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63b5c5a4a0
is_local_maximum() is a wrapper function for peak_local_max() is_local_maximum() runs much faster (~20% of prior runtime, nearly = to peak_local_max()) All tests in .feature and .morphology subpackages pass as written with these changes. Todo: * Fully document API * remove commented-out old algorithm in is_local_maximum() * add new tests for full coverage of new, more complex peak_local_max()
138 lines
4.7 KiB
Python
138 lines
4.7 KiB
Python
import numpy as np
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import scipy.ndimage as ndi
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def peak_local_max(image, min_distance=10, threshold_abs=0, threshold_rel=0.1,
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exclude_border=True, indices=True, num_peaks=np.inf,
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footprint=None, labels=None, **kwargs):
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"""Return coordinates of peaks in an image.
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Peaks are the local maxima in a region of `2 * min_distance + 1`
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(i.e. peaks are separated by at least `min_distance`).
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NOTE: If peaks are flat (i.e. multiple pixels have exact same intensity),
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the coordinates of all pixels are returned.
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Parameters
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----------
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image : ndarray of floats
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Input image.
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min_distance : int
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Minimum number of pixels separating peaks and image boundary.
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threshold : float
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Deprecated. See `threshold_rel`.
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threshold_abs : float
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Minimum intensity of peaks.
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threshold_rel : float
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Minimum intensity of peaks calculated as `max(image) * threshold_rel`.
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num_peaks : int
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Maximum number of peaks. When the number of peaks exceeds `num_peaks`,
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return `num_peaks` coordinates based on peak intensity.
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Returns
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-------
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coordinates : (N, 2) array
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(row, column) coordinates of peaks.
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Notes
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-----
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The peak local maximum function returns the coordinates of local peaks
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(maxima) in a image. A maximum filter is used for finding local maxima.
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This operation dilates the original image. After comparison between
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dilated and original image, peak_local_max function returns the
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coordinates of peaks where dilated image = original.
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Examples
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--------
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>>> im = np.zeros((7, 7))
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>>> im[3, 4] = 1
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>>> im[3, 2] = 1.5
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>>> im
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array([[ 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
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[ 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
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[ 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
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[ 0. , 0. , 1.5, 0. , 1. , 0. , 0. ],
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[ 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
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[ 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
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[ 0. , 0. , 0. , 0. , 0. , 0. , 0. ]])
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>>> peak_local_max(im, min_distance=1)
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array([[3, 2],
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[3, 4]])
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>>> peak_local_max(im, min_distance=2)
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array([[3, 2]])
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"""
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# In the case of labels, recursively build and return an output
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# operating on each label separately; for API compatibility with
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# ..watershed.is_local_maximum()
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if labels is not None:
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label_values = np.unique(labels)
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# Reorder label values to have consecutive integers (no gaps)
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if np.any(np.diff(label_values) != 1):
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mask = labels >= 0
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labels[mask] = rank_order(labels[mask])[0].astype(labels.dtype)
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labels = labels.astype(np.int32)
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out = np.zeros_like(image)
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for label in labels:
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out += peak_local_max(image, min_distance=min_distance,
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threshold_abs=threshold_abs,
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threshold_rel=threshold_rel,
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exclude_border=exclude_border,
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indices=False, num_peaks=np.inf,
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footprint=footprint, labels=None,
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**kwargs)
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if indices is True:
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return np.transpose(out.nonzero())
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else:
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return out
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if np.all(image == image.flat[0]):
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if indices is True:
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return []
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else:
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return np.zeros_like(image)
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image = image.copy()
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# Non maximum filter
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if footprint is not None:
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image_max = ndi.maximum_filter(image, footprint=footprint,
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mode='constant')
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else:
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size = 2 * min_distance + 1
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image_max = ndi.maximum_filter(image, size=size, mode='constant')
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mask = (image == image_max)
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image *= mask
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if exclude_border:
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# Remove the image borders
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image[:min_distance] = 0
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image[-min_distance:] = 0
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image[:, :min_distance] = 0
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image[:, -min_distance:] = 0
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if kwargs.has_key('threshold'):
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threshold_rel = kwargs['threshold']
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# find top peak candidates above a threshold
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peak_threshold = max(np.max(image.ravel()) * threshold_rel, threshold_abs)
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image_t = (image > peak_threshold) * 1
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# get coordinates of peaks
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coordinates = np.transpose(image_t.nonzero())
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if coordinates.shape[0] > num_peaks:
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intensities = image[coordinates[:, 0], coordinates[:, 1]]
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idx_maxsort = np.argsort(intensities)[::-1]
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coordinates = coordinates[idx_maxsort][:num_peaks]
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if indices is True:
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return coordinates
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else:
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out = np.zeros_like(image)
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out[coordinates[:, 0], coordinates[:, 1]] = 1
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return out
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