将逻辑回归从 R 迁移到 rpy2 [英] Migrating a logistic regression from R to rpy2
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问题描述
我正在尝试使用 ryp2 进行逻辑回归.我设法执行它,但不知道如何从结果中提取系数和 p 值.我不想在屏幕上打印值,而是创建一个函数来独立使用它们.
I'm trying to use ryp2 to do a logistic regression. I managed to execute it, but don't know how to extract the coefficients and p-values from the result. I don't want to print the values on the screen bu create a function to use them independently.
import rpy2.robjects as ro
mydata = ro.r['data.frame']
read = ro.r['read.csv']
head = ro.r['head']
summary = ro.r['summary']
mydata = read("http://www.ats.ucla.edu/stat/data/binary.csv")
#cabecalho = head(mydata)
formula = 'admit ~ gre + gpa + rank'
mylogit = ro.r.glm(formula=ro.r(formula), data=mydata,family=ro.r('binomial(link="logit")'))
#What NEXT?
推荐答案
我不知道如何获得 p 值,但对于其他任何人,它应该是这样的:
I don't known how you can get the p-values, but for any others it should be something like this:
In [24]:
#what is stored in mylogit?
mylogit.names
Out[24]:
<StrVector - Python:0x10a01a0e0 / R:0x10353ab20>
['coef..., 'resi..., 'fitt..., ..., 'meth..., 'cont..., 'xlev...]
In [25]:
#looks like the first item is the coefficients
mylogit.names[0]
Out[25]:
'coefficients'
In [26]:
#OK, let's get the the coefficients.
mylogit[0]
Out[26]:
<FloatVector - Python:0x10a01a5f0 / R:0x1028bcc80>
[-3.449548, 0.002294, 0.777014, -0.560031]
In [27]:
#be careful that the index from print is R index, starting with 1. I don't see p values here
print mylogit.names
[1] "coefficients" "residuals" "fitted.values"
[4] "effects" "R" "rank"
[7] "qr" "family" "linear.predictors"
[10] "deviance" "aic" "null.deviance"
[13] "iter" "weights" "prior.weights"
[16] "df.residual" "df.null" "y"
[19] "converged" "boundary" "model"
[22] "call" "formula" "terms"
[25] "data" "offset" "control"
[28] "method" "contrasts" "xlevels"
编辑
每个术语的 P 值:
Edit
The P values for each terms:
In [55]:
#p values:
list(summary(mylogit)[-6])[-4:]
Out[55]:
[0.0023265825120094407,
0.03564051883525258,
0.017659683902155117,
1.0581094283250368e-05]
还有:
In [56]:
#coefficients
list(summary(mylogit)[-6])[:4]
Out[56]:
[-3.449548397668471,
0.0022939595044433334,
0.7770135737198545,
-0.5600313868499897]
In [57]:
#S.E.
list(summary(mylogit)[-6])[4:8]
Out[57]:
[1.1328460085495897,
0.001091839095422917,
0.327483878497867,
0.12713698917130048]
In [58]:
#Z value
list(summary(mylogit)[-6])[8:12]
Out[58]:
[-3.0450285137032984,
2.1010050968680347,
2.3726773277632214,
-4.4049445444662885]
或更一般地说:
In [60]:
import numpy as np
In [62]:
COEF=np.array(summary(mylogit)[-6]) #it has a shape of (number_of_terms, 4)
In [63]:
COEF[:, -1] #p-value
Out[63]:
array([ 2.32658251e-03, 3.56405188e-02, 1.76596839e-02,
1.05810943e-05])
In [66]:
COEF[:, 0] #coefficients
Out[66]:
array([ -3.44954840e+00, 2.29395950e-03, 7.77013574e-01,
-5.60031387e-01])
In [68]:
COEF[:, 1] #S.E.
Out[68]:
array([ 1.13284601e+00, 1.09183910e-03, 3.27483878e-01,
1.27136989e-01])
In [69]:
COEF[:, 2] #Z
Out[69]:
array([-3.04502851, 2.1010051 , 2.37267733, -4.40494454])
你也可以 summary(mylogit).rx2('coefficient')
(或 rx
),如果你知道 coefficient
在汇总向量.
You can also summary(mylogit).rx2('coefficient')
(or rx
), if you know that coefficient
is in the summary vector.
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