How to check the presence of a given numpy array in a larger-shape numpy array?
I guess the title of my question might not be very clear..
I have a small array, say a = ([[0,0,0],[0,0,1],[0,1,1]]). Then I have a bigger array of a higher dimension, say b = ([[[2,2,2],[2,0,1],[2,1,1]],[[0,0,0],[3,3,1],[3,1,1]],[...]]).
I'd like to check if one of the elements of a can be found in b. In this case, I'd find that the first element of a [0,0,0] is indeed in b, and then I'd like to retrieve the corresponding index in b.
I'd like to do that avoiding looping, since from the very little I understood from numpy arrays, they are not meant to be iterated over in a classic way. In other words, I need it to be very fast, because my actual arrays are quite big.
Any idea?
Thanks a lot!
Arnaud.
numpy
edited Nov 19 '18 at 8:09
Aqueous Carlos
360314
asked Nov 19 '18 at 7:51
ArnaudArnaud
12813
add a comment |
I guess the title of my question might not be very clear..
I have a small array, say a = ([[0,0,0],[0,0,1],[0,1,1]]). Then I have a bigger array of a higher dimension, say b = ([[[2,2,2],[2,0,1],[2,1,1]],[[0,0,0],[3,3,1],[3,1,1]],[...]]).
I'd like to check if one of the elements of a can be found in b. In this case, I'd find that the first element of a [0,0,0] is indeed in b, and then I'd like to retrieve the corresponding index in b.
I'd like to do that avoiding looping, since from the very little I understood from numpy arrays, they are not meant to be iterated over in a classic way. In other words, I need it to be very fast, because my actual arrays are quite big.
Any idea?
Thanks a lot!
Arnaud.
numpy
edited Nov 19 '18 at 8:09
Aqueous Carlos
360314
asked Nov 19 '18 at 7:51
ArnaudArnaud
12813
add a comment |
I guess the title of my question might not be very clear..
I have a small array, say a = ([[0,0,0],[0,0,1],[0,1,1]]). Then I have a bigger array of a higher dimension, say b = ([[[2,2,2],[2,0,1],[2,1,1]],[[0,0,0],[3,3,1],[3,1,1]],[...]]).
I'd like to check if one of the elements of a can be found in b. In this case, I'd find that the first element of a [0,0,0] is indeed in b, and then I'd like to retrieve the corresponding index in b.
I'd like to do that avoiding looping, since from the very little I understood from numpy arrays, they are not meant to be iterated over in a classic way. In other words, I need it to be very fast, because my actual arrays are quite big.
Any idea?
Thanks a lot!
Arnaud.
numpy
edited Nov 19 '18 at 8:09
Aqueous Carlos
360314
asked Nov 19 '18 at 7:51
ArnaudArnaud
12813
I guess the title of my question might not be very clear..
I have a small array, say a = ([[0,0,0],[0,0,1],[0,1,1]]). Then I have a bigger array of a higher dimension, say b = ([[[2,2,2],[2,0,1],[2,1,1]],[[0,0,0],[3,3,1],[3,1,1]],[...]]).
I'd like to check if one of the elements of a can be found in b. In this case, I'd find that the first element of a [0,0,0] is indeed in b, and then I'd like to retrieve the corresponding index in b.
I'd like to do that avoiding looping, since from the very little I understood from numpy arrays, they are not meant to be iterated over in a classic way. In other words, I need it to be very fast, because my actual arrays are quite big.
Any idea?
Thanks a lot!
Arnaud.
numpy
numpy
edited Nov 19 '18 at 8:09
Aqueous Carlos
360314
asked Nov 19 '18 at 7:51
ArnaudArnaud
12813
edited Nov 19 '18 at 8:09
Aqueous Carlos
360314
asked Nov 19 '18 at 7:51
ArnaudArnaud
12813
edited Nov 19 '18 at 8:09
Aqueous Carlos
360314
edited Nov 19 '18 at 8:09
Aqueous Carlos
360314
edited Nov 19 '18 at 8:09
Aqueous Carlos
360314
360314
asked Nov 19 '18 at 7:51
ArnaudArnaud
12813
asked Nov 19 '18 at 7:51
ArnaudArnaud
12813
asked Nov 19 '18 at 7:51
ArnaudArnaud
12813
12813
add a comment |
add a comment |
1 Answer
1
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oldest
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I don't know of a direct way, but I here's a function that works around the problem:
import numpy as np
def find_indices(val, arr):
# first take a mean at the lowest level of each array,
# then compare these to eliminate the majority of entries
mb = np.mean(arr, axis=2); ma = np.mean(val)
Y = np.argwhere(mb==ma)
indices =
# Then run a quick loop on the remaining elements to
# eliminate arrays that don't match the order
for i in range(len(Y)):
idx = (Y[i,0],Y[i,1])
if np.array_equal(val, arr[idx]):
indices.append(idx)
return indices
# Sample arrays
a = np.array([[0,0,0],[0,0,1],[0,1,1]])
b = np.array([ [[6,5,4],[0,0,1],[2,3,3]],
[[2,5,4],[6,5,4],[0,0,0]],
[[2,0,2],[3,5,4],[5,4,6]],
[[6,5,4],[0,0,0],[2,5,3]] ])
print(find_indices(a[0], b))
# [(1, 2), (3, 1)]
print(find_indices(a[1], b))
# [(0, 1)]
The idea is to use the mean of each array and compare this with the mean of the input. np.argwhere() is the key here. That way you remove most of the unwanted matches, but I did need to use a loop on the remainder to avoid the unsorted matches (this shouldn't be too memory-consuming). You'll probably want to customise it further, but I hope this helps.
answered Nov 19 '18 at 18:37
John KealyJohn Kealy
808
Thanks for the suggestion! Indeed such kind of workaround is a help!
– Arnaud
Nov 20 '18 at 15:20
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
I don't know of a direct way, but I here's a function that works around the problem:
import numpy as np
def find_indices(val, arr):
# first take a mean at the lowest level of each array,
# then compare these to eliminate the majority of entries
mb = np.mean(arr, axis=2); ma = np.mean(val)
Y = np.argwhere(mb==ma)
indices =
# Then run a quick loop on the remaining elements to
# eliminate arrays that don't match the order
for i in range(len(Y)):
idx = (Y[i,0],Y[i,1])
if np.array_equal(val, arr[idx]):
indices.append(idx)
return indices
# Sample arrays
a = np.array([[0,0,0],[0,0,1],[0,1,1]])
b = np.array([ [[6,5,4],[0,0,1],[2,3,3]],
[[2,5,4],[6,5,4],[0,0,0]],
[[2,0,2],[3,5,4],[5,4,6]],
[[6,5,4],[0,0,0],[2,5,3]] ])
print(find_indices(a[0], b))
# [(1, 2), (3, 1)]
print(find_indices(a[1], b))
# [(0, 1)]
The idea is to use the mean of each array and compare this with the mean of the input. np.argwhere() is the key here. That way you remove most of the unwanted matches, but I did need to use a loop on the remainder to avoid the unsorted matches (this shouldn't be too memory-consuming). You'll probably want to customise it further, but I hope this helps.
answered Nov 19 '18 at 18:37
John KealyJohn Kealy
808
Thanks for the suggestion! Indeed such kind of workaround is a help!
– Arnaud
Nov 20 '18 at 15:20
add a comment |
I don't know of a direct way, but I here's a function that works around the problem:
import numpy as np
def find_indices(val, arr):
# first take a mean at the lowest level of each array,
# then compare these to eliminate the majority of entries
mb = np.mean(arr, axis=2); ma = np.mean(val)
Y = np.argwhere(mb==ma)
indices =
# Then run a quick loop on the remaining elements to
# eliminate arrays that don't match the order
for i in range(len(Y)):
idx = (Y[i,0],Y[i,1])
if np.array_equal(val, arr[idx]):
indices.append(idx)
return indices
# Sample arrays
a = np.array([[0,0,0],[0,0,1],[0,1,1]])
b = np.array([ [[6,5,4],[0,0,1],[2,3,3]],
[[2,5,4],[6,5,4],[0,0,0]],
[[2,0,2],[3,5,4],[5,4,6]],
[[6,5,4],[0,0,0],[2,5,3]] ])
print(find_indices(a[0], b))
# [(1, 2), (3, 1)]
print(find_indices(a[1], b))
# [(0, 1)]
The idea is to use the mean of each array and compare this with the mean of the input. np.argwhere() is the key here. That way you remove most of the unwanted matches, but I did need to use a loop on the remainder to avoid the unsorted matches (this shouldn't be too memory-consuming). You'll probably want to customise it further, but I hope this helps.
answered Nov 19 '18 at 18:37
John KealyJohn Kealy
808
Thanks for the suggestion! Indeed such kind of workaround is a help!
– Arnaud
Nov 20 '18 at 15:20
add a comment |
I don't know of a direct way, but I here's a function that works around the problem:
import numpy as np
def find_indices(val, arr):
# first take a mean at the lowest level of each array,
# then compare these to eliminate the majority of entries
mb = np.mean(arr, axis=2); ma = np.mean(val)
Y = np.argwhere(mb==ma)
indices =
# Then run a quick loop on the remaining elements to
# eliminate arrays that don't match the order
for i in range(len(Y)):
idx = (Y[i,0],Y[i,1])
if np.array_equal(val, arr[idx]):
indices.append(idx)
return indices
# Sample arrays
a = np.array([[0,0,0],[0,0,1],[0,1,1]])
b = np.array([ [[6,5,4],[0,0,1],[2,3,3]],
[[2,5,4],[6,5,4],[0,0,0]],
[[2,0,2],[3,5,4],[5,4,6]],
[[6,5,4],[0,0,0],[2,5,3]] ])
print(find_indices(a[0], b))
# [(1, 2), (3, 1)]
print(find_indices(a[1], b))
# [(0, 1)]
The idea is to use the mean of each array and compare this with the mean of the input. np.argwhere() is the key here. That way you remove most of the unwanted matches, but I did need to use a loop on the remainder to avoid the unsorted matches (this shouldn't be too memory-consuming). You'll probably want to customise it further, but I hope this helps.
answered Nov 19 '18 at 18:37
John KealyJohn Kealy
808
I don't know of a direct way, but I here's a function that works around the problem:
import numpy as np
def find_indices(val, arr):
# first take a mean at the lowest level of each array,
# then compare these to eliminate the majority of entries
mb = np.mean(arr, axis=2); ma = np.mean(val)
Y = np.argwhere(mb==ma)
indices =
# Then run a quick loop on the remaining elements to
# eliminate arrays that don't match the order
for i in range(len(Y)):
idx = (Y[i,0],Y[i,1])
if np.array_equal(val, arr[idx]):
indices.append(idx)
return indices
# Sample arrays
a = np.array([[0,0,0],[0,0,1],[0,1,1]])
b = np.array([ [[6,5,4],[0,0,1],[2,3,3]],
[[2,5,4],[6,5,4],[0,0,0]],
[[2,0,2],[3,5,4],[5,4,6]],
[[6,5,4],[0,0,0],[2,5,3]] ])
print(find_indices(a[0], b))
# [(1, 2), (3, 1)]
print(find_indices(a[1], b))
# [(0, 1)]
The idea is to use the mean of each array and compare this with the mean of the input. np.argwhere() is the key here. That way you remove most of the unwanted matches, but I did need to use a loop on the remainder to avoid the unsorted matches (this shouldn't be too memory-consuming). You'll probably want to customise it further, but I hope this helps.
answered Nov 19 '18 at 18:37
John KealyJohn Kealy
808
answered Nov 19 '18 at 18:37
John KealyJohn Kealy
808
answered Nov 19 '18 at 18:37
John KealyJohn Kealy
808
answered Nov 19 '18 at 18:37
John KealyJohn Kealy
808
808
Thanks for the suggestion! Indeed such kind of workaround is a help!
– Arnaud
Nov 20 '18 at 15:20
add a comment |
Thanks for the suggestion! Indeed such kind of workaround is a help!
– Arnaud
Nov 20 '18 at 15:20
Thanks for the suggestion! Indeed such kind of workaround is a help!
– Arnaud
Nov 20 '18 at 15:20
Thanks for the suggestion! Indeed such kind of workaround is a help!
– Arnaud
Nov 20 '18 at 15:20
add a comment |
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