【发布时间】:2015-10-30 18:13:30
【问题描述】:
我无法使用 rpy2 绘制观察值与预测值。下面的 R 代码有效:
#==============================#
# Set up the data. #
#==============================#
yTmp <- seq(1,20)
y <- NULL
x1Tmp <- seq(1,20)
x1 <- NULL
x2Tmp <- seq(1,20)
x2 <- NULL
for (i in 1:20)
{
y[i] = yTmp[i] + runif(1, 0, 1)
x1[i] = x1Tmp[i] + runif(1, 0, 1)
x2[i] = x2Tmp[i] + runif(1, 0, 1)
}
#==============================#
# Fit the model. #
#==============================#
fittedModel <- lm(y ~ x1 + x2)
#==============================#
# Plot the observed vs predicted
#==============================#
png(filename="fittedModel_R.png")
plot(fittedModel$fitted.values, y)
dev.off()
这会产生一个很好的观察与预测图(我会发布,但我需要至少 10 名声望才能发布图片)。
我曾尝试使用 rpy2 重现这一点,但我无法弄清楚如何让拟合值发挥得很好。下面的代码与上面的 R 代码一样,我可以做到,但不起作用:
#!/usr/bin/env python
#==============================#
# Set up packages. #
#==============================#
import rpy2.robjects as robjects
from rpy2.robjects.packages import importr
import random
import string
stats = importr('stats')
from rpy2.robjects import Formula
lattice = importr('lattice')
rprint = robjects.globalenv.get("print")
grdevices = importr('grDevices')
#==============================#
# Set up the data. #
#==============================#
y = robjects.FloatVector(())
x1 = robjects.FloatVector(())
x2 = robjects.FloatVector(())
for i in range(1,20):
yValue = i + random.random()
y.rx[i] = yValue
x1Value = i + random.random()
x1.rx[i] = x1Value
x2Value = i + random.random()
x2.rx[i] = x2Value
robjects.globalenv["y"] = y
robjects.globalenv["x1"] = x1
robjects.globalenv["x2"] = x2
#==============================#
# Fit the model. #
#==============================#
fittedModel = stats.lm("y ~ x1 + x2")
#==============================#
# Attempt to extract the fitted
# values from the model and put
# on a vector. #
#==============================#
robjects.globalenv['predicted'] = robjects.Vector(fittedModel.rx('fitted.values'))
#==============================#
# Plot the observed vs predicted
#==============================#
grdevices.png(file = "fittedModel_RPY2.png", width = 512, height = 512)
formula = Formula('y ~ predicted')
p = lattice.xyplot(formula)
rprint(p)
grdevices.dev_off()
这会产生错误:
/usr/local/lib/python2.7/dist-packages/rpy2/robjects/functions.py:106: UserWarning:顺序错误(as.numeric(x)):(列表)对象不能 被强制输入'double'
res = super(Function, self).call(*new_args, **new_kwargs) Traceback(最近一次调用最后一次):文件“./testRPY2.py”,第 45 行,在 p = lattice.xyplot(公式)文件“/usr/local/lib/python2.7/dist-packages/rpy2/robjects/functions.py”, 第 178 行,在 调用 return super(SignatureTranslatedFunction, self).call(*args, **kwargs) File "/usr/local/lib/python2.7/dist-packages/rpy2/robjects/functions.py", 第 106 行,在 调用 res = super(Function, self).call(*new_args, **new_kwargs) rpy2.rinterface.RRuntimeError: Error in order(as.numeric(x)) :
(list) 对象不能被强制输入'double'
问题肯定出在预测值上,因为将公式更改为绘制 y 与 y 会生成一个不错的图:
formula = Formula('y ~ y')
我做了很多尝试将 python 中的数据强制转换为可绘图格式,包括在 python 中转换为字符串、操作以及作为浮动向量发送回 rpy2。但我真的不明白为什么它不起作用,必须有更好的方法。非常感谢任何对我的问题的见解和帮助。
【问题讨论】:
标签: rpy2