我正在使用scipy Cookbook下载的ols.py代码（下载是第一段带有粗体OLS），但我需要了解而不是使用随机数据来执行ols功能多元线性回归。
我有一个特定的因变量y和三个解释变量。每当我尝试用变量代替随机变量时，它就会给出错误：
TypeError：此构造函数不带参数。
有人可以帮忙吗？这可能吗？
以下是我尝试使用的ols代码的副本以及我尝试输入的变量
from __future__ import division
from scipy import c_, ones, dot, stats, diff
from scipy.linalg import inv, solve, det
from numpy import log, pi, sqrt, square, diagonal
from numpy.random import randn, seed
import time
class ols:
"""
Author: Vincent Nijs (+ ?)
Email: v-nijs at kellogg.northwestern.edu
Last Modified: Mon Jan 15 17:56:17 CST 2007
Dependencies: See import statement at the top of this file
Doc: Class for multi-variate regression using OLS
Input:
dependent variable
y_varnm = string with the variable label for y
x = independent variables, note that a constant is added by default
x_varnm = string or list of variable labels for the independent variables
Output:
There are no values returned by the class. Summary provides printed output.
All other measures can be accessed as follows:
Step 1: Create an OLS instance by passing data to the class
m = ols(y,x,y_varnm = 'y',x_varnm = ['x1','x2','x3','x4'])
Step 2: Get specific metrics
To print the coefficients:
>>> print m.b
To print the coefficients p-values:
>>> print m.p
"""
y = [29.4, 29.9, 31.4, 32.8, 33.6, 34.6, 35.5, 36.3, 37.2, 37.8, 38.5, 38.8,
38.6, 38.8, 39, 39.7, 40.6, 41.3, 42.5, 43.9, 44.9, 45.3, 45.8, 46.5,
77.1, 48.2, 48.8, 50.5, 51, 51.3, 50.7, 50.7, 50.6, 50.7, 50.6, 50.7]
#tuition
x1 = [376, 407, 438, 432, 433, 479, 512, 543, 583, 635, 714, 798, 891,
971, 1045, 1106, 1218, 1285, 1356, 1454, 1624, 1782, 1942, 2057, 2179,
2271, 2360, 2506, 2562, 2700, 2903, 3319, 3629, 3874, 4102, 4291]
#research and development
x2 = [28740.00, 30952.00, 33359.00, 35671.00, 39435.00, 43338.00, 48719.00, 55379.00, 63224.00,
72292.00, 80748.00, 89950.00, 102244.00, 114671.00, 120249.00, 126360.00, 133881.00, 141891.00,
151993.00, 160876.00, 165350.00, 165730.00, 169207.00, 183625.00, 197346.00, 212152.00, 226402.00,
267298.00, 277366.00, 276022.00, 288324.00, 299201.00, 322104.00, 347048.00, 372535.00,
397629.00]
#one/none parents
x3 = [11610, 12143, 12486, 13015, 13028, 13327, 14074, 14094, 14458, 14878, 15610, 15649,
15584, 16326, 16379, 16923, 17237, 17088, 17634, 18435, 19327, 19712, 21424, 21978,
22684, 22597, 22735, 22217, 22214, 22655, 23098, 23602, 24013, 24003, 21593, 22319]
def __init__(self,y,x1,y_varnm = 'y',x_varnm = ''):
"""
Initializing the ols class.
"""
self.y = y
#self.x1 = c_[ones(x1.shape[0]),x1]
self.y_varnm = y_varnm
if not isinstance(x_varnm,list):
self.x_varnm = ['const'] + list(x_varnm)
else:
self.x_varnm = ['const'] + x_varnm
# Estimate model using OLS
self.estimate()
def estimate(self):
# estimating coefficients, and basic stats
self.inv_xx = inv(dot(self.x.T,self.x))
xy = dot(self.x.T,self.y)
self.b = dot(self.inv_xx,xy) # estimate coefficients
self.nobs = self.y.shape[0] # number of observations
self.ncoef = self.x.shape[1] # number of coef.
self.df_e = self.nobs - self.ncoef # degrees of freedom, error
self.df_r = self.ncoef - 1 # degrees of freedom, regression
self.e = self.y - dot(self.x,self.b) # residuals
self.sse = dot(self.e,self.e)/self.df_e # SSE
self.se = sqrt(diagonal(self.sse*self.inv_xx)) # coef. standard errors
self.t = self.b / self.se # coef. t-statistics
self.p = (1-stats.t.cdf(abs(self.t), self.df_e)) * 2 # coef. p-values
self.R2 = 1 - self.e.var()/self.y.var() # model R-squared
self.R2adj = 1-(1-self.R2)*((self.nobs-1)/(self.nobs-self.ncoef)) # adjusted R-square
self.F = (self.R2/self.df_r) / ((1-self.R2)/self.df_e) # model F-statistic
self.Fpv = 1-stats.f.cdf(self.F, self.df_r, self.df_e) # F-statistic p-value
def dw(self):
"""
Calculates the Durbin-Waston statistic
"""
de = diff(self.e,1)
dw = dot(de,de) / dot(self.e,self.e);
return dw
def omni(self):
"""
Omnibus test for normality
"""
return stats.normaltest(self.e)
def JB(self):
"""
Calculate residual skewness, kurtosis, and do the JB test for normality
"""
# Calculate residual skewness and kurtosis
skew = stats.skew(self.e)
kurtosis = 3 + stats.kurtosis(self.e)
# Calculate the Jarque-Bera test for normality
JB = (self.nobs/6) * (square(skew) + (1/4)*square(kurtosis-3))
JBpv = 1-stats.chi2.cdf(JB,2);
return JB, JBpv, skew, kurtosis
def ll(self):
"""
Calculate model log-likelihood and two information criteria
"""
# Model log-likelihood, AIC, and BIC criterion values
ll = -(self.nobs*1/2)*(1+log(2*pi)) - (self.nobs/2)*log(dot(self.e,self.e)/self.nobs)
aic = -2*ll/self.nobs + (2*self.ncoef/self.nobs)
bic = -2*ll/self.nobs + (self.ncoef*log(self.nobs))/self.nobs
return ll, aic, bic
def summary(self):
"""
Printing model output to screen
"""
# local time & date
t = time.localtime()
# extra stats
ll, aic, bic = self.ll()
JB, JBpv, skew, kurtosis = self.JB()
omni, omnipv = self.omni()
# printing output to screen
print '\n=============================================================================='
print "Dependent Variable: " + self.y_varnm
print "Method: Least Squares"
print "Date: ", time.strftime("%a, %d %b %Y",t)
print "Time: ", time.strftime("%H:%M:%S",t)
print '# obs: %5.0f' % self.nobs
print '# variables: %5.0f' % self.ncoef
print '=============================================================================='
print 'variable coefficient std. Error t-statistic prob.'
print '=============================================================================='
for i in range(len(self.x_varnm)):
print '''% -5s % -5.6f % -5.6f % -5.6f % -5.6f''' % tuple([self.x_varnm[i],self.b[i],self.se[i],self.t[i],self.p[i]])
print '=============================================================================='
print 'Models stats Residual stats'
print '=============================================================================='
print 'R-squared % -5.6f Durbin-Watson stat % -5.6f' % tuple([self.R2, self.dw()])
print 'Adjusted R-squared % -5.6f Omnibus stat % -5.6f' % tuple([self.R2adj, omni])
print 'F-statistic % -5.6f Prob(Omnibus stat) % -5.6f' % tuple([self.F, omnipv])
print 'Prob (F-statistic) % -5.6f JB stat % -5.6f' % tuple([self.Fpv, JB])
print 'Log likelihood % -5.6f Prob(JB) % -5.6f' % tuple([ll, JBpv])
print 'AIC criterion % -5.6f Skew % -5.6f' % tuple([aic, skew])
print 'BIC criterion % -5.6f Kurtosis % -5.6f' % tuple([bic, kurtosis])
print '=============================================================================='
if __name__ == '__main__':
##########################
### testing the ols class
##########################
# intercept is added, by default
m = ols(y,x1,y_varnm = 'y',x_varnm = ['x1','x2','x3'])
m.summary()
答案 0 :(得分：5)
你应区分两种情况：i）你只想解决这个问题。 ii）您还想知道有关您的模型的统计信息。你可以用np.linalg.lstsq做i）;对于ii），你最好使用statsmodels。
下面是两个解决方案的示例示例：
# The standard imports
import numpy as np
import pandas as pd
# For the statistic
from statsmodels.formula.api import ols
def generatedata():
''' Generate and show the data '''
x = np.linspace(-5,5,101)
(X,Y) = np.meshgrid(x,x)
# To get reproducable values, I provide a seed value
np.random.seed(987654321)
Z = -5 + 3*X-0.5*Y+np.random.randn(np.shape(X)[0], np.shape(X)[1])
return (X.flatten(),Y.flatten(),Z.flatten())
def regressionmodel(X,Y,Z):
'''Multilinear regression model, calculating fit, P-values, confidence intervals etc.'''
# Convert the data into a Pandas DataFrame
df = pd.DataFrame({'x':X, 'y':Y, 'z':Z})
# Fit the model
model = ols("z ~ x + y", df).fit()
# Print the summary
print(model.summary())
return model._results.params # should be array([-4.99754526, 3.00250049, -0.50514907])
def linearmodel(X,Y,Z):
'''Just fit the plane'''
M = np.vstack((np.ones(len(X)), X, Y)).T
bestfit = np.linalg.lstsq(M,Z)[0]
print('Best fit plane:', bestfit)
return bestfit
if __name__ == '__main__':
(X,Y,Z) = generatedata()
regressionmodel(X,Y,Z)
linearmodel(X,Y,Z)
答案 1 :(得分：3)
使用http://pypi.python.org/pypi/scikits.statsmodels可能更容易，而且功能更多
import numpy as np
import scikits.statsmodels.api as sm
y = [29.4, 29.9, 31.4, 32.8, 33.6, 34.6, 35.5, 36.3, 37.2, 37.8, 38.5, 38.8,
38.6, 38.8, 39, 39.7, 40.6, 41.3, 42.5, 43.9, 44.9, 45.3, 45.8, 46.5,
77.1, 48.2, 48.8, 50.5, 51, 51.3, 50.7, 50.7, 50.6, 50.7, 50.6, 50.7]
#tuition
x1 = [376, 407, 438, 432, 433, 479, 512, 543, 583, 635, 714, 798, 891,
971, 1045, 1106, 1218, 1285, 1356, 1454, 1624, 1782, 1942, 2057, 2179,
2271, 2360, 2506, 2562, 2700, 2903, 3319, 3629, 3874, 4102, 4291]
#research and development
x2 = [28740.00, 30952.00, 33359.00, 35671.00, 39435.00, 43338.00, 48719.00, 55379.00, 63224.00,
72292.00, 80748.00, 89950.00, 102244.00, 114671.00, 120249.00, 126360.00, 133881.00, 141891.00,
151993.00, 160876.00, 165350.00, 165730.00, 169207.00, 183625.00, 197346.00, 212152.00, 226402.00,
267298.00, 277366.00, 276022.00, 288324.00, 299201.00, 322104.00, 347048.00, 372535.00,
397629.00]
#one/none parents
x3 = [11610, 12143, 12486, 13015, 13028, 13327, 14074, 14094, 14458, 14878, 15610, 15649,
15584, 16326, 16379, 16923, 17237, 17088, 17634, 18435, 19327, 19712, 21424, 21978,
22684, 22597, 22735, 22217, 22214, 22655, 23098, 23602, 24013, 24003, 21593, 22319]
x = np.column_stack((x1,x2,x3)) #stack explanatory variables into an array
x = sm.add_constant(x, prepend=True) #add a constant
res = sm.OLS(y,x).fit() #create a model and fit it
print res.params
print res.bse
print res.summary()
答案 2 :(得分：1)
ols函数将整个独立数据集作为第二个参数。试试
m = ols(y, [x1, x2, x3], ...)
虽然我怀疑你可能需要将它包装在numpy数组中：
x = numpy.ndarray([3, len(x1)])
x[0] = numpy.array(x1)
x[1] = numpy.array(x2)
x[2] = numpy.array(x3)
m = ols(y, x, ...)