我正致力于在scipy中擬合三維分布函數。我有一個numpy數組，在x和y-bin中有計數，我正試圖將其與一個相當復雜的三維分布函數相匹配。數據適合26(！)描述其兩個組成種群形狀的參數。

我在這里了解到，當我調用leatsq時，必須將x和y坐標作為“args”傳遞。unutbu提供的代碼是為我編寫的，但是當我試圖將其應用于我的特定情況時，會出現錯誤“TypeError:leastsq()為關鍵字參數“args”獲取多個值”

這是我的代碼(對不起，長度太長了)：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)

這是我的數據，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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