rpy2问题，nls将list（）作为参数从python传递给R

我试图用numpy数组中的rpy2拟合一条非线性曲线，但由于我不知道如何在R端传递“start”参数而被卡住了。我使用R2.12.1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

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主要错误是：

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

可变职业在哪里？就业呢？ 似乎它们没有一个起始值，也许你必须改变/转换一些R

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明白吗？

嗯，是的。。。我很抱歉，但是职业和总体就业都很重要！！！你是对的，我所要做的就是让wrao mu numpy用robjects.IntVectorprofessions等等点亮！在这里，更明确地说明您的解决方案会很有用，比如给出正确的代码版本。

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