【问题标题】:rpy2: how to get the return values from calling a R functionrpy2:如何从调用 R 函数中获取返回值
【发布时间】:2017-11-05 06:53:25
【问题描述】:

我想在Python 中使用以下R 脚本:

> library(bfast)
> apple <- read.csv("/Users/nskalis/Downloads/R/apple.csv", sep = ";", header=TRUE)
> data = apple
# data$in_bps: is vector of double numbers
> data.ts <- ts(data$in_bps, frequency=1)
> data.fit <- bfast(data.ts, h=0.1, season="none", max.iter=1)
> data.fit$output[[1]]$Tt
> data.fit$output[[1]]$Vt.bp
> data.fit$output[[1]]$ci.Vt
> data.fit$output[[1]]$ci.Vt$confint

因此我使用rpy2 并且我做了以下事情:

from rpy2.robjects.packages import importr
import rpy2.robjects as robjects
importr("bfast")
data = range(1,100)
data = robjects.FloatVector(data)
data = robjects.r.ts(data, frequency=1)
x = robjects.r.bfast(data, h=0.1, season="none", max_iter=1)

结果变量x等于

In [42]: x
Out[42]: 
R object with classes: ('bfast',) mapped to:
<ListVector - Python:0x7f234f7ad6c8 / R:0x76a2d60>
[Float..., ListV..., ListV..., ..., Float..., BoolV..., ListV...]
  Yt: <class 'rpy2.robjects.vectors.FloatVector'>
  R object with classes: ('ts',) mapped to:
<FloatVector - Python:0x7f234fd22dc8 / R:0x7605740>
[1.000000, 2.000000, 3.000000, ..., 97.000000, 98.000000, 99.000000]
R object with classes: ('bfast',) mapped to:
<ListVector - Python:0x7f234f7ad6c8 / R:0x76a2d60>
[Float..., ListV..., ListV..., ..., Float..., BoolV..., ListV...]
R object with classes: ('bfast',) mapped to:
<ListVector - Python:0x7f234f7ad6c8 / R:0x76a2d60>
[Float..., ListV..., ListV..., ..., Float..., BoolV..., ListV...]
  ...
  Yt: <class 'rpy2.robjects.vectors.FloatVector'>
  R object with classes: ('numeric',) mapped to:
<FloatVector - Python:0x7f234c053388 / R:0x586b668>
[0.000000]
  output: <class 'rpy2.robjects.vectors.BoolVector'>
  R object with classes: ('logical',) mapped to:
<BoolVector - Python:0x7f234c04eac8 / R:0x57ee518>
[NA]
R object with classes: ('bfast',) mapped to:
<ListVector - Python:0x7f234f7ad6c8 / R:0x76a2d60>
[Float..., ListV..., ListV..., ..., Float..., BoolV..., ListV...]

能否请您告知如何获取变量data.fit$output[[1]]$Vt.bp

PS:这是我第一次使用rpy2,所以如果我已经做错了什么,请随时告诉我。

【问题讨论】:

标签: python r rpy2


【解决方案1】:

正如所见,bfast 方法显然返回了一个嵌套对象,其中包含数据项的深度层。此外,返回的 Python 对象属于外部类 &lt;class 'rpy2.robjects. vectors.ListVector'&gt;,具有嵌套的未命名元素。

作为通过整数位置挖掘 Python 对象的替代方法,考虑使用匿名包导入工具STAP 导入用户定义的 R 函数,它允许您保留 R 代码并专门返回您需要使用的变量R 的命名元素命名法:

from rpy2.robjects.packages import STAP

r_fct_string ='''   
bfast_out <- function(path){
    data <- read.csv(path, sep = ";", header=TRUE)

    data.ts <- ts(data$in_bps, frequency=1)
    data.fit <- bfast(data.ts, h=0.1, season="none", max.iter=1)

    data.fit$output[[1]]$Vt.bp    
}
'''

r_pkg = STAP(r_fct_string, "r_pkg")

Vt_bp = r_pkg.bfast_out("/Users/nskalis/Downloads/R/apple.csv")

print(Vt_bp)

【讨论】:

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