我正在使用rpy2在py中对r进行一些非线性回归。
import rpy2.robjects as robjects
from rpy2.robjects import DataFrame, Formula
from rpy2.robjects import r
import rpy2.robjects.numpy2ri as npr
import numpy as np
from rpy2.robjects.packages import importr
r.nls(rates * 1-(1/(10^(a * count ^ (b-1)))), weights=count, start=list(a=a, b=b))
我有以下错误：
LookupError: 'nls' not found
AttributeError: 'R' object has no attribute 'nls'
它还将'〜'称为无效语法（我将其更改为*以通过它，但我确实需要它为'〜'）
关于出了什么问题的任何想法？
代码在R中工作正常。
这是在R：
中正常工作的完整代码#This recipe assumes that the data is in a csv file called 'ratedata.csv' and that the values are in columns titled:
#Entity, Trials and Successes
#Data must be sorted in order of number of applications (i.e. the 'Trials' column) highest to lowest.
data <- read.csv("ratedata.csv") #get the data
count <- data$Trials #define count as the number of trials
rates <- data$Successes / data$Trials #define rate as the success rate for each entity
a <- .05 #set initial values for a and b to generate predicted rates
b <- 1.1 #these values need to be reasonably sensible otherwise the later estimate will not converge sensibly
fit <- nls(rates ~ 1-(1/(10^(a * count ^ (b-1)))), weights=count, start=list(a=a, b=b)) #non-linear least squares fit of data, weighted by count (weighting is optional but helps if it won't converge sensibly)
summary(fit) #to show estimates of a and b
coef <- as.vector(coef(fit)) #extract the coefficients into a vector for re-use
a <- coef[[1]] # extract the calculated coefficient for a
b <- coef[[2]] # extract the calculated coefficient for b
confidence <- confint(fit)
intervals <- as.vector(confidence[c(2,4)])
predopt <- 1-(1/(10^(a * count ^ (b-1)))) #predict rate by count with optimised coefficients
se <- sqrt(( predopt* (1-predopt))/count) #calculate standard error for predicted rate
upper95 <- predopt + 2*se #upper 95% limit - roughly speaking. Wald interval is appropriate in this case.
lower95 <- predopt - 2*se #lower 95% limit
upper99 <- predopt + 3*se #upper 99% limit
lower99 <- predopt - 3*se #lower 99% limit
xlim <- range(count + 10) #setup plot
ylim <- range(c(upper99, 0)) #lower limit truncated at zero
main <- plot(count, rates, pch = 21, col = "navajowhite4", bg = "mistyrose4") #plot rates by organisation
lines(count, predopt, type="l", xlim=xlim, ylim=ylim, xlab="Trials", ylab="Predicted rate", col = "red") #plot predicted rate
lines (count, upper95, lty="dashed") #plot upper limit
lines (count, lower95, lty="dashed") #plot lower limit
lines (count, upper99, lty="dotted") #plot upper limit
lines (count, lower99, lty="dotted") #plot lower limit
cat("The least-squares values of a and b are", coef[[1]], "and", coef[[2]], "respectively", "\n")
print(confint(fit))
if (intervals[[1]] < 1 & intervals [[2]] > 1)
{
message ("There is probably no relationship between success rate and number of trials")
} else
{
message ("There is probably a relationship between success rate and number of trials")
}
列Trials和Successes只是两列48个整数（它们可以是任何东西。试验范围从129到2359，成功范围从8到365
更新问题2018年1月25日19.40 pm
目前的代码是：
import rpy2.robjects as ro
from rpy2.robjects.packages import importr
count = ro.IntVector([1,2,3,4,5])
rates = ro.IntVector([1,2,3,4,5])
a = ro.FloatVector([0.5])
b = ro.FloatVector([1.1])
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)
我收到错误：
---------------------------------------------------------------------------
RRuntimeError Traceback (most recent call last)
<ipython-input-2-3f7fcd7d7851> 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
我猜我的计数和费率变量不是列表？或者是其他东西？我试过乱搞并转换它们但无济于事。任何帮助非常感谢！
这是我为数据帧制作的代码：
dataf = ro.DataFrame({})
d = {'count': ro.IntVector((1,2,3,4,5)),'rates': ro.IntVector((1,2,3,4,5))}
dataf = ro.DataFrame(d)
count = dataf.rx(True, 'count')
rates = dataf.rx(True, 'rates')
答案 0 :(得分：1)
考虑导入R＆lt; stat> stats 和 base 库，然后复制所需的调用。并使用as_formula将公式的字符串表示形式转换为实际的公式对象。由于这些是默认的R库，因此找出哪个方法属于哪个包，如stats::nls()和base::list()。
另外请注意，为了与Python的语法规则保持一致，R名称中的任何句点都会转换为下划线。其他一些方法被重命名，以避免与Python自己的方法发生冲突。
...
import rpy2.robjects as ro
from rpy2.robjects.packages import importr
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)