rpy2、rpart在Python与R之间传递数据时遇到问题

0 投票
1 回答
1735 浏览
提问于 2025-04-16 08:29

我正在尝试通过RPY2在Python 2.6.5和R 10.0中运行rpart。

我在Python中创建了一个数据框,然后把它传过去,但我遇到了一个错误,错误信息是:

Error in function (x)  : binary operation on non-conformable arrays
Traceback (most recent call last):
  File "partitioningSANDBOX.py", line 86, in <module>
    model=r.rpart(**rpart_params)
  File "build/bdist.macosx-10.3-fat/egg/rpy2/robjects/functions.py", line 83, in __call__
  File "build/bdist.macosx-10.3-fat/egg/rpy2/robjects/functions.py", line 35, in __call__
rpy2.rinterface.RRuntimeError: Error in function (x)  : binary operation on non-conformable arrays

有没有人能帮我看看我哪里做错了,导致出现这个错误?

我代码中相关的部分是:

import numpy as np
import rpy2
import rpy2.robjects as rob
import rpy2.robjects.numpy2ri


#Fire up the interface to R
r = rob.r
r.library("rpart")

datadict = dict(zip(['responsev','predictorv'],[cLogEC,csplitData]))
Rdata = r['data.frame'](**datadict)
Rformula = r['as.formula']('responsev ~.')
#Generate an RPART model in R.
Rpcontrol = r['rpart.control'](minsplit=10, xval=10)
rpart_params = {'formula' : Rformula, \
       'data' : Rdata,
       'control' : Rpcontrol}
model=r.rpart(**rpart_params)

这两个变量cLogEC和csplitData是浮点类型的numpy数组。

另外,我的数据框看起来是这样的:

In [2]: print Rdata
------> print(Rdata)
   responsev predictorv
1  0.6020600        312
2  0.3010300        300
3  0.4771213        303
4  0.4771213        249
5  0.9242793        239
6  1.1986571        297
7  0.7075702        287
8  1.8115750        270
9  0.6020600        296
10 1.3856063        248
11 0.6127839        295
12 0.3010300        283
13 1.1931246        345
14 0.3010300        270
15 0.3010300        251
16 0.3010300        246
17 0.3010300        273
18 0.7075702        252
19 0.4771213        252
20 0.9294189        223
21 0.6127839        252
22 0.7075702        267
23 0.9294189        252
24 0.3010300        378
25 0.3010300        282

而公式看起来是这样的:

In [3]: print Rformula
------> print(Rformula)
responsev ~ .

1 个回答

5

这个问题和R语言中的rpart代码有关,具体来说,是下面这段代码,特别是最后一行:

m <- match.call(expand.dots = FALSE)
m$model <- m$method <- m$control <- NULL
m$x <- m$y <- m$parms <- m$... <- NULL
m$cost <- NULL
m$na.action <- na.action
m[[1L]] <- as.name("model.frame")
m <- eval(m, parent.frame())

解决这个问题的一种方法是避免进入那段代码(见下文),或者可以尝试从Python构建一个嵌套的评估(这样parent.frame()就能正常工作)。这并不像人们希望的那么简单,但也许我将来会找时间让它变得更简单。

from rpy2.robjects import DataFrame, Formula
import rpy2.robjects.numpy2ri as npr
import numpy as np
from rpy2.robjects.packages import importr
rpart = importr('rpart')
stats = importr('stats')

cLogEC = np.random.uniform(size=10)
csplitData = np.array(range(10), 'i')

dataf = DataFrame({'responsev': cLogEC,
                   'predictorv': csplitData})
formula = Formula('responsev ~.')
rpart.rpart(formula=formula, data=dataf, 
            control=rpart.rpart_control(minsplit = 10, xval = 10),
            model = stats.model_frame(formula, data=dataf))

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