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python - numpy中的多项式拟合问题

转载 作者:太空宇宙 更新时间:2023-11-04 10:45:35 25 4
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我试图实现一个标准的多项式拟合程序,遇到了一个我无法理解的问题。这是我有样本 x 和 y 数据的代码,我使用正规方程拟合这些数据,还使用 ​​numpy 中的 polyfit 函数。

def mytest(N):
print "\nPolynomial order (N): {}".format(N)
mVals = [5, 10, 15, 20]
a_orig = [198.764, 13.5, 0.523]
for M in mVals:
x = arange(-M, M+1)
y = matrix(a_orig[0]+ a_orig[1]*x + a_orig[2]*x**2).T

# Code implementing the solution from the normal equations
nArray = arange(N+1)
A = matrix([[n**i for i in nArray] for n in x])
B = (A.T*A).I
a_myfit = B*(A.T*y)

# numpy's polyfit
a_polyfit = polyfit(x, y, N)

print "M: {}".format(M)
print ["{0:0.3f}".format(i) for i in a_orig]
print ["{0:0.3f}".format(i) for i in array(a_myfit)[:,0]]
print ["{0:0.3f}".format(i) for i in list(array(a_polyfit)[:,0])[::-1]]

mytest(N=5)
mytest(N=6)

这是输出

Polynomial order (N): 5
M: 5
['198.764', '13.500', '0.523']
['198.764', '13.500', '0.523', '0.000', '0.000', '-0.000']
['198.764', '13.500', '0.523', '-0.000', '0.000', '0.000']
M: 10
['198.764', '13.500', '0.523']
['198.764', '13.500', '0.523', '-0.000', '0.000', '-0.000']
['198.764', '13.500', '0.523', '0.000', '0.000', '0.000']
M: 15
['198.764', '13.500', '0.523']
['198.764', '13.500', '0.523', '0.000', '0.000', '-0.000']
['198.764', '13.500', '0.523', '-0.000', '-0.000', '0.000']
M: 20
['198.764', '13.500', '0.523']
['198.764', '13.500', '0.523', '0.000', '-0.000', '-0.000']
['198.764', '13.500', '0.523', '0.000', '0.000', '0.000']

Polynomial order (N): 6
M: 5
['198.764', '13.500', '0.523']
['198.764', '13.500', '0.523', '0.000', '0.000', '-0.000', '-0.000']
['198.764', '13.500', '0.523', '-0.000', '0.000', '0.000', '-0.000']
M: 10
['198.764', '13.500', '0.523']
['198.764', '13.500', '0.523', '-0.000', '-0.000', '-0.000', '-0.000']
['198.764', '13.500', '0.523', '0.000', '0.000', '-0.000', '-0.000']
M: 15
['198.764', '13.500', '0.523']
['294.451', '13.500', '-0.061', '0.000', '-0.001', '-0.000', '-0.000']
['198.764', '13.500', '0.523', '0.000', '0.000', '0.000', '-0.000']
M: 20
['198.764', '13.500', '0.523']
['369.135', '13.500', '-0.046', '0.000', '-0.000', '-0.000', '-0.000']
['198.764', '13.500', '0.523', '0.000', '0.000', '-0.000', '-0.000']

多项式拟合的值给出了 N > 5 和 M > 13 的错误值。

我哪里错了? polyfit 实现有何不同?

最佳答案

A 是 dtype int32

可表示的最大值为:

In [10]: np.iinfo(np.dtype('int32')).max
Out[10]: 2147483647

当指数非常大时,n**i 可能会大于 2147483647。此时您会得到不正确的结果。

NumPy does not check for arithmetic overflows (在对数组执行操作时)因为这会影响性能。您可以选择避免算术溢出的数据类型。

要解决这个问题,请声明一个可以表示更大数字的不同数据类型:

A = np.matrix([[n**i for i in nArray] for n in x], dtype='float64')

请注意,问题仍然存在,它只会在 N 的值大得多时发生。

关于python - numpy中的多项式拟合问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17613851/

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