python 三维曲线拟合_基于三维数据和参数的Scipy曲线拟合
我正致力于在scipy中擬合三維分布函數(shù)。我有一個(gè)numpy數(shù)組,在x和y-bin中有計(jì)數(shù),我正試圖將其與一個(gè)相當(dāng)復(fù)雜的三維分布函數(shù)相匹配。數(shù)據(jù)適合26(!)描述其兩個(gè)組成種群形狀的參數(shù)。
我在這里了解到,當(dāng)我調(diào)用leatsq時(shí),必須將x和y坐標(biāo)作為“args”傳遞。unutbu提供的代碼是為我編寫的,但是當(dāng)我試圖將其應(yīng)用于我的特定情況時(shí),會(huì)出現(xiàn)錯(cuò)誤“TypeError:leastsq()為關(guān)鍵字參數(shù)“args”獲取多個(gè)值”
這是我的代碼(對(duì)不起,長(zhǎng)度太長(zhǎng)了):import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize as spopt
from textwrap import wrap
import collections
cl = 0.5
ch = 3.5
rl = -23.5
rh = -18.5
mbins = 10
cbins = 10
def hist_data(mixed_data, mbins, cbins):
import numpy as np
H, xedges, yedges = np.histogram2d(mixed_data[:,1], mixed_data[:,2], bins = (mbins, cbins), weights = mixed_data[:,3])
x, y = 0.5 * (xedges[:-1] + xedges[1:]), 0.5 * (yedges[:-1] + yedges[1:])
return H.T, x, y
def gauss(x, s, mu, a):
import numpy as np
return a * np.exp(-((x - mu)**2. / (2. * s**2.)))
def tanhlin(x, p0, p1, q0, q1, q2):
import numpy as np
return p0 + p1 * (x + 20.) + q0 * np.tanh((x - q1)/q2)
def func3d(p, x, y):
import numpy as np
from sys import exit
rsp0, rsp1, rsq0, rsq1, rsq2, rmp0, rmp1, rmq0, rmq1, rmq2, rs, rm, ra, bsp0, bsp1, bsq0, bsq1, bsq2, bmp0, bmp1, bmq0, bmq1, bmq2, bs, bm, ba = p
x, y = np.meshgrid(coords[0], coords[1])
rs = tanhlin(x, rsp0, rsp1, rsq0, rsq1, rsq2)
rm = tanhlin(x, rmp0, rmp1, rmq0, rmq1, rmq2)
ra = schechter(x, rap, raa, ram) # unused
bs = tanhlin(x, bsp0, bsp1, bsq0, bsq1, bsq2)
bm = tanhlin(x, bmp0, bmp1, bmq0, bmq1, bmq2)
ba = schechter(x, bap, baa, bam) # unused
red_dist = ra / (rs * np.sqrt(2 * np.pi)) * gauss(y, rs, rm, ra)
blue_dist = ba / (bs * np.sqrt(2 * np.pi)) * gauss(y, bs, bm, ba)
result = red_dist + blue_dist
return result
def residual(p, coords, data):
import numpy as np
model = func3d(p, coords)
res = (model.flatten() - data.flatten())
# can put parameter restrictions in here
return res
def poiss_err(data):
import numpy as np
return np.where(np.sqrt(H) > 0., np.sqrt(H), 2.)
# =====
H, x, y = hist_data(mixed_data, mbins, cbins)
data = H
coords = x, y
# x and y will be the projected coordinates of the data H onto the plane z = 0
# x has bins of width 0.5, with centers at -23.25, -22.75, ... , -19.25, -18.75
# y has bins of width 0.3, with centers at 0.65, 0.95, ... , 3.05, 3.35
Param = collections.namedtuple('Param', 'rsp0 rsp1 rsq0 rsq1 rsq2 rmp0 rmp1 rmq0 rmq1 rmq2 rs rm ra bsp0 bsp1 bsq0 bsq1 bsq2 bmp0 bmp1 bmq0 bmq1 bmq2 bs bm ba')
p_guess = Param(rsp0 = 0.152, rsp1 = 0.008, rsq0 = 0.044, rsq1 = -19.91, rsq2 = 0.94, rmp0 = 2.279, rmp1 = -0.037, rmq0 = -0.108, rmq1 = -19.81, rmq2 = 0.96, rs = 1., rm = -20.5, ra = 10000., bsp0 = 0.298, bsp1 = 0.014, bsq0 = -0.067, bsq1 = -19.90, bsq2 = 0.58, bmp0 = 1.790, bmp1 = -0.053, bmq0 = -0.363, bmq1 = -20.75, bmq2 = 1.12, bs = 1., bm = -20., ba = 2000.)
opt, cov, infodict, mesg, ier = spopt.leastsq(residual, p_guess, poiss_err(H), args = coords, maxfev = 100000, full_output = True)
這是我的數(shù)據(jù),只有更少的箱子:[[ 1.00000000e+01 1.10000000e+01 2.10000000e+01 1.90000000e+01
1.70000000e+01 2.10000000e+01 2.40000000e+01 1.90000000e+01
2.80000000e+01 1.90000000e+01]
[ 1.40000000e+01 4.50000000e+01 6.00000000e+01 6.80000000e+01
1.34000000e+02 1.97000000e+02 2.23000000e+02 2.90000000e+02
3.23000000e+02 3.03000000e+02]
[ 3.00000000e+01 1.17000000e+02 3.78000000e+02 9.74000000e+02
1.71900000e+03 2.27700000e+03 2.39000000e+03 2.25500000e+03
1.85600000e+03 1.31000000e+03]
[ 1.52000000e+02 9.32000000e+02 2.89000000e+03 5.23800000e+03
6.66200000e+03 6.19100000e+03 4.54900000e+03 3.14600000e+03
2.09000000e+03 1.33800000e+03]
[ 5.39000000e+02 2.58100000e+03 6.51300000e+03 8.89900000e+03
8.52900000e+03 6.22900000e+03 3.55000000e+03 2.14300000e+03
1.19000000e+03 6.92000000e+02]
[ 1.49600000e+03 4.49200000e+03 8.77200000e+03 1.07610000e+04
9.76700000e+03 7.04900000e+03 4.23200000e+03 2.47200000e+03
1.41500000e+03 7.02000000e+02]
[ 2.31800000e+03 7.01500000e+03 1.28870000e+04 1.50840000e+04
1.35590000e+04 8.55600000e+03 4.15600000e+03 1.77100000e+03
6.57000000e+02 2.55000000e+02]
[ 1.57500000e+03 3.79300000e+03 5.20900000e+03 4.77800000e+03
3.26600000e+03 1.44700000e+03 5.31000000e+02 1.85000000e+02
9.30000000e+01 4.90000000e+01]
[ 7.01000000e+02 1.21600000e+03 1.17600000e+03 7.93000000e+02
4.79000000e+02 2.02000000e+02 8.80000000e+01 3.90000000e+01
2.30000000e+01 1.90000000e+01]
[ 2.93000000e+02 3.93000000e+02 2.90000000e+02 1.97000000e+02
1.18000000e+02 6.40000000e+01 4.10000000e+01 1.20000000e+01
1.10000000e+01 4.00000000e+00]]
非常感謝!
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