Question

我试图使用来自numpy数组的rpy2来拟合非线性曲线，但是由于我不知道如何在R侧传递'start'参数而卡住了。我使用R 2.12.1和python 2.6.6

Error in function (formula, data = parent.frame(), start, control = nls.control(),  : 
parameters without starting value in 'data': responsev, predictorv
Traceback (most recent call last):
File "./employmentsHoro.py", line 279, in <module>
nls.nls2(formula=formula, data=dataf, start=mylist)
File "/usr/lib/python2.6/dist-packages/rpy2/robjects/functions.py", line 83, in __call__
return super(SignatureTranslatedFunction, self).__call__(*args, **kwargs)
File "/usr/lib/python2.6/dist-packages/rpy2/robjects/functions.py", line 35, in __call__
res = super(Function, self).__call__(*new_args, **new_kwargs)
rpy2.rinterface.RRuntimeError: Error in function (formula, data = parent.frame(),start, control = nls.control(),  : 
parameters without starting value in 'data': responsev, predictorv

有人可以帮我确定如何将list（）对象传递给nls公式吗？

我的代码的相关部分是：

import rpy2.robjects as robjects
from rpy2.robjects import DataFrame, Formula
import rpy2.robjects.numpy2ri as npr
import numpy as np
from rpy2.robjects.packages import importr
nls = importr('nls2')
stats = importr('stats')

mylist = robjects.r('list(a=700,b=0.8,c=200000)')

dataf = DataFrame({'responsev': professions, 'predictorv': totalEmployment})
starter= DataFrame({'a':700,'b':0.80,'c':200000})
formula = Formula('responsev ~I( a*(predictorv/c)^b )/( 1+( predictorv/c )^b )')
nls.nls2(formula=formula, data=dataf, start=starter)

Answer 1

主要错误是：

Error in function (formula, data = parent.frame(), start, control = 
nls.control(),  : parameters without starting value in 
    'data': responsev, predictorv

哪里宣布变量专业？和DataEmployment？似乎他们没有起始值，也许你必须改变/转换R的东西理解？

rpy2问题，nls传递list（）作为从python到R的参数

1 个答案: