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Updated scaling factors for proper geometric representation of uncertainty ellipses #1067

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Apr 12, 2022
Merged

Updated scaling factors for proper geometric representation of uncertainty ellipses #1067

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d9a00de
Modified the scaling values for plotting uncertainty ellipses to prop…
magicbycalvin Jan 25, 2022
1b81776
Changed all instances of the Sigma value, k, to 5 for plotting the co…
magicbycalvin Jan 28, 2022
e524e28
Updated comments to reflect standard deviations instead of variances …
magicbycalvin Jan 28, 2022
df7fb47
Merge branch 'fix/ellipses' into develop
dellaert Apr 12, 2022
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42 changes: 27 additions & 15 deletions python/gtsam/utils/plot.py
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Original file line number Diff line number Diff line change
Expand Up @@ -75,8 +75,9 @@ def plot_covariance_ellipse_3d(axes,
Plots a Gaussian as an uncertainty ellipse

Based on Maybeck Vol 1, page 366
k=2.296 corresponds to 1 std, 68.26% of all probability
k=11.82 corresponds to 3 std, 99.74% of all probability
For the 3D case:
k = 3.527 corresponds to 1 std, 68.26% of all probability
k = 14.157 corresponds to 3 std, 99.74% of all probability
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Do you have an online reference?

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@senselessDev senselessDev Jan 26, 2022 •
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With code from other comment:

pct_to_sigma(sigma_to_pct(1, 1), 3) ** 2 --> 3.5267403802617303
pct_to_sigma(sigma_to_pct(3, 1), 3) ** 2 --> 14.156413609126677

So seems about right if we want this behavior.

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I think these (3.5 and 14.1) should still be changed to standard deviation


Args:
axes (matplotlib.axes.Axes): Matplotlib axes.
Expand All @@ -87,7 +88,7 @@ def plot_covariance_ellipse_3d(axes,
n: Defines the granularity of the ellipse. Higher values indicate finer ellipses.
alpha: Transparency value for the plotted surface in the range [0, 1].
"""
k = 11.82
k = np.sqrt(14.157)
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above = 14.157 and here it is sqrt(14.157)

Did you take a look at the PR #1063 as I suggested?

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I took a look at PR #1063 but am not sure how it is relevant. Perhaps I am missing something? I am new to contributing to open source projects so perhaps there was a file or something I needed to look at. As far as I can tell, the discussion there is regarding removing a negative sign in equations 5.3 and 5.6 in the math.pdf document.

As for the k values I chose, they come from the Chi Squared distribution's p-table for different dimensions. E.g., for a single dimension, k = 9 corresponds to 3 standard deviations, for 2D we have k = 11.820, for 3D we have k = 14.157, and for 4D k = 16.251 (all values taken from the table on page 366 of the Maybeck Vol 1 textbook). Note that I am choosing k to be the variance rather than the standard deviation, which is why you see the square root. I chose to write the comments as I did simply because I wanted to keep them as similar as possible to the previous comments pointing out the k values.

The most useful online reference I could find for choosing these k values can be found here and the Maybeck textbook can be found here.

Please let me know if I can answer any other questions.

Thanks!

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Take a look at my note at the top of plot.py. I added some python code for 2D.

  1. In 2D I chose to leave the k=5, because it is consistent with what you said in the comments, and it is referenced in the documentation.
  2. I don't think we should change the semantics of k in the middle of the game.

Given the note and 1 and 2 above, do you agree that code was already good as it was? You would then only have to bring 3D in line.

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Just for reference, we can easily calculate that in python with scipy (even though I guess we won't include that as a dependency, we can put the code in the comment to regenerate the values):

def pct_to_sigma(pct, dof):
    return np.sqrt(scipy.stats.chi2.ppf(pct / 100., df=dof))

def sigma_to_pct(sigma, dof):
    return scipy.stats.chi2.cdf(sigma**2, df=dof) * 100.

for dim in range(0, 4):
    print("{}D".format(dim), end="")
    for n_sigma in range(1, 6):
        if dim == 0: print("\t    {}    ".format(n_sigma), end="")
        else: print("\t{:.5f}".format(sigma_to_pct(n_sigma, dim)), end="")
    print()
0D	    1    	    2    	    3    	    4    	    5    
1D	68.26895	95.44997	99.73002	99.99367	99.99994
2D	39.34693	86.46647	98.88910	99.96645	99.99963
3D	19.87480	73.85359	97.07091	99.88660	99.99846

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Cool, this code is more general than the one I put in plot.py (which I assume you looked at) and it agrees with it in the 2D case. So I propose:

  • undo the changes to 2D plotting
  • we keep k=5 for 2D
  • k == sigma, not variance
  • use k = 5 for 3D as well, which is 99.99846 of all probability.
  • make sure it multiplies sigmas, not variances.

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Just for reference, we can easily calculate that in python with scipy (even though I guess we won't include that as a dependency, we can put the code in the comment to regenerate the values):

Actually, maybe we should include this code, and comment on some of its output values, just like @senselessDev did below. I think that it makes sense for users to specify an integer number of standard deviations, and we can document in 2D and 3D how much percentage 1,2,3,4,5 stddevs corresponds to. k can be an input argument defaulting to 5. But we could provide another optional argument, percentage that, when given, calculates k using pct_to_sigma.

0D 1 2 3 4 5
1D 68.26895 95.44997 99.73002 99.99367 99.99994
2D 39.34693 86.46647 98.88910 99.96645 99.99963
3D 19.87480 73.85359 97.07091 99.88660 99.99846

U, S, _ = np.linalg.svd(P)

radii = k * np.sqrt(S)
Expand All @@ -113,7 +114,14 @@ def plot_point2_on_axes(axes,
linespec: str,
P: Optional[np.ndarray] = None) -> None:
"""
Plot a 2D point on given axis `axes` with given `linespec`.
Plot a 2D point and its corresponding uncertainty ellipse on given axis
`axes` with given `linespec`.

Based on Stochastic Models, Estimation, and Control Vol 1 by Maybeck,
page 366
For the 2D case:
k = 2.296 corresponds to 1 std, 68.26% of all probability
k = 11.820 corresponds to 3 std, 99.74% of all probability
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@senselessDev senselessDev Jan 26, 2022 •
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pct_to_sigma(sigma_to_pct(1, 1), 2) ** 2 --> 2.295748928898636
pct_to_sigma(sigma_to_pct(3, 1), 2) ** 2 --> 11.829158081900795

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same


Args:
axes (matplotlib.axes.Axes): Matplotlib axes.
Expand All @@ -125,16 +133,15 @@ def plot_point2_on_axes(axes,
if P is not None:
w, v = np.linalg.eig(P)

# "Sigma" value for drawing the uncertainty ellipse. 5 sigma corresponds
# to a 99.9999% confidence, i.e. assuming the estimation has been
# computed properly, there is a 99.999% chance that the true position
# of the point will lie within the uncertainty ellipse.
k = 5.0
# Scaling value for the uncertainty ellipse, we multiply by 2 because
# matplotlib takes the diameter and not the radius of the major and
# minor axes of the ellipse.
k = 2*np.sqrt(11.820)

angle = np.arctan2(v[1, 0], v[0, 0])
e1 = patches.Ellipse(point,
np.sqrt(w[0] * k),
np.sqrt(w[1] * k),
np.sqrt(w[0]) * k,
np.sqrt(w[1]) * k,
np.rad2deg(angle),
fill=False)
axes.add_patch(e1)
Expand Down Expand Up @@ -178,6 +185,12 @@ def plot_pose2_on_axes(axes,
"""
Plot a 2D pose on given axis `axes` with given `axis_length`.

Based on Stochastic Models, Estimation, and Control Vol 1 by Maybeck,
page 366
For the 2D case:
k = 2.296 corresponds to 1 std, 68.26% of all probability
k = 11.820 corresponds to 3 std, 99.74% of all probability
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same?


Args:
axes (matplotlib.axes.Axes): Matplotlib axes.
pose: The pose to be plotted.
Expand Down Expand Up @@ -205,13 +218,12 @@ def plot_pose2_on_axes(axes,

w, v = np.linalg.eig(gPp)

# k = 2.296
k = 5.0
k = 2*np.sqrt(11.820)

angle = np.arctan2(v[1, 0], v[0, 0])
e1 = patches.Ellipse(origin,
np.sqrt(w[0] * k),
np.sqrt(w[1] * k),
np.sqrt(w[0]) * k,
np.sqrt(w[1]) * k,
np.rad2deg(angle),
fill=False)
axes.add_patch(e1)
Expand Down