我试图比较python的statsmodels和R中的逻辑回归实现。
Python版本：
import statsmodels.api as sm
import pandas as pd
import pylab as pl
import numpy as np
df = pd.read_csv("http://www.ats.ucla.edu/stat/data/binary.csv")
df.columns = list(df.columns)[:3] + ["prestige"]
# df.hist()
# pl.show()
dummy_ranks = pd.get_dummies(df["prestige"], prefix="prestige")
cols_to_keep = ["admit", "gre", "gpa"]
data = df[cols_to_keep].join(dummy_ranks.ix[:, "prestige_2":])
data["intercept"] = 1.0
train_cols = data.columns[1:]
logit = sm.Logit(data["admit"], data[train_cols])
result = logit.fit()
result.summary2()
结果：
Results: Logit
=================================================================
Model: Logit Pseudo R-squared: 0.083
Dependent Variable: admit AIC: 470.5175
Date: 2014-12-19 01:11 BIC: 494.4663
No. Observations: 400 Log-Likelihood: -229.26
Df Model: 5 LL-Null: -249.99
Df Residuals: 394 LLR p-value: 7.5782e-08
Converged: 1.0000 Scale: 1.0000
No. Iterations: 6.0000
------------------------------------------------------------------
Coef. Std.Err. z P>|z| [0.025 0.975]
------------------------------------------------------------------
gre 0.0023 0.0011 2.0699 0.0385 0.0001 0.0044
gpa 0.8040 0.3318 2.4231 0.0154 0.1537 1.4544
prestige_2 -0.6754 0.3165 -2.1342 0.0328 -1.2958 -0.0551
prestige_3 -1.3402 0.3453 -3.8812 0.0001 -2.0170 -0.6634
prestige_4 -1.5515 0.4178 -3.7131 0.0002 -2.3704 -0.7325
intercept -3.9900 1.1400 -3.5001 0.0005 -6.2242 -1.7557
=================================================================
R版：
data = read.csv("http://www.ats.ucla.edu/stat/data/binary.csv", head=T)
require(reshape2)
data1 = dcast(data, admit + gre + gpa ~ rank)
require(dplyr)
names(data1)[4:7] = paste("rank", 1:4, sep="")
data1 = data1[, -4]
summary(glm(admit ~ gre + gpa + rank2 + rank3 + rank4, family=binomial, data=data1))
结果：
Call:
glm(formula = admit ~ gre + gpa + rank2 + rank3 + rank4, family = binomial,
data = data1)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.5133 -0.8661 -0.6573 1.1808 2.0629
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -4.184029 1.162421 -3.599 0.000319 ***
gre 0.002358 0.001112 2.121 0.033954 *
gpa 0.770591 0.343908 2.241 0.025046 *
rank2 -0.369711 0.310342 -1.191 0.233535
rank3 -1.015012 0.335147 -3.029 0.002457 **
rank4 -1.249251 0.414416 -3.014 0.002574 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 466.13 on 377 degrees of freedom
Residual deviance: 434.12 on 372 degrees of freedom
AIC: 446.12
Number of Fisher Scoring iterations: 4
结果完全不同，例如，rank_2的p值分别为0.03和0.2。我想知道这种差异的原因是什么？请注意，我为这两个版本创建了虚拟变量，并为python版本创建了一个常量列，它在R中自动处理。
此外，似乎python的速度提高了2倍：
##################################################
# python timing
def test():
for i in range(5000):
logit = sm.Logit(data["admit"], data[train_cols])
result = logit.fit(disp=0)
import time
start = time.time()
test()
print(time.time() - start)
10.099738836288452
##################################################
# R timing
> f = function() for(i in 1:5000) {mod = glm(admit ~ gre + gpa + rank2 + rank3 + rank4, family=binomial, data=data1)}
> system.time(f())
user system elapsed
17.505 0.021 17.526
答案 0 :(得分：6)
不确定您的数据操作是什么，但它们似乎在R运行中丢失了信息。如果我保留所有等级信息，那么我就会在原始数据对象上得到这个（并且结果在它们重叠的区域看起来非常相似。）（可能性只估计到任意常数，因此您只能比较差异log-likelihood。即使有这个警告，偏差应该是负对数似然的两倍，所以这些结果也是可比的。）
> summary(glm(admit ~ gre + gpa +as.factor( rank), family=binomial,
data=data)) # notice that I'm using your original data-object
Call:
glm(formula = admit ~ gre + gpa + as.factor(rank), family = binomial,
data = data)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.6268 -0.8662 -0.6388 1.1490 2.0790
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -3.989979 1.139951 -3.500 0.000465 ***
gre 0.002264 0.001094 2.070 0.038465 *
gpa 0.804038 0.331819 2.423 0.015388 *
as.factor(rank)2 -0.675443 0.316490 -2.134 0.032829 *
as.factor(rank)3 -1.340204 0.345306 -3.881 0.000104 ***
as.factor(rank)4 -1.551464 0.417832 -3.713 0.000205 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 499.98 on 399 degrees of freedom
Residual deviance: 458.52 on 394 degrees of freedom
AIC: 470.52
Number of Fisher Scoring iterations: 4
答案 1 :(得分：0)
我像这样重写R部分：
makeDummy = function(x, x1) { ifelse(is.na(x), NA, ifelse(x == x1, 1, 0)) }
data = read.csv("http://www.ats.ucla.edu/stat/data/binary.csv", head=T)
data$rank2 = makeDummy(data$rank, 2)
data$rank3 = makeDummy(data$rank, 3)
data$rank4 = makeDummy(data$rank, 4)
summary(glm(admit ~ gre + gpa + rank2 + rank3 + rank4, family=binomial, data=data))
结果与
完全相同Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -3.989979 1.139951 -3.500 0.000465 ***
gre 0.002264 0.001094 2.070 0.038465 *
gpa 0.804038 0.331819 2.423 0.015388 *
rank2 -0.675443 0.316490 -2.134 0.032829 *
rank3 -1.340204 0.345306 -3.881 0.000104 ***
rank4 -1.551464 0.417832 -3.713 0.000205 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 499.98 on 399 degrees of freedom
Residual deviance: 458.52 on 394 degrees of freedom
AIC: 470.52
我猜我要么以错误的方式使用dplyr::dcast，要么dcast出现问题。
答案 2 :(得分：0)
我只能回答，不能在接受的答案中添加评论。在python中，通常必须删除一个虚拟类以使其成为参考类，但是我认为您不需要为R做这个，因为glm会为您设置参考类。基本上，如果我正确理解您的代码，则不需要此行...
data1 = data1[, -4]
尝试直接按原样放置声望，但首先使用as.factor（）将其设置为一个因数。