Python rpy2 - nls 回归 RRuntimeError

Python rpy2 - nls regression RRuntimeError

我正在尝试在 Python 中使用 R 进行一些 nls 回归。我被 RRuntimeError 困住了,现在我已经超出了我的专业知识范围,并且已经努力了几天才能让它发挥作用,因此希望得到一些帮助。

这是我的 csv 数据: http://www.sharecsv.com/s/4cdd4f832b606d6616260f9dc0eedf38/ratedata.csv

这是我的代码:

import pandas as pd
import rpy2.robjects as ro
from rpy2.robjects.packages import importr
from rpy2.robjects import pandas2ri
pandas2ri.activate()

dfData = pd.read_csv('C:\Users\nick\Desktop\ratedata.csv')
rdf = pandas2ri.py2ri(dfData)

a = 0.5
b = 1.1
count = rdf.rx(True, 'Trials')
rates = rdf.rx(True, 'Successes')

base = importr('base', robject_translations={'with': '_with'})
stats = importr('stats', robject_translations={'format_perc': '_format_perc'})

my_formula = stats.as_formula('rates ~ 1-(1/(10^(a * count ^ (b-1))))')

d = ro.ListVector({'a': a, 'b': b})

fit = stats.nls(my_formula, weights=count, start=d)

一切都在编译,除了:

fit = stats.nls(my_formula, weights=count, start=d)

我得到以下追溯:

---------------------------------------------------------------------------
RRuntimeError                             Traceback (most recent call last)
<ipython-input-12-3f7fcd7d7851> in <module>()
      6 d = ro.ListVector({'a': a, 'b': b})
      7 
----> 8 fit = stats.nls(my_formula, weights=count, start=d)

~\AppData\Local\Continuum\anaconda3\lib\site-packages\rpy2\robjects\functions.py in __call__(self, *args, **kwargs)
    176                 v = kwargs.pop(k)
    177                 kwargs[r_k] = v
--> 178         return super(SignatureTranslatedFunction, self).__call__(*args, **kwargs)
    179 
    180 pattern_link = re.compile(r'\link\{(.+?)\}')

~\AppData\Local\Continuum\anaconda3\lib\site-packages\rpy2\robjects\functions.py in __call__(self, *args, **kwargs)
    104         for k, v in kwargs.items():
    105             new_kwargs[k] = conversion.py2ri(v)
--> 106         res = super(Function, self).__call__(*new_args, **new_kwargs)
    107         res = conversion.ri2ro(res)
    108         return res

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

如果有人能看出我哪里错了,或者能提供建议,我将永远感激不尽。我想要的只是 Python 中那个公式中的两个数字,这样我就可以用它们来构建一些置信区间。

谢谢

考虑将所有公式变量合并到一个数据框中并使用 data 参数。 as_formula 调用在 R 环境中查找,但 ratescount 在 Python 范围内。因此,将所有项目包含在同一对象中。然后 运行 您的 nls 使用 Pandas 数据框或 R 数据框:

import pandas as pd
import rpy2.robjects as ro
from rpy2.robjects.packages import importr
from rpy2.robjects import pandas2ri

base = importr('base', robject_translations={'with': '_with'})
stats = importr('stats', robject_translations={'format_perc': '_format_perc'})

a = 0.05
b = 1.1
d = ro.ListVector({'a': a, 'b': b})

dfData = pd.read_csv('Input.csv')
dfData['count'] = dfData['Trials'].astype('float')
dfData['rates'] = dfData['Successes'] / dfData['Trials']
dfData['a'] = a
dfData['b'] = b

pandas2ri.activate()

rdf = pandas2ri.py2ri(dfData)

my_formula = stats.as_formula('rates ~ 1-(1/(10^(a * count ^ (b-1))))')

# WITH PANDAS DATAFRAME
fit = stats.nls(formula=my_formula, data=dfData, weights=dfData['count'], start=d)
print(fit)

# WITH R DATAFRAME
fit = stats.nls(formula=my_formula, data=rdf, weights=rdf.rx(True, 'count'), start=d)
print(fit)

或者,您可以使用 robjects.globalenv 而不是 data 参数:

ro.globalenv['rates'] = dfData['rates']
ro.globalenv['count'] = dfData['count']
ro.globalenv['a'] = dfData['a']
ro.globalenv['b'] = dfData['b']

fit = stats.nls(formula=my_formula, weights=dfData['count'], start=d)
print(fit)

# Nonlinear regression model    
#   model: rates ~ 1 - (1/(10^(a * count^(b - 1))))    
#    data: parent.frame()

#       a       b     
# 0.01043 1.24943     
#  weighted residual sum-of-squares: 14.37       

# Number of iterations to convergence: 6     
# Achieved convergence tolerance: 9.793e-07

# To return parameters    
num = fit.rx('m')[0].names.index('getPars')
obj = fit.rx('m')[0][num]()

print(obj[0])
# 0.010425686223717435

print(obj[1])
# 1.2494303314553932

在 R 中等效:

dfData <- read.csv('Input.csv')

a <- .05
b <- 1.1  
d <- list(a=a, b=b)

dfData$count <- dfData$Trials
dfData$rates <- dfData$Successes / dfData$Trials
dfData$a <- a
dfData$b <- b

my_formula <- stats::as.formula("rates ~ 1-(1/(10^(a * count ^ (b-1))))")

fit <- stats::nls(my_formula, data=dfData, weights=dfData$count, start=d)
print(fit)

# Nonlinear regression model
#   model: rates ~ 1 - (1/(10^(a * count^(b - 1))))
#    data: dfData
#       a       b 
# 0.01043 1.24943 
#  weighted residual sum-of-squares: 14.37

# Number of iterations to convergence: 6 
# Achieved convergence tolerance: 9.793e-07

# To return parameters  
fit$m$getPars()['a']
# 0.01042569 

fit$m$getPars()['b']
# 1.24943