JD *_*ong 88 regression r linear-regression lm
我想使用lm()函数在R中进行线性回归.我的数据是一年一度的时间序列,一年(22年),另一个州(50个州).我想为每个状态拟合一个回归,以便最后我有一个lm响应的向量.我可以想象为每个状态做循环然后在循环内进行回归并将每个回归的结果添加到向量.但是,这似乎不像R一样.在SAS中我会做一个'by'语句,在SQL中我会做'group by'.R的做法是什么？
had*_*ley 57
这是使用plyr包的方法:
d <- data.frame(
state = rep(c('NY', 'CA'), 10),
year = rep(1:10, 2),
response= rnorm(20)
)
library(plyr)
# Break up d by state, then fit the specified model to each piece and
# return a list
models <- dlply(d, "state", function(df)
lm(response ~ year, data = df))
# Apply coef to each model and return a data frame
ldply(models, coef)
# Print the summary of each model
l_ply(models, summary, .print = TRUE)
ars*_*ars 45
这是使用lme4包的一种方式.
library(lme4)
d <- data.frame(state=rep(c('NY', 'CA'), c(10, 10)),
year=rep(1:10, 2),
response=c(rnorm(10), rnorm(10)))
xyplot(response ~ year, groups=state, data=d, type='l')
fits <- lmList(response ~ year | state, data=d)
fits
#------------
Call: lmList(formula = response ~ year | state, data = d)
Coefficients:
(Intercept) year
CA -1.34420990 0.17139963
NY 0.00196176 -0.01852429
Degrees of freedom: 20 total; 16 residual
Residual standard error: 0.8201316
Pau*_*tra 44
自2009年以来,dplyr已经发布,实际上提供了一种非常好的方式来进行这种分组,非常类似于SAS所做的.
library(dplyr)
d <- data.frame(state=rep(c('NY', 'CA'), c(10, 10)),
year=rep(1:10, 2),
response=c(rnorm(10), rnorm(10)))
fitted_models = d %>% group_by(state) %>% do(model = lm(response ~ year, data = .))
# Source: local data frame [2 x 2]
# Groups: <by row>
#
# state model
# (fctr) (chr)
# 1 CA <S3:lm>
# 2 NY <S3:lm>
fitted_models$model
# [[1]]
#
# Call:
# lm(formula = response ~ year, data = .)
#
# Coefficients:
# (Intercept) year
# -0.06354 0.02677
#
#
# [[2]]
#
# Call:
# lm(formula = response ~ year, data = .)
#
# Coefficients:
# (Intercept) year
# -0.35136 0.09385
要检索系数和Rsquared/p.value,可以使用该broom包.该套餐提供:
三个S3泛型:整洁,它总结了模型的统计结果,如回归系数; augment,它为原始数据添加了列,例如预测,残差和集群分配; 和glance,它提供了模型级统计的一行摘要.
library(broom)
fitted_models %>% tidy(model)
# Source: local data frame [4 x 6]
# Groups: state [2]
#
# state term estimate std.error statistic p.value
# (fctr) (chr) (dbl) (dbl) (dbl) (dbl)
# 1 CA (Intercept) -0.06354035 0.83863054 -0.0757668 0.9414651
# 2 CA year 0.02677048 0.13515755 0.1980687 0.8479318
# 3 NY (Intercept) -0.35135766 0.60100314 -0.5846187 0.5749166
# 4 NY year 0.09385309 0.09686043 0.9689519 0.3609470
fitted_models %>% glance(model)
# Source: local data frame [2 x 12]
# Groups: state [2]
#
# state r.squared adj.r.squared sigma statistic p.value df
# (fctr) (dbl) (dbl) (dbl) (dbl) (dbl) (int)
# 1 CA 0.004879969 -0.119510035 1.2276294 0.0392312 0.8479318 2
# 2 NY 0.105032068 -0.006838924 0.8797785 0.9388678 0.3609470 2
# Variables not shown: logLik (dbl), AIC (dbl), BIC (dbl), deviance (dbl),
# df.residual (int)
fitted_models %>% augment(model)
# Source: local data frame [20 x 10]
# Groups: state [2]
#
# state response year .fitted .se.fit .resid .hat
# (fctr) (dbl) (int) (dbl) (dbl) (dbl) (dbl)
# 1 CA 0.4547765 1 -0.036769875 0.7215439 0.4915464 0.3454545
# 2 CA 0.1217003 2 -0.009999399 0.6119518 0.1316997 0.2484848
# 3 CA -0.6153836 3 0.016771076 0.5146646 -0.6321546 0.1757576
# 4 CA -0.9978060 4 0.043541551 0.4379605 -1.0413476 0.1272727
# 5 CA 2.1385614 5 0.070312027 0.3940486 2.0682494 0.1030303
# 6 CA -0.3924598 6 0.097082502 0.3940486 -0.4895423 0.1030303
# 7 CA -0.5918738 7 0.123852977 0.4379605 -0.7157268 0.1272727
# 8 CA 0.4671346 8 0.150623453 0.5146646 0.3165112 0.1757576
# 9 CA -1.4958726 9 0.177393928 0.6119518 -1.6732666 0.2484848
# 10 CA 1.7481956 10 0.204164404 0.7215439 1.5440312 0.3454545
# 11 NY -0.6285230 1 -0.257504572 0.5170932 -0.3710185 0.3454545
# 12 NY 1.0566099 2 -0.163651479 0.4385542 1.2202614 0.2484848
# 13 NY -0.5274693 3 -0.069798386 0.3688335 -0.4576709 0.1757576
# 14 NY 0.6097983 4 0.024054706 0.3138637 0.5857436 0.1272727
# 15 NY -1.5511940 5 0.117907799 0.2823942 -1.6691018 0.1030303
# 16 NY 0.7440243 6 0.211760892 0.2823942 0.5322634 0.1030303
# 17 NY 0.1054719 7 0.305613984 0.3138637 -0.2001421 0.1272727
# 18 NY 0.7513057 8 0.399467077 0.3688335 0.3518387 0.1757576
# 19 NY -0.1271655 9 0.493320170 0.4385542 -0.6204857 0.2484848
# 20 NY 1.2154852 10 0.587173262 0.5170932 0.6283119 0.3454545
# Variables not shown: .sigma (dbl), .cooksd (dbl), .std.resid (dbl)
Thi*_*rry 23
在我看来,混合线性模型是这类数据的更好方法.以下代码给出了固定效应的整体趋势.随机效应表明每个州的趋势与全球趋势有何不同.相关结构考虑了时间自相关.看看Pinheiro&Bates(S和S-Plus中的混合效果模型).
library(nlme)
lme(response ~ year, random = ~year|state, correlation = corAR1(~year))
Fra*_*Nut 13
使用一个很好的解决方案data.table张贴在这里通过@Zach交叉验证.我只想补充一点,也可以迭代获得回归系数r ^ 2:
## make fake data
library(data.table)
set.seed(1)
dat <- data.table(x=runif(100), y=runif(100), grp=rep(1:2,50))
##calculate the regression coefficient r^2
dat[,summary(lm(y~x))$r.squared,by=grp]
grp V1
1: 1 0.01465726
2: 2 0.02256595
以及来自的所有其他输出summary(lm):
dat[,list(r2=summary(lm(y~x))$r.squared , f=summary(lm(y~x))$fstatistic[1] ),by=grp]
grp r2 f
1: 1 0.01465726 0.714014
2: 2 0.02256595 1.108173
## make fake data
ngroups <- 2
group <- 1:ngroups
nobs <- 100
dta <- data.frame(group=rep(group,each=nobs),y=rnorm(nobs*ngroups),x=runif(nobs*ngroups))
head(dta)
#--------------------
group y x
1 1 0.6482007 0.5429575
2 1 -0.4637118 0.7052843
3 1 -0.5129840 0.7312955
4 1 -0.6612649 0.9028034
5 1 -0.5197448 0.1661308
6 1 0.4240346 0.8944253
#------------
## function to extract the results of one model
foo <- function(z) {
## coef and se in a data frame
mr <- data.frame(coef(summary(lm(y~x,data=z))))
## put row names (predictors/indep variables)
mr$predictor <- rownames(mr)
mr
}
## see that it works
foo(subset(dta,group==1))
#=========
Estimate Std..Error t.value Pr...t.. predictor
(Intercept) 0.2176477 0.1919140 1.134090 0.2595235 (Intercept)
x -0.3669890 0.3321875 -1.104765 0.2719666 x
#----------
## one option: use command by
res <- by(dta,dta$group,foo)
res
#=========
dta$group: 1
Estimate Std..Error t.value Pr...t.. predictor
(Intercept) 0.2176477 0.1919140 1.134090 0.2595235 (Intercept)
x -0.3669890 0.3321875 -1.104765 0.2719666 x
------------------------------------------------------------
dta$group: 2
Estimate Std..Error t.value Pr...t.. predictor
(Intercept) -0.04039422 0.1682335 -0.2401081 0.8107480 (Intercept)
x 0.06286456 0.3020321 0.2081387 0.8355526 x
## using package plyr is better
library(plyr)
res <- ddply(dta,"group",foo)
res
#----------
group Estimate Std..Error t.value Pr...t.. predictor
1 1 0.21764767 0.1919140 1.1340897 0.2595235 (Intercept)
2 1 -0.36698898 0.3321875 -1.1047647 0.2719666 x
3 2 -0.04039422 0.1682335 -0.2401081 0.8107480 (Intercept)
4 2 0.06286456 0.3020321 0.2081387 0.8355526 x
小智 6
我现在的答案有点迟了,但我正在寻找类似的功能.看起来R中的内置函数'by'也可以轻松地进行分组:
？by包含以下示例,该示例适合每个组并使用sapply提取系数:
require(stats)
## now suppose we want to extract the coefficients by group
tmp <- with(warpbreaks,
by(warpbreaks, tension,
function(x) lm(breaks ~ wool, data = x)))
sapply(tmp, coef)
我认为值得purrr::map为该问题添加方法。
library(tidyverse)
d <- data.frame(state=rep(c('NY', 'CA'), c(10, 10)),
year=rep(1:10, 2),
response=c(rnorm(10), rnorm(10)))
d %>%
group_by(state) %>%
nest() %>%
mutate(model = map(data, ~lm(response ~ year, data = .)))
有关将broom软件包与这些结果结合使用的更多想法，请参见@Paul Hiemstra的答案。
小智 5
lm()上面的函数是一个简单的例子。顺便说一句，我想您的数据库具有以下形式的列：
年份状态 var1 var2 y...
在我看来，您可以使用以下代码：
require(base)
library(base)
attach(data) # data = your data base
#state is your label for the states column
modell<-by(data, data$state, function(data) lm(y~I(1/var1)+I(1/var2)))
summary(modell)