Use Euclidean distance in peak_local_max as new default

[patquem]

Description:

Hi,
It took me a while to understand why some maximum points were not detected in my case (diffraction images of a crystal).
I finally found the reason: the default choice for the distance calculation with p-norm=np.inf.
My personal representation of a distance in space tends naturally towards Euclidean distance.
So why such a choice (p_norm = inf) ? p-norm=2 wouldn't be more appropriate ?
Patrick

[stefanv]

With p_norm=inf you get the Chebychev distance, which can be read "largest difference along any axis". I don't know why that was chosen as the default. Perhaps @rfezzani, who implemented it, can comment?

[rfezzani]

Hello @patquem, and sorry for the late answer 🙏.
This choice was made for retro-compatibility reason! Previous implementation was searching for the local minima in square region around each localization, which corresponds to Chebychev distance...
@scikit-image/core, maybe this default choice can revised for the future major version?

[patquem]

Hello,
Thank you for your answers.
I understand the reason of retro-compatibility.
Hope you can revisit this :)

Below is an illustration of the confusion that the current choice can cause.
The crosses in the squares are the Maxima detected with p_norm=np.inf.
In the second figure, only 1 maximum is detected while the 2nd spot is further away than in the first figure (sqrt(50)~7 px vs 6 px)

import numpy as np
import matplotlib.pyplot as plt
from skimage.feature import peak_local_max

arr0 = np.zeros((10, 10))
arr1 = np.zeros((10, 10))

arr0[2, 2] = 1
arr0[2, 8] = 1

arr1[2, 2] = 1
arr1[7, 7] = 1


def plot(ax, arr):
    coords = peak_local_max(arr,
                            min_distance=6,
                            exclude_border=False)
    ax.imshow(arr, origin='lower')
    for i, (y, x) in enumerate(coords):
        ax.plot(x, y, 'rx', ms=10, markerfacecolor="None")


fig, (ax0, ax1) = plt.subplots(1, 2)
plot(ax0, arr0)
plot(ax1, arr1)
plt.show()

[lagru]

I'd also advocate p_norm=2 / Euclidean distance as the least surprising default here.

https://github.com/orgs/scikit-image/teams/core, maybe this default choice can revised for the future major version?

That would probably be easiest. Alternative would be a deprecation, perhaps using a sentinel value to signal a changing default. But tagging this for the v2 API for now, so we don't forget it.

[lagru]

@patquem I took the liberty to make the title a bit more concise but that may be subjective. 😉

[rfezzani]

@patquem

... while the 2nd spot is further away than in the first figure (sqrt(50)~7 px vs 6 px)

This is true if you consider the Euclidean 2-norm, but with the $\infty$-norm the corresponding distances are respectively 5 vs 6 😉.

[github-actions]

Hey, there hasn't been any activity on this issue for more than 180 days. For now, we have marked it as "dormant" until there is some new activity. You are welcome to reach out to people by mentioning them here or on our forum if you need more feedback! If you think that this issue is no longer relevant, you may close it by yourself; otherwise, we may do it at some point (either way, it will be done manually). In any case, thank you for your contributions so far!

[lagru]

I added it to our informal todo list for skimage2 so it isn't forgotten. 😉

[lagru]

Note that I moved the list to our Wiki https://github.com/scikit-image/scikit-image/wiki/API-changes-for-skimage2.