rpy2：如何從調用一個R函數

得到的返回值我想用下面R腳本Python：rpy2：如何從調用一個R函數

> 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，我也做了以下內容：

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)

結果變量x等於

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...]

您能否告知如何獲取變量data.fit$output[[1]]$Vt.bp？附：

PS：這是我第一次與rpy2，所以請隨時告訴我，如果我已經做了任何錯誤。

來源

2017-06-04 iamsterdam

我想有對結果沒有引用，如一切是一個向量和所期望的數據分別位於：'列表（x [1] [0] [4] [0] ）' – iamsterdam

你有沒有看過文檔？ http://rpy2.readthedocs.io/en/version_2.8.x/vector.html#extracting-items – lgautier

如圖所示，bfast方法顯然返回一個嵌套的對象與深層的數據項。此外，返回的Python對象是具有嵌套的未命名元素的外部類<class 'rpy2.robjects. vectors.ListVector'>。

作爲通過整數位置挖掘Python對象的替代方案，請考慮使用匿名包導入工具STAP導入用戶定義的R函數，該工具允許您保留R代碼並專門返回使用R的所需變量命名元素命名法：

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)

來源

2017-06-04 17:08:52 Parfait

rpy2：如何從調用一個R函數

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