我有一些按時間引數化的軌跡資料點，我想知道每個點到擬合它們的曲線的最短距離。似乎有幾種方法可以解決這個問題（例如這里或這里），但我選擇了scipy.optimize.fminbound，正如這里所建議的（因為它似乎是最巧妙的方法，因為我實際上讓它作業了）：
import numpy as np
import pandas as pd
from scipy.optimize import fminbound
data = np.array(
[
(212.82275865, 1650.40828168, 0., 0),
(214.22056952, 1649.99898924, 10.38, 0),
(212.86786868, 1644.25228805, 116.288, 0),
(212.78680031, 1643.87461108, 122.884, 0),
(212.57489485, 1642.01124032, 156.313, 0),
(212.53483954, 1641.61858242, 162.618, 0),
(212.43922274, 1639.58782771, 196.314, 0),
(212.53726315, 1639.13842423, 202.619, 0),
(212.2888428, 1637.24641296, 236.306, 0),
(212.2722447, 1636.92307229, 242.606, 0),
(212.15559302, 1635.0529813, 276.309, 0),
(212.17535631, 1634.60618711, 282.651, 0),
(212.02545613, 1632.72139574, 316.335, 0),
(211.99988779, 1632.32053329, 322.634, 0),
(211.33419846, 1631.07592039, 356.334, 0),
(211.58972239, 1630.21971902, 362.633, 0),
(211.70332122, 1628.2088542, 396.316, 0),
(211.74610735, 1627.67591368, 402.617, 0),
(211.88819518, 1625.67310022, 436.367, 0),
(211.90709414, 1625.39410321, 442.673, 0),
(212.00090348, 1623.42655008, 476.332, 0),
(211.9249017, 1622.94540583, 482.63, 0),
(212.34321938, 1616.32949197, 597.329, 0),
(213.09638942, 1615.2869643, 610.4, 0),
(219.41313491, 1580.22234313, 1197.332, 0),
(220.38660128, 1579.20043302, 1210.37, 0),
(236.35472669, 1542.30863041, 1798.267, 0),
(237.41755384, 1541.41679119, 1810.383, 0),
(264.08373622, 1502.66620597, 2398.244, 0),
(265.65655239, 1501.64308908, 2410.443, 0),
(304.66999824, 1460.94068336, 2997.263, 0),
(306.22550945, 1459.75817211, 3010.38, 0),
(358.88879764, 1416.472238, 3598.213, 0),
(361.14046402, 1415.40942931, 3610.525, 0),
(429.96379858, 1369.7972467, 4198.282, 0),
(432.06565776, 1368.22265539, 4210.505, 0),
(519.30493383, 1319.01141844, 4798.277, 0),
(522.12134083, 1317.68234967, 4810.4, 0),
(630.00294242, 1265.05368942, 5398.236, 0),
(633.67624272, 1263.63633508, 5410.431, 0),
(766.29767476, 1206.91262814, 5997.266, 0),
(770.78300649, 1205.48393374, 6010.489, 0),
(932.92308019, 1145.87780431, 6598.279, 0),
(937.54373403, 1141.55438694, 6609.525, 0),
], dtype=[
('x', 'f8'), ('y', 'f8'), ('t', 'f8'), ('dmin', 'f8'),
]
)
# fyi my data comes as a structured array; unfortunately, simply passing
# data[['x', 'y']] to np.polyfit does not work. using
# pd.DataFrame(data[['x', y']]).values seems to be the easiest solution:
# https://stackoverflow.com/a/26175750/5472354
coeffs = np.polyfit(
data['t'], pd.DataFrame(data[['x', 'y']]).values, 3
)
def curve(t):
# this can probably also be done in one statement, but I don't know how
x = np.polyval(coeffs[:, 0], t)
y = np.polyval(coeffs[:, 1], t)
return x, y
def f(t, p):
x, y = curve(t)
return np.hypot(x - p['x'], y - p['y'])
# instead of this:
for point in data:
tmin = fminbound(f, -50, 6659.525, args=(point, ))
point['dmin'] = f(tmin, point)
# do something like this:
# tmin = fminbound(f, -50, 6659.525, args=(data, ))
# data['dmin'] = f(tmin, data)
但是正如您所看到的，我使用for-loop 來計算每個資料點的最短距離，這顯著減慢了我的程式，因為這會執行數千次。因此我想對操作進行矢量化/提高性能，但還沒有找到方法。有與此相關的帖子（例如此處或此處），但我不知道建議的解決方案如何適用于我的情況。
uj5u.com熱心網友回復：
不，不可能進行矢量化，fminbound因為它需要一個變數的標量函式。但是，您仍然可以通過重新制定基礎數學優化問題來對回圈進行矢量化：
該N標量優化問題
min f_1(t) s.t. t_l <= t <= t_u
min f_2(t) s.t. t_l <= t <= t_u
.
.
.
min f_N(t) s.t. t_l <= t <= t_u
對于標量函式f_i等價于N變數中的一個優化問題：
min f_1(t_1)**2 ... f_N(t_N)**2 s.t. t_l <= t_i <= t_u for all i = 1, .., N
可以通過 解決scipy.optimize.minimize。根據您的整個演算法，您可以使用這種方法進一步消除更多回圈，即您只解決一個大規模優化問題而不是數千個標量優化問題。
清理代碼后，可以按如下方式完成：
import numpy as np
from scipy.optimize import minimize
data = np.array([
(212.82275865, 1650.40828168, 0., 0),
(214.22056952, 1649.99898924, 10.38, 0),
(212.86786868, 1644.25228805, 116.288, 0),
(212.78680031, 1643.87461108, 122.884, 0),
(212.57489485, 1642.01124032, 156.313, 0),
(212.53483954, 1641.61858242, 162.618, 0),
(212.43922274, 1639.58782771, 196.314, 0),
(212.53726315, 1639.13842423, 202.619, 0),
(212.2888428, 1637.24641296, 236.306, 0),
(212.2722447, 1636.92307229, 242.606, 0),
(212.15559302, 1635.0529813, 276.309, 0),
(212.17535631, 1634.60618711, 282.651, 0),
(212.02545613, 1632.72139574, 316.335, 0),
(211.99988779, 1632.32053329, 322.634, 0),
(211.33419846, 1631.07592039, 356.334, 0),
(211.58972239, 1630.21971902, 362.633, 0),
(211.70332122, 1628.2088542, 396.316, 0),
(211.74610735, 1627.67591368, 402.617, 0),
(211.88819518, 1625.67310022, 436.367, 0),
(211.90709414, 1625.39410321, 442.673, 0),
(212.00090348, 1623.42655008, 476.332, 0),
(211.9249017, 1622.94540583, 482.63, 0),
(212.34321938, 1616.32949197, 597.329, 0),
(213.09638942, 1615.2869643, 610.4, 0),
(219.41313491, 1580.22234313, 1197.332, 0),
(220.38660128, 1579.20043302, 1210.37, 0),
(236.35472669, 1542.30863041, 1798.267, 0),
(237.41755384, 1541.41679119, 1810.383, 0),
(264.08373622, 1502.66620597, 2398.244, 0),
(265.65655239, 1501.64308908, 2410.443, 0),
(304.66999824, 1460.94068336, 2997.263, 0),
(306.22550945, 1459.75817211, 3010.38, 0),
(358.88879764, 1416.472238, 3598.213, 0),
(361.14046402, 1415.40942931, 3610.525, 0),
(429.96379858, 1369.7972467, 4198.282, 0),
(432.06565776, 1368.22265539, 4210.505, 0),
(519.30493383, 1319.01141844, 4798.277, 0),
(522.12134083, 1317.68234967, 4810.4, 0),
(630.00294242, 1265.05368942, 5398.236, 0),
(633.67624272, 1263.63633508, 5410.431, 0),
(766.29767476, 1206.91262814, 5997.266, 0),
(770.78300649, 1205.48393374, 6010.489, 0),
(932.92308019, 1145.87780431, 6598.279, 0),
(937.54373403, 1141.55438694, 6609.525, 0)])
# the coefficients
coeffs = np.polyfit(data[:, 2], data[:, 0:2], 3)
# the points
points = data[:, :2]
# vectorized version of your objective function
# i.e. evaluates f_1, ..., f_N
def f(t, points):
poly_x = np.polyval(coeffs[:, 0], t)
poly_y = np.polyval(coeffs[:, 1], t)
return np.hypot(poly_x - points[:, 0], poly_y - points[:, 1])
# the scalar objective function we want to minimize
def obj_vec(t, points):
vals = f(t, points)
return np.sum(vals**2)
# variable bounds
bnds = [(-50, 6659.525)]*len(points)
# solve the optimization problem
res = minimize(lambda t: obj_vec(t, points), x0=np.zeros(len(points)), bounds=bnds)
dmins = f(res.x, points)
為了進一步加速優化，強烈建議將目標函式的精確梯度傳遞給minimize。目前，梯度是通過有限差分來近似的，這很慢。由于可以通過鏈式法則輕松計算梯度，因此我將其作為練習留給讀者:)。
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