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Local maxima 3D array python
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I'm trying to find the local maxima in a 3D numpy array, but I can't seem to find an easy way to do that using numpy, scipy, or anything else.
For now I implemented it using scipy.signal.argrelexrema. But it's very long to process large arrays, and only works on separated axes.
import numpy as np
from scipy.signal import argrelextrema
def local_maxima_3D(data, order=1):
"""Detects local maxima in a 3D array
Parameters
---------
data : 3d ndarray
order : int
How many points on each side to use for the comparison
Returns
-------
coordinates : ndarray
coordinates of the local maxima
values : ndarray
values of the local maxima
"""
# Coordinates of local maxima along each axis
peaks0 = np.array(argrelextrema(data, np.greater, axis=0, order=order))
peaks1 = np.array(argrelextrema(data, np.greater, axis=1, order=order))
peaks2 = np.array(argrelextrema(data, np.greater, axis=2, order=order))
# Stack all coordinates
stacked = np.vstack((peaks0.transpose(), peaks1.transpose(),
peaks2.transpose()))
# We keep coordinates that appear three times (once for each axis)
elements, counts = np.unique(stacked, axis=0, return_counts=True)
coords = elements[np.where(counts == 3)[0]]
# Compute values at filtered coordinates
values = data[coords[:, 0], coords[:, 1], coords[:, 2]]
return coords, values
I know this solution is far from optimal and only works with order=1. Is there any better way to find local maxima in a 3D array in Python?
python numpy scipy
edited yesterday
Jamal♦
30.6k11121227
asked Apr 1 at 13:35
theobdttheobdt
111
New contributor
theobdt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm trying to find the local maxima in a 3D numpy array, but I can't seem to find an easy way to do that using numpy, scipy, or anything else.
For now I implemented it using scipy.signal.argrelexrema. But it's very long to process large arrays, and only works on separated axes.
import numpy as np
from scipy.signal import argrelextrema
def local_maxima_3D(data, order=1):
"""Detects local maxima in a 3D array
Parameters
---------
data : 3d ndarray
order : int
How many points on each side to use for the comparison
Returns
-------
coordinates : ndarray
coordinates of the local maxima
values : ndarray
values of the local maxima
"""
# Coordinates of local maxima along each axis
peaks0 = np.array(argrelextrema(data, np.greater, axis=0, order=order))
peaks1 = np.array(argrelextrema(data, np.greater, axis=1, order=order))
peaks2 = np.array(argrelextrema(data, np.greater, axis=2, order=order))
# Stack all coordinates
stacked = np.vstack((peaks0.transpose(), peaks1.transpose(),
peaks2.transpose()))
# We keep coordinates that appear three times (once for each axis)
elements, counts = np.unique(stacked, axis=0, return_counts=True)
coords = elements[np.where(counts == 3)[0]]
# Compute values at filtered coordinates
values = data[coords[:, 0], coords[:, 1], coords[:, 2]]
return coords, values
I know this solution is far from optimal and only works with order=1. Is there any better way to find local maxima in a 3D array in Python?
python numpy scipy
edited yesterday
Jamal♦
30.6k11121227
asked Apr 1 at 13:35
theobdttheobdt
111
New contributor
theobdt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Do you know anything about the data? If they're determined by a formula, then you should consider getting an analytical answer. Otherwise, the data being continuous, derivable and/or single-maximum would allow for your use of an optimizer, especially if you can provide a gradient function.
$endgroup$
– Reinderien
Apr 1 at 13:54
$begingroup$
As opposed tounique(stacked(...))[where(counts==3)], have you tried set intersection, likeset(peaks0) & set(peaks1) & set(peaks2). Of course, thepeaks#coordinates would need to be converted into something hashable (tuples) in order to create the sets. I don’t know if this would be faster, so this is an experiment not an answer ...
$endgroup$
– AJNeufeld
Apr 1 at 14:28
add a comment |
$begingroup$
I'm trying to find the local maxima in a 3D numpy array, but I can't seem to find an easy way to do that using numpy, scipy, or anything else.
For now I implemented it using scipy.signal.argrelexrema. But it's very long to process large arrays, and only works on separated axes.
import numpy as np
from scipy.signal import argrelextrema
def local_maxima_3D(data, order=1):
"""Detects local maxima in a 3D array
Parameters
---------
data : 3d ndarray
order : int
How many points on each side to use for the comparison
Returns
-------
coordinates : ndarray
coordinates of the local maxima
values : ndarray
values of the local maxima
"""
# Coordinates of local maxima along each axis
peaks0 = np.array(argrelextrema(data, np.greater, axis=0, order=order))
peaks1 = np.array(argrelextrema(data, np.greater, axis=1, order=order))
peaks2 = np.array(argrelextrema(data, np.greater, axis=2, order=order))
# Stack all coordinates
stacked = np.vstack((peaks0.transpose(), peaks1.transpose(),
peaks2.transpose()))
# We keep coordinates that appear three times (once for each axis)
elements, counts = np.unique(stacked, axis=0, return_counts=True)
coords = elements[np.where(counts == 3)[0]]
# Compute values at filtered coordinates
values = data[coords[:, 0], coords[:, 1], coords[:, 2]]
return coords, values
I know this solution is far from optimal and only works with order=1. Is there any better way to find local maxima in a 3D array in Python?
python numpy scipy
edited yesterday
Jamal♦
30.6k11121227
asked Apr 1 at 13:35
theobdttheobdt
111
New contributor
theobdt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm trying to find the local maxima in a 3D numpy array, but I can't seem to find an easy way to do that using numpy, scipy, or anything else.
For now I implemented it using scipy.signal.argrelexrema. But it's very long to process large arrays, and only works on separated axes.
import numpy as np
from scipy.signal import argrelextrema
def local_maxima_3D(data, order=1):
"""Detects local maxima in a 3D array
Parameters
---------
data : 3d ndarray
order : int
How many points on each side to use for the comparison
Returns
-------
coordinates : ndarray
coordinates of the local maxima
values : ndarray
values of the local maxima
"""
# Coordinates of local maxima along each axis
peaks0 = np.array(argrelextrema(data, np.greater, axis=0, order=order))
peaks1 = np.array(argrelextrema(data, np.greater, axis=1, order=order))
peaks2 = np.array(argrelextrema(data, np.greater, axis=2, order=order))
# Stack all coordinates
stacked = np.vstack((peaks0.transpose(), peaks1.transpose(),
peaks2.transpose()))
# We keep coordinates that appear three times (once for each axis)
elements, counts = np.unique(stacked, axis=0, return_counts=True)
coords = elements[np.where(counts == 3)[0]]
# Compute values at filtered coordinates
values = data[coords[:, 0], coords[:, 1], coords[:, 2]]
return coords, values
I know this solution is far from optimal and only works with order=1. Is there any better way to find local maxima in a 3D array in Python?
python numpy scipy
python numpy scipy
edited yesterday
Jamal♦
30.6k11121227
asked Apr 1 at 13:35
theobdttheobdt
111
New contributor
theobdt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Jamal♦
30.6k11121227
asked Apr 1 at 13:35
theobdttheobdt
111
New contributor
theobdt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Jamal♦
30.6k11121227
edited yesterday
Jamal♦
30.6k11121227
edited yesterday
Jamal♦
30.6k11121227
30.6k11121227
asked Apr 1 at 13:35
theobdttheobdt
111
New contributor
theobdt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 1 at 13:35
theobdttheobdt
111
asked Apr 1 at 13:35
theobdttheobdt
111
111
New contributor
theobdt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
theobdt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
theobdt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Do you know anything about the data? If they're determined by a formula, then you should consider getting an analytical answer. Otherwise, the data being continuous, derivable and/or single-maximum would allow for your use of an optimizer, especially if you can provide a gradient function.
$endgroup$
– Reinderien
Apr 1 at 13:54
$begingroup$
As opposed tounique(stacked(...))[where(counts==3)], have you tried set intersection, likeset(peaks0) & set(peaks1) & set(peaks2). Of course, thepeaks#coordinates would need to be converted into something hashable (tuples) in order to create the sets. I don’t know if this would be faster, so this is an experiment not an answer ...
$endgroup$
– AJNeufeld
Apr 1 at 14:28
add a comment |
2
$begingroup$
Do you know anything about the data? If they're determined by a formula, then you should consider getting an analytical answer. Otherwise, the data being continuous, derivable and/or single-maximum would allow for your use of an optimizer, especially if you can provide a gradient function.
$endgroup$
– Reinderien
Apr 1 at 13:54
$begingroup$
As opposed tounique(stacked(...))[where(counts==3)], have you tried set intersection, likeset(peaks0) & set(peaks1) & set(peaks2). Of course, thepeaks#coordinates would need to be converted into something hashable (tuples) in order to create the sets. I don’t know if this would be faster, so this is an experiment not an answer ...
$endgroup$
– AJNeufeld
Apr 1 at 14:28
2
2
$begingroup$
Do you know anything about the data? If they're determined by a formula, then you should consider getting an analytical answer. Otherwise, the data being continuous, derivable and/or single-maximum would allow for your use of an optimizer, especially if you can provide a gradient function.
$endgroup$
– Reinderien
Apr 1 at 13:54
$begingroup$
Do you know anything about the data? If they're determined by a formula, then you should consider getting an analytical answer. Otherwise, the data being continuous, derivable and/or single-maximum would allow for your use of an optimizer, especially if you can provide a gradient function.
$endgroup$
– Reinderien
Apr 1 at 13:54
$begingroup$
As opposed to
unique(stacked(...))[where(counts==3)], have you tried set intersection, like set(peaks0) & set(peaks1) & set(peaks2). Of course, the peaks# coordinates would need to be converted into something hashable (tuples) in order to create the sets. I don’t know if this would be faster, so this is an experiment not an answer ...$endgroup$
– AJNeufeld
Apr 1 at 14:28
$begingroup$
As opposed to
unique(stacked(...))[where(counts==3)], have you tried set intersection, like set(peaks0) & set(peaks1) & set(peaks2). Of course, the peaks# coordinates would need to be converted into something hashable (tuples) in order to create the sets. I don’t know if this would be faster, so this is an experiment not an answer ...$endgroup$
– AJNeufeld
Apr 1 at 14:28
add a comment |
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$begingroup$
Do you know anything about the data? If they're determined by a formula, then you should consider getting an analytical answer. Otherwise, the data being continuous, derivable and/or single-maximum would allow for your use of an optimizer, especially if you can provide a gradient function.
$endgroup$
– Reinderien
Apr 1 at 13:54
$begingroup$
As opposed to
unique(stacked(...))[where(counts==3)], have you tried set intersection, likeset(peaks0) & set(peaks1) & set(peaks2). Of course, thepeaks#coordinates would need to be converted into something hashable (tuples) in order to create the sets. I don’t know if this would be faster, so this is an experiment not an answer ...$endgroup$
– AJNeufeld
Apr 1 at 14:28