回答編集履歴
3
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test
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```python
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import cv2
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import numpy
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import numpy as np
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import matplotlib.pyplot as plt
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D2 = numpy.mat(numpy.vstack([x, y, numpy.ones(len(x))])).T
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# forming scatter matrix [eqn. 17] from (*)
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S1 = D1.T * D1
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S2 = D1.T * D2
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S3 = D2.T * D2
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# Constraint matrix [eqn. 18]
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C1 = numpy.mat("0. 0. 2.; 0. -1. 0.; 2. 0. 0.")
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# Reduced scatter matrix [eqn. 29]
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M = C1.I * (S1 - S2 * S3.I * S2.T)
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# M*|a b c >=l|a b c >. Find eigenvalues and eigenvectors from this equation [eqn. 28]
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eval, evec = numpy.linalg.eig(M)
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# eigenvector must meet constraint 4ac - b^2 to be valid.
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cond = 4 * numpy.multiply(evec[0, :], evec[2, :]) - numpy.power(evec[1, :], 2)
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a1 = evec[:, numpy.nonzero(cond.A > 0)[1]]
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# |d f g> = -S3^(-1)*S2^(T)*|a b c> [eqn. 24]
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a2 = -S3.I * S2.T * a1
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# eigenvectors |a b c d f g>
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self.coef = numpy.vstack([a1, a2])
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self._save_parameters()
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def _save_parameters(self):
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"""finds the important parameters of the fitted ellipse
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Theory taken form http://mathworld.wolfram
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Args
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-----
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coef (list): list of the coefficients describing an ellipse
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[a,b,c,d,f,g] corresponding to ax**2+2bxy+cy**2+2dx+2fy+g
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Returns
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_______
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center (List): of the form [x0, y0]
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width (float): major axis
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height (float): minor axis
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phi (float): rotation of major axis form the x-axis in radians
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"""
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# eigenvectors are the coefficients of an ellipse in general form
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# a*x^2 + 2*b*x*y + c*y^2 + 2*d*x + 2*f*y + g = 0 [eqn. 15) from (**) or (***)
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a = self.coef[0, 0]
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b = self.coef[1, 0] / 2.0
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c = self.coef[2, 0]
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d = self.coef[3, 0] / 2.0
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f = self.coef[4, 0] / 2.0
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g = self.coef[5, 0]
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# finding center of ellipse [eqn.19 and 20] from (**)
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x0 = (c * d - b * f) / (b ** 2.0 - a * c)
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y0 = (a * f - b * d) / (b ** 2.0 - a * c)
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# Find the semi-axes lengths [eqn. 21 and 22] from (**)
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numerator = 2 * (a * f * f + c * d * d + g * b * b - 2 * b * d * f - a * c * g)
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denominator1 = (b * b - a * c) * (
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(c - a) * numpy.sqrt(1 + 4 * b * b / ((a - c) * (a - c))) - (c + a)
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)
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denominator2 = (b * b - a * c) * (
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(a - c) * numpy.sqrt(1 + 4 * b * b / ((a - c) * (a - c))) - (c + a)
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)
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width = numpy.sqrt(numerator / denominator1)
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height = numpy.sqrt(numerator / denominator2)
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# angle of counterclockwise rotation of major-axis of ellipse to x-axis [eqn. 23] from (**)
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# or [eqn. 26] from (***).
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phi = 0.5 * numpy.arctan((2.0 * b) / (a - c))
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self._center = [x0, y0]
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self._width = width
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self._height = height
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self._phi = phi
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@property
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def center(self):
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return self._center
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@property
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def width(self):
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return self._width
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@property
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def height(self):
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return self._height
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@property
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def phi(self):
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"""angle of counterclockwise rotation of major-axis of ellipse to x-axis
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[eqn. 23] from (**)
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"""
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return self._phi
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def parameters(self):
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return self.center, self.width, self.height, self.phi
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def make_test_ellipse(center=[1, 1], width=1, height=0.6, phi=3.14 / 5):
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"""Generate Elliptical data with noise
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----
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center (list:float): (<x_location>, <y_location>)
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width (float): semimajor axis. Horizontal dimension of the ellipse (**)
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height (float): semiminor axis. Vertical dimension of the ellipse (**)
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phi (float:radians): tilt of the ellipse, the angle the semimajor axis
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makes with the x-axis
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Returns
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-------
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data (list:list:float): list of two lists containing the x and y data of the
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ellipse. of the form [[x1, x2, ..., xi],[y1, y2, ..., yi]]
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noisy_ellipse = cv2.ellipse2Poly((0, 0), (50, 70), 45, 0, 360, 5)
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# Extract x coords and y coords of the ellipse as column vectors
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X = noisy_ellipse[:, 0:1]
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Y = noisy_ellipse[:, 1:]
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# Formulate and solve the least squares problem ||Ax - b ||^2
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A = np.hstack([X ** 2, X * Y, Y ** 2])
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b = np.ones_like(X)
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x = np.linalg.lstsq(A, b)[0].squeeze()
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# Print the equation of the ellipse in standard form
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print(
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"The ellipse is given by {0:.3}x^2 + {1:.3}xy+{2:.3}y^2 = 1".format(
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x[0], x[1], x[2]
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def plot_ellipse_by_equation(ax, a, b, c):
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"""a x^2 + b xy + c y^2 = 1 で表される楕円を描画する。
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x = -np.linspace(-100, 100, 1000)
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y = np.linspace(-100, 100, 1000)
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X, Y = np.meshgrid(x, y)
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eqn = a * X ** 2 + b * X * Y + c * Y ** 2
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Z = 1
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# 楕円の式をそのまま描画できる関数は matplotlib にないので、
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# f = a * X ** 2 + b * X * Y + c * Y ** 2 の f = 1 の等高線を描画する形で
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# 楕円を描画する。
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ax.contour(X, Y, eqn, levels=[Z], colors=["r"])
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# Plot the noisy data
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fig, ax = plt.subplots()
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ax.grid()
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ax.scatter(X, Y, label="Data Points", color="g")
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plot_ellipse_by_equation(ax, x[0], x[1], x[2])
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```
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## 結果
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![イメージ説明](6330b456050f23b7c2b6c56d9d8c4824.png)
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緑の点がデータ、赤の線が近似した楕円
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## 追記
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長軸 (major axis) と x 軸の角度、短軸 (minor axis) と x 軸の角度が考えられますが、楕円の式を行列表記して、固有値問題を問き、出てきた固有ベクトル2つが長軸、短軸になります。(小さい固有値の固有ベクトルが長軸、大きい固有値の固有ベクトルが短軸に対応する)
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[二次形式の意味,微分,標準形など | 高校数学の美しい物語](https://mathtrain.jp/quadraticform)
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あとは逆正弦関数で x 軸と長軸、または x 軸と短軸の角度を求めればよいです。
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```python
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import matplotlib.pyplot as plt
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import numpy as np
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alpha = 5
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beta = 3
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N = 500
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DIM = 2
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np.random.seed(2)
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# Generate random points on the unit circle by sampling uniform angles
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theta = np.random.uniform(0, 2 * np.pi, (N, 1))
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fig, ax = plt.subplots(figsize=(6, 6))
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# Stretch and rotate circle to an ellipse with random linear tranformation
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B = np.random.randint(-3, 3, (DIM, DIM))
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noisy_ellipse = cv2.ellipse2Poly((0, 0), (50, 70), 45, 0, 360, 5)
|
410
|
-
|
411
|
-
|
412
|
-
|
413
|
-
# Extract x coords and y coords of the ellipse as column vectors
|
414
|
-
|
415
|
-
X = noisy_ellipse[:, 0:1]
|
416
|
-
|
417
|
-
Y = noisy_ellipse[:, 1:]
|
418
|
-
|
419
|
-
|
420
|
-
|
421
|
-
# Formulate and solve the least squares problem ||Ax - b ||^2
|
422
|
-
|
423
|
-
A = np.hstack([X ** 2, X * Y, Y ** 2])
|
424
|
-
|
425
|
-
b = np.ones_like(X)
|
426
|
-
|
427
|
-
x = np.linalg.lstsq(A, b)[0].squeeze()
|
428
|
-
|
429
|
-
|
430
|
-
|
431
|
-
# Print the equation of the ellipse in standard form
|
432
|
-
|
433
|
-
|
185
|
+
ax.grid()
|
434
|
-
|
435
|
-
|
186
|
+
|
436
|
-
|
187
|
+
|
188
|
+
|
437
|
-
|
189
|
+
a, b, c = 10, 12, 10
|
438
|
-
|
439
|
-
)
|
440
|
-
|
441
|
-
)
|
442
190
|
|
443
191
|
|
444
192
|
|
@@ -450,9 +198,9 @@
|
|
450
198
|
|
451
199
|
"""
|
452
200
|
|
453
|
-
x =
|
201
|
+
x = np.linspace(-1, 1, 100)
|
454
|
-
|
202
|
+
|
455
|
-
y = np.linspace(-1
|
203
|
+
y = np.linspace(-1, 1, 100)
|
456
204
|
|
457
205
|
X, Y = np.meshgrid(x, y)
|
458
206
|
|
@@ -472,28 +220,60 @@
|
|
472
220
|
|
473
221
|
|
474
222
|
|
475
|
-
|
223
|
+
# 固有値問題を解く
|
224
|
+
|
476
|
-
|
225
|
+
Q = np.array([[a, b / 2], [b / 2, c]])
|
226
|
+
|
227
|
+
eig_val, eig_vec = np.linalg.eig(Q)
|
228
|
+
|
477
|
-
|
229
|
+
print(eig_val, eig_vec)
|
230
|
+
|
231
|
+
|
232
|
+
|
478
|
-
|
233
|
+
# 長軸、短軸を求める。
|
234
|
+
|
479
|
-
|
235
|
+
idx1, idx2 = eig_val.argsort()
|
236
|
+
|
480
|
-
|
237
|
+
major_axis = eig_vec[:, idx1] # 固有値が小さいほうの固有ベクトルが major axis
|
238
|
+
|
239
|
+
minor_axis = eig_vec[:, idx2] # 固有値が大きいほうの固有ベクトルが minor axis
|
240
|
+
|
241
|
+
|
242
|
+
|
481
|
-
ax.
|
243
|
+
ax.annotate(
|
482
|
-
|
244
|
+
|
483
|
-
ax
|
245
|
+
s="", xy=major_axis, xytext=(0, 0), arrowprops=dict(facecolor="r"),
|
246
|
+
|
484
|
-
|
247
|
+
)
|
248
|
+
|
249
|
+
ax.annotate(
|
250
|
+
|
251
|
+
s="", xy=minor_axis, xytext=(0, 0), arrowprops=dict(facecolor="b"),
|
252
|
+
|
253
|
+
)
|
254
|
+
|
255
|
+
|
256
|
+
|
485
|
-
plot_ellipse_by_equation(ax,
|
257
|
+
plot_ellipse_by_equation(ax, a, b, c)
|
258
|
+
|
259
|
+
|
260
|
+
|
261
|
+
# 角度を求める
|
262
|
+
|
263
|
+
theta1 = np.rad2deg(np.arctan(major_axis[1] / major_axis[0]))
|
264
|
+
|
265
|
+
theta2 = np.rad2deg(np.arctan(minor_axis[1] / minor_axis[0]))
|
266
|
+
|
267
|
+
print(f"angle between major axis and x axis= {theta1:.2f}")
|
268
|
+
|
269
|
+
print(f"angle between minor axis and x axis= {theta2:.2f}")
|
270
|
+
|
271
|
+
# angle between major axis and x axis= -45.00
|
272
|
+
|
273
|
+
# angle between minor axis and x axis= 45.00
|
486
274
|
|
487
275
|
```
|
488
276
|
|
489
277
|
|
490
278
|
|
491
|
-
## 結果
|
492
|
-
|
493
|
-
|
494
|
-
|
495
|
-
![イメージ説明](
|
279
|
+
![イメージ説明](b145bd531d08d205d36f4fde20ad4cfc.png)
|
496
|
-
|
497
|
-
|
498
|
-
|
499
|
-
緑の点がデータ、赤の線が近似した楕円
|
2
d
test
CHANGED
@@ -485,3 +485,15 @@
|
|
485
485
|
plot_ellipse_by_equation(ax, x[0], x[1], x[2])
|
486
486
|
|
487
487
|
```
|
488
|
+
|
489
|
+
|
490
|
+
|
491
|
+
## 結果
|
492
|
+
|
493
|
+
|
494
|
+
|
495
|
+
![イメージ説明](6330b456050f23b7c2b6c56d9d8c4824.png)
|
496
|
+
|
497
|
+
|
498
|
+
|
499
|
+
緑の点がデータ、赤の線が近似した楕円
|
1
d
test
CHANGED
@@ -46,6 +46,10 @@
|
|
46
46
|
|
47
47
|
|
48
48
|
|
49
|
+
楕円の点を作成するのに OpenCV を使っていますが、本題には関係ないので、その部分は CSV から読み込むように置き換えてください。
|
50
|
+
|
51
|
+
|
52
|
+
|
49
53
|
```python
|
50
54
|
|
51
55
|
import numpy
|