Render visible voxel in 3D array without iterate it all - python

I'm doing a voxel engine and my data are in a 3D array per chunk. To render voxels, I need to know which one can be visible, so they will have Air block around it (0 in my data structure).
So I need to find all number which have a 0 beside in left, right, front, back, top and bot direction, but I really don't know how to do it.
I know I can do it with loop, but my goal is to achieve this task the fastest as I can.
For exemple (in 2D):
[[0, 0, 0, 0, 0,],
[0, 1, 1, 1, 0],
[0, 1, 1, 1, 0],
[0, 1, 1, 1, 1],
[0, 1, 1, 1, 0]]
Must return me these indexes:
(1,1) (1,2) (1,3)
(2,1) (2,3)
(3,1) (3,4)
(4,1) (4,2) (4,3)
I want to have the index of the numbers which are at the edge of the array even if there is no 0 around, because I can then check in beside chunk if there is Air data.
I don't know if it is possible, and if it is not, how can I render visible voxel without iterate over all the array?
Thanks.
(Sorry for my english, send comment if you don't understand all)
EDIT
After reading #Mad_Physicist comment, I post my uncomplete method to achieve the task with a loop (I didn't finish it because I was searching for a fastest way instead):
matrix = numpy.zeros((10,10,10), dtype="uint8")
# Filling with some random data
for y in range(10):
for z in range(10):
for x in range(10):
matrix[y, z, x] = random.choice((0,1,1)
result = numpy.transpose(numpy.where(matrix == 0))
for num in result.tolist():
u = matrix.item(*num)
# TODO: Check each direction
The weakness of this method is that I sometime pass throught useless data (0 with 0 in all directions).


You could use scipy.ndimage.morphology.binary_erosion:
from scipy.ndimage.morphology import binary_erosion
np.argwhere(arr & ~binary_erosion(arr))
Output:
array([[1, 1],
[1, 2],
[1, 3],
[2, 1],
[2, 3],
[3, 1],
[3, 4],
[4, 1],
[4, 2],
[4, 3]])

Related

Diagonal array in numpy

If I have the array [[1,0,0],[0,1,0],[0,0,1]] (let's call it So) which is done as numpy.eye(3).
How can I get that the elements below the diagonal are only 2 and 3 like this [[1,0,0],[2,1,0],[3,2,1]] ?? How can I assign vectors of an array to a different set of values?
I know I could use numpy.concatenate to join 3 vectors and I know how to change rows/columns but I can't figure out how to change diagonals below the main diagonal.
I tried to do np.diagonal(So,-1)=2*np.diagonal(So,-1) to change the diagonal right below the main diagonal but I get the error message cannot assign to function call.

I would not start from numpy.eye but rather numpy.ones and use numpy.tril+cumsum to compute the next numbers on the lower triangle:
import numpy as np
np.tril(np.ones((3,3))).cumsum(axis=0).astype(int)
output:
array([[1, 0, 0],
[2, 1, 0],
[3, 2, 1]])
reversed output (from comment)
Assuming the array is square
n = 3
a = np.tril(np.ones((n,n)))
(a*(n+2)-np.eye(n)*n-a.cumsum(axis=0)).astype(int)
Output:
array([[1, 0, 0],
[3, 1, 0],
[2, 3, 1]])
Output for n=5:
array([[1, 0, 0, 0, 0],
[5, 1, 0, 0, 0],
[4, 5, 1, 0, 0],
[3, 4, 5, 1, 0],
[2, 3, 4, 5, 1]])

You can use np.fill_diagonal and index the matrix so the principal diagonal of your matrix is the one you want. This suposing you want to put other values than 2 and 3 is the a good solution:
import numpy as np
q = np.eye(3)
#if you want the first diagonal below the principal
# you can call q[1:,:] (this is not a 3x3 or 2x3 matrix but it'll work)
val =2
np.fill_diagonal(q[1:,:], val)
#note that here you can use an unique value 'val' or
# an array with values of corresponding size
#np.fill_diagonal(q[1:,:], [2, 2])
#then you can do the same on the last one column
np.fill_diagonal(q[2:,:], 3)

You could follow this approach:
def func(n):
... return np.array([np.array(list(range(i, 0, -1)) + [0,] * (n - i)) for i in range(1, n + 1)])
func(3)
OUTPUT
array([[1, 0, 0],
[2, 1, 0],
[3, 2, 1]])

Searching values in matrix in python

I am trying to resolve one task with matrix. I have function: fill_area(matrix, coordinates, value). If I would input coordinates for example (1,0) and value 3, it should write 3 in position [1][0] and rewrite the value to 3. If near numbers have the same value, it should also rewrite them. I would like to use an stack or queue for numbers, which have to change an value.
But I have no idea, how to check all nearest position.
matrix = [[2, 0, 1],
[0, 0, 1],
[0, 1, 1]]
fill_area(matrix, (1, 0), 3)
matrix = [[2, 0, 1],
[3, 3, 1],
[3, 1, 1]]

The preferred way to handle matrices is numpy but there seems to be no obvious way to handle this special case while avoiding loops over items in numpy. I take it from your example that you define "nearness" as the patch around the given coordinates but in coordinate-direction only. I have two versions, one for the entire coordinate-patch around the given coordinates, the other in ij-direction only (per your example). Both are list comprehensions, for better readability you might want to change that to regular loops.
def fill_area(mat, ij, val):
'''set patch around ij in mat to val for all positions where the original mat value == mat val at ij, full patch'''
return [[val if abs(i-ij[0])<2 and abs(j-ij[1])<2 and mat[i][j] == mat[ij[0]][ij[1]]
else mat[i][j] for j in range(len(mat[1]))] for i in range(len(mat))]
def fill_area1(mat, ij, val):
'''set patch around ij in mat to val for all positions where the original mat value == mat val at ij, patch ij-dir'''
return [[val if ((abs(i-ij[0])<2 and j==ij[1]) or (abs(j-ij[1])<2 and i==ij[0])) and mat[i][j] == mat[ij[0]][ij[1]]
else mat[i][j] for j in range(len(mat[1]))] for i in range(len(mat))]
matrix = [[2, 0, 1],
[0, 0, 1],
[0, 1, 1]]
print('entire patch: ', fill_area(matrix, (1, 0), 3))
print('ij-directional: ', fill_area1(matrix, (1, 0), 3))
which produces
entire patch: [[2, 3, 1], [3, 3, 1], [3, 1, 1]]
ij-directional: [[2, 0, 1], [3, 3, 1], [3, 1, 1]]

Compare numpy arrays of different sizes and find index that matches in Python 3 [duplicate]

I have an array X:
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
And I wish to find the index of the row of several values in this array:
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
For this example I would like a result like:
[0,3,4]
I have a code doing this, but I think it is overly complicated:
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
result = []
for s in searched_values:
idx = np.argwhere([np.all((X-s)==0, axis=1)])[0][1]
result.append(idx)
print(result)
I found this answer for a similar question but it works only for 1d arrays.
Is there a way to do what I want in a simpler way?

Approach #1
One approach would be to use NumPy broadcasting, like so -
np.where((X==searched_values[:,None]).all(-1))[1]
Approach #2
A memory efficient approach would be to convert each row as linear index equivalents and then using np.in1d, like so -
dims = X.max(0)+1
out = np.where(np.in1d(np.ravel_multi_index(X.T,dims),\
np.ravel_multi_index(searched_values.T,dims)))[0]
Approach #3
Another memory efficient approach using np.searchsorted and with that same philosophy of converting to linear index equivalents would be like so -
dims = X.max(0)+1
X1D = np.ravel_multi_index(X.T,dims)
searched_valuesID = np.ravel_multi_index(searched_values.T,dims)
sidx = X1D.argsort()
out = sidx[np.searchsorted(X1D,searched_valuesID,sorter=sidx)]
Please note that this np.searchsorted method assumes there is a match for each row from searched_values in X.
How does np.ravel_multi_index work?
This function gives us the linear index equivalent numbers. It accepts a 2D array of n-dimensional indices, set as columns and the shape of that n-dimensional grid itself onto which those indices are to be mapped and equivalent linear indices are to be computed.
Let's use the inputs we have for the problem at hand. Take the case of input X and note the first row of it. Since, we are trying to convert each row of X into its linear index equivalent and since np.ravel_multi_index assumes each column as one indexing tuple, we need to transpose X before feeding into the function. Since, the number of elements per row in X in this case is 2, the n-dimensional grid to be mapped onto would be 2D. With 3 elements per row in X, it would had been 3D grid for mapping and so on.
To see how this function would compute linear indices, consider the first row of X -
In [77]: X
Out[77]:
array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
We have the shape of the n-dimensional grid as dims -
In [78]: dims
Out[78]: array([10, 7])
Let's create the 2-dimensional grid to see how that mapping works and linear indices get computed with np.ravel_multi_index -
In [79]: out = np.zeros(dims,dtype=int)
In [80]: out
Out[80]:
array([[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]])
Let's set the first indexing tuple from X, i.e. the first row from X into the grid -
In [81]: out[4,2] = 1
In [82]: out
Out[82]:
array([[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]])
Now, to see the linear index equivalent of the element just set, let's flatten and use np.where to detect that 1.
In [83]: np.where(out.ravel())[0]
Out[83]: array([30])
This could also be computed if row-major ordering is taken into account.
Let's use np.ravel_multi_index and verify those linear indices -
In [84]: np.ravel_multi_index(X.T,dims)
Out[84]: array([30, 66, 61, 24, 41])
Thus, we would have linear indices corresponding to each indexing tuple from X, i.e. each row from X.
Choosing dimensions for np.ravel_multi_index to form unique linear indices
Now, the idea behind considering each row of X as indexing tuple of a n-dimensional grid and converting each such tuple to a scalar is to have unique scalars corresponding to unique tuples, i.e. unique rows in X.
Let's take another look at X -
In [77]: X
Out[77]:
array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
Now, as discussed in the previous section, we are considering each row as indexing tuple. Within each such indexing tuple, the first element would represent the first axis of the n-dim grid, second element would be the second axis of the grid and so on until the last element of each row in X. In essence, each column would represent one dimension or axis of the grid. If we are to map all elements from X onto the same n-dim grid, we need to consider the maximum stretch of each axis of such a proposed n-dim grid. Assuming we are dealing with positive numbers in X, such a stretch would be the maximum of each column in X + 1. That + 1 is because Python follows 0-based indexing. So, for example X[1,0] == 9 would map to the 10th row of the proposed grid. Similarly, X[4,1] == 6 would go to the 7th column of that grid.
So, for our sample case, we had -
In [7]: dims = X.max(axis=0) + 1 # Or simply X.max(0) + 1
In [8]: dims
Out[8]: array([10, 7])
Thus, we would need a grid of at least a shape of (10,7) for our sample case. More lengths along the dimensions won't hurt and would give us unique linear indices too.
Concluding remarks : One important thing to be noted here is that if we have negative numbers in X, we need to add proper offsets along each column in X to make those indexing tuples as positive numbers before using np.ravel_multi_index.

Another alternative is to use asvoid (below) to view each row as a single
value of void dtype. This reduces a 2D array to a 1D array, thus allowing you to use np.in1d as usual:
import numpy as np
def asvoid(arr):
"""
Based on http://stackoverflow.com/a/16973510/190597 (Jaime, 2013-06)
View the array as dtype np.void (bytes). The items along the last axis are
viewed as one value. This allows comparisons to be performed which treat
entire rows as one value.
"""
arr = np.ascontiguousarray(arr)
if np.issubdtype(arr.dtype, np.floating):
""" Care needs to be taken here since
np.array([-0.]).view(np.void) != np.array([0.]).view(np.void)
Adding 0. converts -0. to 0.
"""
arr += 0.
return arr.view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[-1])))
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
idx = np.flatnonzero(np.in1d(asvoid(X), asvoid(searched_values)))
print(idx)
# [0 3 4]

The numpy_indexed package (disclaimer: I am its author) contains functionality for performing such operations efficiently (also uses searchsorted under the hood). In terms of functionality, it acts as a vectorized equivalent of list.index:
import numpy_indexed as npi
result = npi.indices(X, searched_values)
Note that using the 'missing' kwarg, you have full control over behavior of missing items, and it works for nd-arrays (fi; stacks of images) as well.
Update: using the same shapes as #Rik X=[520000,28,28] and searched_values=[20000,28,28], it runs in 0.8064 secs, using missing=-1 to detect and denote entries not present in X.

Here is a pretty fast solution that scales up well using numpy and hashlib. It can handle large dimensional matrices or images in seconds. I used it on 520000 X (28 X 28) array and 20000 X (28 X 28) in 2 seconds on my CPU
Code:
import numpy as np
import hashlib
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
#hash using sha1 appears to be efficient
xhash=[hashlib.sha1(row).digest() for row in X]
yhash=[hashlib.sha1(row).digest() for row in searched_values]
z=np.in1d(xhash,yhash)
##Use unique to get unique indices to ind1 results
_,unique=np.unique(np.array(xhash)[z],return_index=True)
##Compute unique indices by indexing an array of indices
idx=np.array(range(len(xhash)))
unique_idx=idx[z][unique]
print('unique_idx=',unique_idx)
print('X[unique_idx]=',X[unique_idx])
Output:
unique_idx= [4 3 0]
X[unique_idx]= [[5 6]
[3 3]
[4 2]]

X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
S = np.array([[4, 2],
[3, 3],
[5, 6]])
result = [[i for i,row in enumerate(X) if (s==row).all()] for s in S]
or
result = [i for s in S for i,row in enumerate(X) if (s==row).all()]
if you want a flat list (assuming there is exactly one match per searched value).

Another way is to use cdist function from scipy.spatial.distance like this:
np.nonzero(cdist(X, searched_values) == 0)[0]
Basically, we get row numbers of X which have distance zero to a row in searched_values, meaning they are equal. Makes sense if you look on rows as coordinates.

I had similar requirement and following worked for me:
np.argwhere(np.isin(X, searched_values).all(axis=1))

Here's what worked out for me:
def find_points(orig: np.ndarray, search: np.ndarray) -> np.ndarray:
equals = [np.equal(orig, p).all(1) for p in search]
exists = np.max(equals, axis=1)
indices = np.argmax(equals, axis=1)
indices[exists == False] = -1
return indices
test:
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6],
[0, 0]])
find_points(X, searched_values)
output:
[0,3,4,-1]

Finding common rows between two 2D NumPy arrays [duplicate]

I have an array X:
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
And I wish to find the index of the row of several values in this array:
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
For this example I would like a result like:
[0,3,4]
I have a code doing this, but I think it is overly complicated:
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
result = []
for s in searched_values:
idx = np.argwhere([np.all((X-s)==0, axis=1)])[0][1]
result.append(idx)
print(result)
I found this answer for a similar question but it works only for 1d arrays.
Is there a way to do what I want in a simpler way?

Approach #1
One approach would be to use NumPy broadcasting, like so -
np.where((X==searched_values[:,None]).all(-1))[1]
Approach #2
A memory efficient approach would be to convert each row as linear index equivalents and then using np.in1d, like so -
dims = X.max(0)+1
out = np.where(np.in1d(np.ravel_multi_index(X.T,dims),\
np.ravel_multi_index(searched_values.T,dims)))[0]
Approach #3
Another memory efficient approach using np.searchsorted and with that same philosophy of converting to linear index equivalents would be like so -
dims = X.max(0)+1
X1D = np.ravel_multi_index(X.T,dims)
searched_valuesID = np.ravel_multi_index(searched_values.T,dims)
sidx = X1D.argsort()
out = sidx[np.searchsorted(X1D,searched_valuesID,sorter=sidx)]
Please note that this np.searchsorted method assumes there is a match for each row from searched_values in X.
How does np.ravel_multi_index work?
This function gives us the linear index equivalent numbers. It accepts a 2D array of n-dimensional indices, set as columns and the shape of that n-dimensional grid itself onto which those indices are to be mapped and equivalent linear indices are to be computed.
Let's use the inputs we have for the problem at hand. Take the case of input X and note the first row of it. Since, we are trying to convert each row of X into its linear index equivalent and since np.ravel_multi_index assumes each column as one indexing tuple, we need to transpose X before feeding into the function. Since, the number of elements per row in X in this case is 2, the n-dimensional grid to be mapped onto would be 2D. With 3 elements per row in X, it would had been 3D grid for mapping and so on.
To see how this function would compute linear indices, consider the first row of X -
In [77]: X
Out[77]:
array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
We have the shape of the n-dimensional grid as dims -
In [78]: dims
Out[78]: array([10, 7])
Let's create the 2-dimensional grid to see how that mapping works and linear indices get computed with np.ravel_multi_index -
In [79]: out = np.zeros(dims,dtype=int)
In [80]: out
Out[80]:
array([[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]])
Let's set the first indexing tuple from X, i.e. the first row from X into the grid -
In [81]: out[4,2] = 1
In [82]: out
Out[82]:
array([[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]])
Now, to see the linear index equivalent of the element just set, let's flatten and use np.where to detect that 1.
In [83]: np.where(out.ravel())[0]
Out[83]: array([30])
This could also be computed if row-major ordering is taken into account.
Let's use np.ravel_multi_index and verify those linear indices -
In [84]: np.ravel_multi_index(X.T,dims)
Out[84]: array([30, 66, 61, 24, 41])
Thus, we would have linear indices corresponding to each indexing tuple from X, i.e. each row from X.
Choosing dimensions for np.ravel_multi_index to form unique linear indices
Now, the idea behind considering each row of X as indexing tuple of a n-dimensional grid and converting each such tuple to a scalar is to have unique scalars corresponding to unique tuples, i.e. unique rows in X.
Let's take another look at X -
In [77]: X
Out[77]:
array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
Now, as discussed in the previous section, we are considering each row as indexing tuple. Within each such indexing tuple, the first element would represent the first axis of the n-dim grid, second element would be the second axis of the grid and so on until the last element of each row in X. In essence, each column would represent one dimension or axis of the grid. If we are to map all elements from X onto the same n-dim grid, we need to consider the maximum stretch of each axis of such a proposed n-dim grid. Assuming we are dealing with positive numbers in X, such a stretch would be the maximum of each column in X + 1. That + 1 is because Python follows 0-based indexing. So, for example X[1,0] == 9 would map to the 10th row of the proposed grid. Similarly, X[4,1] == 6 would go to the 7th column of that grid.
So, for our sample case, we had -
In [7]: dims = X.max(axis=0) + 1 # Or simply X.max(0) + 1
In [8]: dims
Out[8]: array([10, 7])
Thus, we would need a grid of at least a shape of (10,7) for our sample case. More lengths along the dimensions won't hurt and would give us unique linear indices too.
Concluding remarks : One important thing to be noted here is that if we have negative numbers in X, we need to add proper offsets along each column in X to make those indexing tuples as positive numbers before using np.ravel_multi_index.

Another alternative is to use asvoid (below) to view each row as a single
value of void dtype. This reduces a 2D array to a 1D array, thus allowing you to use np.in1d as usual:
import numpy as np
def asvoid(arr):
"""
Based on http://stackoverflow.com/a/16973510/190597 (Jaime, 2013-06)
View the array as dtype np.void (bytes). The items along the last axis are
viewed as one value. This allows comparisons to be performed which treat
entire rows as one value.
"""
arr = np.ascontiguousarray(arr)
if np.issubdtype(arr.dtype, np.floating):
""" Care needs to be taken here since
np.array([-0.]).view(np.void) != np.array([0.]).view(np.void)
Adding 0. converts -0. to 0.
"""
arr += 0.
return arr.view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[-1])))
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
idx = np.flatnonzero(np.in1d(asvoid(X), asvoid(searched_values)))
print(idx)
# [0 3 4]

The numpy_indexed package (disclaimer: I am its author) contains functionality for performing such operations efficiently (also uses searchsorted under the hood). In terms of functionality, it acts as a vectorized equivalent of list.index:
import numpy_indexed as npi
result = npi.indices(X, searched_values)
Note that using the 'missing' kwarg, you have full control over behavior of missing items, and it works for nd-arrays (fi; stacks of images) as well.
Update: using the same shapes as #Rik X=[520000,28,28] and searched_values=[20000,28,28], it runs in 0.8064 secs, using missing=-1 to detect and denote entries not present in X.

Here is a pretty fast solution that scales up well using numpy and hashlib. It can handle large dimensional matrices or images in seconds. I used it on 520000 X (28 X 28) array and 20000 X (28 X 28) in 2 seconds on my CPU
Code:
import numpy as np
import hashlib
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
#hash using sha1 appears to be efficient
xhash=[hashlib.sha1(row).digest() for row in X]
yhash=[hashlib.sha1(row).digest() for row in searched_values]
z=np.in1d(xhash,yhash)
##Use unique to get unique indices to ind1 results
_,unique=np.unique(np.array(xhash)[z],return_index=True)
##Compute unique indices by indexing an array of indices
idx=np.array(range(len(xhash)))
unique_idx=idx[z][unique]
print('unique_idx=',unique_idx)
print('X[unique_idx]=',X[unique_idx])
Output:
unique_idx= [4 3 0]
X[unique_idx]= [[5 6]
[3 3]
[4 2]]

X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
S = np.array([[4, 2],
[3, 3],
[5, 6]])
result = [[i for i,row in enumerate(X) if (s==row).all()] for s in S]
or
result = [i for s in S for i,row in enumerate(X) if (s==row).all()]
if you want a flat list (assuming there is exactly one match per searched value).

Another way is to use cdist function from scipy.spatial.distance like this:
np.nonzero(cdist(X, searched_values) == 0)[0]
Basically, we get row numbers of X which have distance zero to a row in searched_values, meaning they are equal. Makes sense if you look on rows as coordinates.

I had similar requirement and following worked for me:
np.argwhere(np.isin(X, searched_values).all(axis=1))

Here's what worked out for me:
def find_points(orig: np.ndarray, search: np.ndarray) -> np.ndarray:
equals = [np.equal(orig, p).all(1) for p in search]
exists = np.max(equals, axis=1)
indices = np.argmax(equals, axis=1)
indices[exists == False] = -1
return indices
test:
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6],
[0, 0]])
find_points(X, searched_values)
output:
[0,3,4,-1]

Quick method to find cross index between two numpy arrays [duplicate]

I have an array X:
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
And I wish to find the index of the row of several values in this array:
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
For this example I would like a result like:
[0,3,4]
I have a code doing this, but I think it is overly complicated:
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
result = []
for s in searched_values:
idx = np.argwhere([np.all((X-s)==0, axis=1)])[0][1]
result.append(idx)
print(result)
I found this answer for a similar question but it works only for 1d arrays.
Is there a way to do what I want in a simpler way?

Approach #1
One approach would be to use NumPy broadcasting, like so -
np.where((X==searched_values[:,None]).all(-1))[1]
Approach #2
A memory efficient approach would be to convert each row as linear index equivalents and then using np.in1d, like so -
dims = X.max(0)+1
out = np.where(np.in1d(np.ravel_multi_index(X.T,dims),\
np.ravel_multi_index(searched_values.T,dims)))[0]
Approach #3
Another memory efficient approach using np.searchsorted and with that same philosophy of converting to linear index equivalents would be like so -
dims = X.max(0)+1
X1D = np.ravel_multi_index(X.T,dims)
searched_valuesID = np.ravel_multi_index(searched_values.T,dims)
sidx = X1D.argsort()
out = sidx[np.searchsorted(X1D,searched_valuesID,sorter=sidx)]
Please note that this np.searchsorted method assumes there is a match for each row from searched_values in X.
How does np.ravel_multi_index work?
This function gives us the linear index equivalent numbers. It accepts a 2D array of n-dimensional indices, set as columns and the shape of that n-dimensional grid itself onto which those indices are to be mapped and equivalent linear indices are to be computed.
Let's use the inputs we have for the problem at hand. Take the case of input X and note the first row of it. Since, we are trying to convert each row of X into its linear index equivalent and since np.ravel_multi_index assumes each column as one indexing tuple, we need to transpose X before feeding into the function. Since, the number of elements per row in X in this case is 2, the n-dimensional grid to be mapped onto would be 2D. With 3 elements per row in X, it would had been 3D grid for mapping and so on.
To see how this function would compute linear indices, consider the first row of X -
In [77]: X
Out[77]:
array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
We have the shape of the n-dimensional grid as dims -
In [78]: dims
Out[78]: array([10, 7])
Let's create the 2-dimensional grid to see how that mapping works and linear indices get computed with np.ravel_multi_index -
In [79]: out = np.zeros(dims,dtype=int)
In [80]: out
Out[80]:
array([[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]])
Let's set the first indexing tuple from X, i.e. the first row from X into the grid -
In [81]: out[4,2] = 1
In [82]: out
Out[82]:
array([[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]])
Now, to see the linear index equivalent of the element just set, let's flatten and use np.where to detect that 1.
In [83]: np.where(out.ravel())[0]
Out[83]: array([30])
This could also be computed if row-major ordering is taken into account.
Let's use np.ravel_multi_index and verify those linear indices -
In [84]: np.ravel_multi_index(X.T,dims)
Out[84]: array([30, 66, 61, 24, 41])
Thus, we would have linear indices corresponding to each indexing tuple from X, i.e. each row from X.
Choosing dimensions for np.ravel_multi_index to form unique linear indices
Now, the idea behind considering each row of X as indexing tuple of a n-dimensional grid and converting each such tuple to a scalar is to have unique scalars corresponding to unique tuples, i.e. unique rows in X.
Let's take another look at X -
In [77]: X
Out[77]:
array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
Now, as discussed in the previous section, we are considering each row as indexing tuple. Within each such indexing tuple, the first element would represent the first axis of the n-dim grid, second element would be the second axis of the grid and so on until the last element of each row in X. In essence, each column would represent one dimension or axis of the grid. If we are to map all elements from X onto the same n-dim grid, we need to consider the maximum stretch of each axis of such a proposed n-dim grid. Assuming we are dealing with positive numbers in X, such a stretch would be the maximum of each column in X + 1. That + 1 is because Python follows 0-based indexing. So, for example X[1,0] == 9 would map to the 10th row of the proposed grid. Similarly, X[4,1] == 6 would go to the 7th column of that grid.
So, for our sample case, we had -
In [7]: dims = X.max(axis=0) + 1 # Or simply X.max(0) + 1
In [8]: dims
Out[8]: array([10, 7])
Thus, we would need a grid of at least a shape of (10,7) for our sample case. More lengths along the dimensions won't hurt and would give us unique linear indices too.
Concluding remarks : One important thing to be noted here is that if we have negative numbers in X, we need to add proper offsets along each column in X to make those indexing tuples as positive numbers before using np.ravel_multi_index.

Another alternative is to use asvoid (below) to view each row as a single
value of void dtype. This reduces a 2D array to a 1D array, thus allowing you to use np.in1d as usual:
import numpy as np
def asvoid(arr):
"""
Based on http://stackoverflow.com/a/16973510/190597 (Jaime, 2013-06)
View the array as dtype np.void (bytes). The items along the last axis are
viewed as one value. This allows comparisons to be performed which treat
entire rows as one value.
"""
arr = np.ascontiguousarray(arr)
if np.issubdtype(arr.dtype, np.floating):
""" Care needs to be taken here since
np.array([-0.]).view(np.void) != np.array([0.]).view(np.void)
Adding 0. converts -0. to 0.
"""
arr += 0.
return arr.view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[-1])))
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
idx = np.flatnonzero(np.in1d(asvoid(X), asvoid(searched_values)))
print(idx)
# [0 3 4]

The numpy_indexed package (disclaimer: I am its author) contains functionality for performing such operations efficiently (also uses searchsorted under the hood). In terms of functionality, it acts as a vectorized equivalent of list.index:
import numpy_indexed as npi
result = npi.indices(X, searched_values)
Note that using the 'missing' kwarg, you have full control over behavior of missing items, and it works for nd-arrays (fi; stacks of images) as well.
Update: using the same shapes as #Rik X=[520000,28,28] and searched_values=[20000,28,28], it runs in 0.8064 secs, using missing=-1 to detect and denote entries not present in X.

Here is a pretty fast solution that scales up well using numpy and hashlib. It can handle large dimensional matrices or images in seconds. I used it on 520000 X (28 X 28) array and 20000 X (28 X 28) in 2 seconds on my CPU
Code:
import numpy as np
import hashlib
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6]])
#hash using sha1 appears to be efficient
xhash=[hashlib.sha1(row).digest() for row in X]
yhash=[hashlib.sha1(row).digest() for row in searched_values]
z=np.in1d(xhash,yhash)
##Use unique to get unique indices to ind1 results
_,unique=np.unique(np.array(xhash)[z],return_index=True)
##Compute unique indices by indexing an array of indices
idx=np.array(range(len(xhash)))
unique_idx=idx[z][unique]
print('unique_idx=',unique_idx)
print('X[unique_idx]=',X[unique_idx])
Output:
unique_idx= [4 3 0]
X[unique_idx]= [[5 6]
[3 3]
[4 2]]

X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
S = np.array([[4, 2],
[3, 3],
[5, 6]])
result = [[i for i,row in enumerate(X) if (s==row).all()] for s in S]
or
result = [i for s in S for i,row in enumerate(X) if (s==row).all()]
if you want a flat list (assuming there is exactly one match per searched value).

Another way is to use cdist function from scipy.spatial.distance like this:
np.nonzero(cdist(X, searched_values) == 0)[0]
Basically, we get row numbers of X which have distance zero to a row in searched_values, meaning they are equal. Makes sense if you look on rows as coordinates.

I had similar requirement and following worked for me:
np.argwhere(np.isin(X, searched_values).all(axis=1))

Here's what worked out for me:
def find_points(orig: np.ndarray, search: np.ndarray) -> np.ndarray:
equals = [np.equal(orig, p).all(1) for p in search]
exists = np.max(equals, axis=1)
indices = np.argmax(equals, axis=1)
indices[exists == False] = -1
return indices
test:
X = np.array([[4, 2],
[9, 3],
[8, 5],
[3, 3],
[5, 6]])
searched_values = np.array([[4, 2],
[3, 3],
[5, 6],
[0, 0]])
find_points(X, searched_values)
output:
[0,3,4,-1]

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