使用 3d 数据和参数在 Scipy 中进行曲线拟合

Question

我正在研究在 scipy 中拟合 3d 分布函数。我有一个在 x- 和 y-bins 中有计数的 numpy 数组，我正试图将它拟合到一个相当复杂的 3-d 分布函数中。数据适合 26 个 (!) 参数，这些参数描述了两个组成种群的形状。

我在这里了解到，当我调用 minimumsq 时，我必须将 x 和 y 坐标作为“args”传递。 unutbu 提供的代码按照为我编写的代码工作，但是当我尝试将其应用于我的特定情况时，我收到错误“TypeError: leastsq() got multiple values for keyword argument 'args'”

这是我的代码（抱歉，篇幅较长）：

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)

这是我的数据，只是垃圾箱更少：

[[  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]]

非常感谢！

Answer 1

那么leastsq 所做的就是尝试：

“最小化一组方程的平方和” -scipy docs

正如它所说，它正在最小化一组函数，因此如果您查看参数here，实际上并不会以最简单的方式获取任何 x 或 y 数据输入，因此您可以随心所欲地进行操作并传递残差但是，使用 curve_fit 会更容易，它会为您完成它:) 并创建必要的方程式

对于拟合，您应该使用：curve_fit 如果您对他们使用的通用残差没问题，这实际上是您通过自身传递的函数res = leastsq(func, p0, args=args, full_output=1, **kw) 如果您查看code here.

例如如果我在 2d 中拟合 Rosenbrock 函数并猜测 y 参数：

from scipy.optimize import curve_fit
from itertools import imap
import numpy as np
# use only an even number of arguments
def rosen2d(x,a):
    return (1-x)**2 + 100*(a - (x**2))**2
#generate some random data slightly off

datax = np.array([.01*x for x in range(-10,10)])
datay = 2.3
dataz = np.array(map(lambda x: rosen2d(x,datay), datax))
optimalparams, covmatrix = curve_fit(rosen2d, datax, dataz)
print 'opt:',optimalparams

在 4d 中拟合 colville 函数：

from scipy.optimize import curve_fit
import numpy as np

# 4 dimensional colville function
# definition from http://www.sfu.ca/~ssurjano/colville.html
def colville(x,x3,x4):
    x1,x2 = x[:,0],x[:,1]
    return 100*(x1**2 - x2)**2 + (x1-1)**2 + (x3-1)**2 + \
            90*(x3**2 - x4)**2 + \
            10.1*((x2 - 1)**2 + (x4 - 1)**2) + \
            19.8*(x2 - 1)*(x4 - 1)
#generate some random data slightly off

datax = np.array([[x,x] for x in range(-10,10)])
#add gaussian noise
datax+= np.random.rand(*datax.shape)
#set 2 of the 4 parameters to constants
x3 = 3.5
x4 = 4.5
#calculate the function
dataz = colville(datax, x3, x4)
#fit the function
optimalparams, covmatrix = curve_fit(colville, datax, dataz)
print 'opt:',optimalparams

使用自定义残差函数：

from scipy.optimize import leastsq
import numpy as np

# 4 dimensional colville function
# definition from http://www.sfu.ca/~ssurjano/colville.html
def colville(x,x3,x4):
    x1,x2 = x[:,0],x[:,1]
    return 100*(x1**2 - x2)**2 + (x1-1)**2 + (x3-1)**2 + \
            90*(x3**2 - x4)**2 + \
            10.1*((x2 - 1)**2 + (x4 - 1)**2) + \
            19.8*(x2 - 1)*(x4 - 1)
#generate some random data slightly off


datax = np.array([[x,x] for x in range(-10,10)])
#add gaussian noise
datax+= np.random.rand(*datax.shape)
#set 2 of the 4 parameters to constants
x3 = 3.5
x4 = 4.5

def residual(p, x, y):
    return y - colville(x,*p)
#calculate the function
dataz = colville(datax, x3, x4)
#guess some initial parameter values
p0 = [0,0]
#calculate a minimization of the residual
optimalparams = leastsq(residual, p0, args=(datax, dataz))[0]
print 'opt:',optimalparams

编辑：您同时使用了args 的位置和关键字arg：如果您查看docs，您会看到它使用位置3，但也可以用作关键字参数。您使用了 both，这意味着该功能符合预期，令人困惑。