Question

我需要用直线拟合来自不同数据集的一些点。从每个数据集我想要一条线。所以我得到了描述i-line的参数ai和bi：ai + bi * x。问题是我想要强加每个ai是平等的，因为我想要相同的拦截。我在这里找到了一个教程：http://www.scipy.org/Cookbook/FittingData#head-a44b49d57cf0165300f765e8f1b011876776502f。不同之处在于我不知道priopri有多少数据集。我的代码是这样的：

from numpy import *
from scipy import optimize

# here I have 3 dataset, but in general I don't know how many dataset are they
ypoints = [array([0, 2.1, 2.4]),    # first dataset, 3 points
           array([0.1, 2.1, 2.9]),  # second dataset
           array([-0.1, 1.4])]      # only 2 points

xpoints = [array([0, 2, 2.5]),      # first dataset
           array([0, 2, 3]),        # second, also x coordinates are different
           array([0, 1.5])]         # the first coordinate is always 0

fitfunc = lambda a, b, x: a + b * x
errfunc = lambda p, xs, ys: array([ yi - fitfunc(p[0], p[i+1], xi) 
                                    for i, (xi,yi) in enumerate(zip(xs, ys)) ])


p_arrays = [r_[0.]] * len(xpoints)
pinit = r_[[ypoints[0][0]] + p_arrays]
fit_parameters, success = optimize.leastsq(errfunc, pinit, args = (xpoints, ypoints))

我得到了

Traceback (most recent call last):
  File "prova.py", line 19, in <module>
    fit_parameters, success = optimize.leastsq(errfunc, pinit, args = (xpoints,    ypoints))
  File "/usr/lib64/python2.6/site-packages/scipy/optimize/minpack.py", line 266, in  leastsq
    m = check_func(func,x0,args,n)[0]
  File "/usr/lib64/python2.6/site-packages/scipy/optimize/minpack.py", line 12, in  check_func
    res = atleast_1d(thefunc(*((x0[:numinputs],)+args)))
  File "prova.py", line 14, in <lambda>
    for i, (xi,yi) in enumerate(zip(xs, ys)) ])
ValueError: setting an array element with a sequence.

Answer 1

（旁注：使用def，而不是lambda分配给某个名称 - 这完全是愚蠢的，只有缺点，lambda唯一的用途就是匿名功能！）。

你的errfunc应该返回一个浮点数的序列（数组或其他），但事实并非如此，因为你试图把数组中的项作为差异{{1 }（请记住，y又名ypoints是一个数组列表！）和拟合函数的结果。因此，您需要将表达式ys“折叠”为单个浮点数，例如yi - fitfunc(p[0], p[i+1], xi)。

Answer 2

如果您只需要线性拟合，那么最好使用线性回归而不是非线性优化器来估计它。可以使用scikits.statsmodels来获得更合适的统计数据。

import numpy as np
from numpy import array

ypoints = np.r_[array([0, 2.1, 2.4]),    # first dataset, 3 points
           array([0.1, 2.1, 2.9]),  # second dataset
           array([-0.1, 1.4])]      # only 2 points

xpoints = [array([0, 2, 2.5]),      # first dataset
           array([0, 2, 3]),        # second, also x coordinates are different
           array([0, 1.5])]         # the first coordinate is always 0

xp = np.hstack(xpoints)
indicator = []
for i,a in enumerate(xpoints):
    indicator.extend([i]*len(a))

indicator = np.array(indicator)


x = xp[:,None]*(indicator[:,None]==np.arange(3)).astype(int)
x = np.hstack((np.ones((xp.shape[0],1)),x))

print np.dot(np.linalg.pinv(x), ypoints)
# [ 0.01947973  0.98656987  0.98481549  0.92034684]

回归量矩阵有一个共同的截距，但每个数据集的列不同：

>>> x
array([[ 1. ,  0. ,  0. ,  0. ],
       [ 1. ,  2. ,  0. ,  0. ],
       [ 1. ,  2.5,  0. ,  0. ],
       [ 1. ,  0. ,  0. ,  0. ],
       [ 1. ,  0. ,  2. ,  0. ],
       [ 1. ,  0. ,  3. ,  0. ],
       [ 1. ,  0. ,  0. ,  0. ],
       [ 1. ,  0. ,  0. ,  1.5]])

scipy智能优化

2 个答案: