0
1 1 1.0
1 2 1.0
1 3 1.0
2 0 -1.0
2 1 1.0
2 2 1.0
2 3 -1.0

Upvotes: 0

hpaulj
hpaulj

Reputation: 231335

If you have scipy installed along with numpy, you can use its sparse module

from scipy import sparse
x = np.array([[3,6,7],[1,-1,3]])
M=sparse.coo_matrix(x.T)
M.data
# array([ 3,  1,  6, -1,  7,  3])
M.col
# array([0, 1, 0, 1, 0, 1])
M.row
# array([0, 0, 1, 1, 2, 2])

Looking at coo_matrix I see that nonzero() is used to get the indices:

row,col = x.T.nonzero()
data = x.T[row,col]

If there could be zeros in x, you'd have use a work around like np.nonzero(np.ones(x.T.shape)).

Upvotes: 3

Bi Rico
Bi Rico

Reputation: 25813

How about something like:

x, y = np.indices(array.shape)
x = x.ravel(order='F')
y = y.ravel(order='F')
values = array.ravel(order='F')

Upvotes: 5

ali_m
ali_m

Reputation: 74154

def xyval(A):
    x, y = np.indices(A.shape)
    return x.ravel(), y.ravel(), A.ravel()

Upvotes: 3

wildwilhelm
wildwilhelm

Reputation: 5019

Your values array can be attained by using the flatten method of a 2D array (matrix). Then it's just a matter of constructing the x and y arrays to contain the proper indices.

import numpy as np
import itertools

a = np.array([[3, 6, 7],
              [1,-1, 3]])

idxs   = np.array(list(itertools.product(*map(range, a.shape))))
x      = idxs[:,0]
y      = idxs[:,1]
values = a.flatten()

Testing this:

>>> x
array([0, 0, 0, 1, 1, 1])
>>> y
array([0, 1, 2, 0, 1, 2])
>>> values
array([ 3,  6,  7,  1, -1,  3])
>>> a[0,0]
3
>>> a[0,1]
6
>>> a[0,2]
7
>>> a[1,0]
1
>>> a[1,1]
-1
>>> a[1,2]
3

Upvotes: 1

Akavall
Akavall

Reputation: 86128

What about this:

import numpy as np

a = np.array([[3, 6, 7],
              [1,-1, 3]])

n_rows, n_cols = a.shape


temp_x = np.repeat(np.arange(n_rows)[np.newaxis, :], n_cols, axis=0) 
# construction of x can probably be simplified
x = temp_x.ravel()    
y = np.repeat(np.arange(n_cols), n_rows)        
values = a.ravel(order='F')

Result:

>>> x
array([0, 1, 0, 1, 0, 1])
>>> y
array([0, 0, 1, 1, 2, 2])
>>> values
array([ 3,  1,  6, -1,  7,  3])

Upvotes: 1

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