Update numpy at each index based on the previous index
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Is there a vectorized way of updating each arbitrary index in a numpy array based on the indices before it? For example, in pseudo code, if I have the matrix
1 2 3
3 1 4
1 3 2
And for every index (i, j), i want to do:
m[i,j] += max(m[i, j-1], m[i-1, j])
Now I know I can do this iteratively, but I want to know if there's a vectorized way to do this, since it would be more efficient than taking it out of the numpy data space over and over again.
Also, I know this is a fence posting problem, since m[0, 0] doesn't have a previous element. This is easily fixed by prepending an extra row and column of 0's to the matrix.
python numpy
asked Nov 23 '18 at 20:35
Mauricio MartinezMauricio Martinez
206
add a comment |
Is there a vectorized way of updating each arbitrary index in a numpy array based on the indices before it? For example, in pseudo code, if I have the matrix
1 2 3
3 1 4
1 3 2
And for every index (i, j), i want to do:
m[i,j] += max(m[i, j-1], m[i-1, j])
Now I know I can do this iteratively, but I want to know if there's a vectorized way to do this, since it would be more efficient than taking it out of the numpy data space over and over again.
Also, I know this is a fence posting problem, since m[0, 0] doesn't have a previous element. This is easily fixed by prepending an extra row and column of 0's to the matrix.
python numpy
asked Nov 23 '18 at 20:35
Mauricio MartinezMauricio Martinez
206
consider usingnumba
– juanpa.arrivillaga
Nov 23 '18 at 20:45
add a comment |
Is there a vectorized way of updating each arbitrary index in a numpy array based on the indices before it? For example, in pseudo code, if I have the matrix
1 2 3
3 1 4
1 3 2
And for every index (i, j), i want to do:
m[i,j] += max(m[i, j-1], m[i-1, j])
Now I know I can do this iteratively, but I want to know if there's a vectorized way to do this, since it would be more efficient than taking it out of the numpy data space over and over again.
Also, I know this is a fence posting problem, since m[0, 0] doesn't have a previous element. This is easily fixed by prepending an extra row and column of 0's to the matrix.
python numpy
asked Nov 23 '18 at 20:35
Mauricio MartinezMauricio Martinez
206
Is there a vectorized way of updating each arbitrary index in a numpy array based on the indices before it? For example, in pseudo code, if I have the matrix
1 2 3
3 1 4
1 3 2
And for every index (i, j), i want to do:
m[i,j] += max(m[i, j-1], m[i-1, j])
Now I know I can do this iteratively, but I want to know if there's a vectorized way to do this, since it would be more efficient than taking it out of the numpy data space over and over again.
Also, I know this is a fence posting problem, since m[0, 0] doesn't have a previous element. This is easily fixed by prepending an extra row and column of 0's to the matrix.
python numpy
python numpy
asked Nov 23 '18 at 20:35
Mauricio MartinezMauricio Martinez
206
asked Nov 23 '18 at 20:35
Mauricio MartinezMauricio Martinez
206
asked Nov 23 '18 at 20:35
Mauricio MartinezMauricio Martinez
206
asked Nov 23 '18 at 20:35
Mauricio MartinezMauricio Martinez
206
asked Nov 23 '18 at 20:35
Mauricio MartinezMauricio Martinez
206
206
consider usingnumba
– juanpa.arrivillaga
Nov 23 '18 at 20:45
add a comment |
consider usingnumba
– juanpa.arrivillaga
Nov 23 '18 at 20:45
consider using
numba– juanpa.arrivillaga
Nov 23 '18 at 20:45
consider using
numba– juanpa.arrivillaga
Nov 23 '18 at 20:45
add a comment |
2 Answers
2
active
oldest
votes
Here's a way you could vectorize it:
arr = np.array([[1, 2, 3],[3, 1, 4],[1, 3, 2]])
arr_A = np.roll(arr, 1, axis=0)
arr_B = np.roll(arr, 1, axis=1)
max_val = np.maximum(arr_A, arr_B)
output = arr + max_val
>>> [[4 5 5]
[7 4 7]
[4 4 6]]
Note that this gives a different answer to your code above because the way you have it written means the values are updated after every loop. If you want that, then you are tied to using the for loops.
>>> [[ 4 6 9] # Output after updating the matrix in each loop.
[ 7 8 13]
[ 8 11 15]]
If you are looking for a similar kind of algorithm rather than trying to recover this exact output then np.roll() should work to speed things up.
answered Nov 23 '18 at 21:00
berkelemberkelem
9632721
add a comment |
You can use numpy.roll to created shifted versions of the matrix:
m += np.maximum(np.roll(m, 1, axis=0), np.roll(m, 1, axis=1))
This creates two new copies though. Zero padding is required because roll re-introduces elements that "rolled" beyond the boundary:
p = np.pad(m, [(1, 1), (1, 1)], 'constant')
m += np.maximum(np.roll(p, 1, axis=0), np.roll(p, 1, axis=1))[1:-1, 1:-1]
answered Nov 23 '18 at 21:02
a_guesta_guest
7,56231345
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's a way you could vectorize it:
arr = np.array([[1, 2, 3],[3, 1, 4],[1, 3, 2]])
arr_A = np.roll(arr, 1, axis=0)
arr_B = np.roll(arr, 1, axis=1)
max_val = np.maximum(arr_A, arr_B)
output = arr + max_val
>>> [[4 5 5]
[7 4 7]
[4 4 6]]
Note that this gives a different answer to your code above because the way you have it written means the values are updated after every loop. If you want that, then you are tied to using the for loops.
>>> [[ 4 6 9] # Output after updating the matrix in each loop.
[ 7 8 13]
[ 8 11 15]]
If you are looking for a similar kind of algorithm rather than trying to recover this exact output then np.roll() should work to speed things up.
answered Nov 23 '18 at 21:00
berkelemberkelem
9632721
add a comment |
Here's a way you could vectorize it:
arr = np.array([[1, 2, 3],[3, 1, 4],[1, 3, 2]])
arr_A = np.roll(arr, 1, axis=0)
arr_B = np.roll(arr, 1, axis=1)
max_val = np.maximum(arr_A, arr_B)
output = arr + max_val
>>> [[4 5 5]
[7 4 7]
[4 4 6]]
Note that this gives a different answer to your code above because the way you have it written means the values are updated after every loop. If you want that, then you are tied to using the for loops.
>>> [[ 4 6 9] # Output after updating the matrix in each loop.
[ 7 8 13]
[ 8 11 15]]
If you are looking for a similar kind of algorithm rather than trying to recover this exact output then np.roll() should work to speed things up.
answered Nov 23 '18 at 21:00
berkelemberkelem
9632721
add a comment |
Here's a way you could vectorize it:
arr = np.array([[1, 2, 3],[3, 1, 4],[1, 3, 2]])
arr_A = np.roll(arr, 1, axis=0)
arr_B = np.roll(arr, 1, axis=1)
max_val = np.maximum(arr_A, arr_B)
output = arr + max_val
>>> [[4 5 5]
[7 4 7]
[4 4 6]]
Note that this gives a different answer to your code above because the way you have it written means the values are updated after every loop. If you want that, then you are tied to using the for loops.
>>> [[ 4 6 9] # Output after updating the matrix in each loop.
[ 7 8 13]
[ 8 11 15]]
If you are looking for a similar kind of algorithm rather than trying to recover this exact output then np.roll() should work to speed things up.
answered Nov 23 '18 at 21:00
berkelemberkelem
9632721
Here's a way you could vectorize it:
arr = np.array([[1, 2, 3],[3, 1, 4],[1, 3, 2]])
arr_A = np.roll(arr, 1, axis=0)
arr_B = np.roll(arr, 1, axis=1)
max_val = np.maximum(arr_A, arr_B)
output = arr + max_val
>>> [[4 5 5]
[7 4 7]
[4 4 6]]
Note that this gives a different answer to your code above because the way you have it written means the values are updated after every loop. If you want that, then you are tied to using the for loops.
>>> [[ 4 6 9] # Output after updating the matrix in each loop.
[ 7 8 13]
[ 8 11 15]]
If you are looking for a similar kind of algorithm rather than trying to recover this exact output then np.roll() should work to speed things up.
answered Nov 23 '18 at 21:00
berkelemberkelem
9632721
answered Nov 23 '18 at 21:00
berkelemberkelem
9632721
answered Nov 23 '18 at 21:00
berkelemberkelem
9632721
answered Nov 23 '18 at 21:00
berkelemberkelem
9632721
9632721
add a comment |
add a comment |
You can use numpy.roll to created shifted versions of the matrix:
m += np.maximum(np.roll(m, 1, axis=0), np.roll(m, 1, axis=1))
This creates two new copies though. Zero padding is required because roll re-introduces elements that "rolled" beyond the boundary:
p = np.pad(m, [(1, 1), (1, 1)], 'constant')
m += np.maximum(np.roll(p, 1, axis=0), np.roll(p, 1, axis=1))[1:-1, 1:-1]
answered Nov 23 '18 at 21:02
a_guesta_guest
7,56231345
add a comment |
You can use numpy.roll to created shifted versions of the matrix:
m += np.maximum(np.roll(m, 1, axis=0), np.roll(m, 1, axis=1))
This creates two new copies though. Zero padding is required because roll re-introduces elements that "rolled" beyond the boundary:
p = np.pad(m, [(1, 1), (1, 1)], 'constant')
m += np.maximum(np.roll(p, 1, axis=0), np.roll(p, 1, axis=1))[1:-1, 1:-1]
answered Nov 23 '18 at 21:02
a_guesta_guest
7,56231345
add a comment |
You can use numpy.roll to created shifted versions of the matrix:
m += np.maximum(np.roll(m, 1, axis=0), np.roll(m, 1, axis=1))
This creates two new copies though. Zero padding is required because roll re-introduces elements that "rolled" beyond the boundary:
p = np.pad(m, [(1, 1), (1, 1)], 'constant')
m += np.maximum(np.roll(p, 1, axis=0), np.roll(p, 1, axis=1))[1:-1, 1:-1]
answered Nov 23 '18 at 21:02
a_guesta_guest
7,56231345
You can use numpy.roll to created shifted versions of the matrix:
m += np.maximum(np.roll(m, 1, axis=0), np.roll(m, 1, axis=1))
This creates two new copies though. Zero padding is required because roll re-introduces elements that "rolled" beyond the boundary:
p = np.pad(m, [(1, 1), (1, 1)], 'constant')
m += np.maximum(np.roll(p, 1, axis=0), np.roll(p, 1, axis=1))[1:-1, 1:-1]
answered Nov 23 '18 at 21:02
a_guesta_guest
7,56231345
edited Nov 23 '18 at 21:09
answered Nov 23 '18 at 21:02
a_guesta_guest
7,56231345
answered Nov 23 '18 at 21:02
a_guesta_guest
7,56231345
answered Nov 23 '18 at 21:02
a_guesta_guest
7,56231345
7,56231345
add a comment |
add a comment |
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consider using
numba– juanpa.arrivillaga
Nov 23 '18 at 20:45