R/plm按指数提取残差_R_Plm_Pseries

R/plm按指数提取残差

R/plm按指数提取残差,r,plm,pseries,R,Plm,Pseries,我使用以下方法创建了一个plm对象： require(plm) plm1 <- plm(Sepal.Length ~ Petal.Length + Petal.Width, data = iris, index = "Species") 如果我有这样的东西就好了： > df1 <- data.frame(time = rep(1:10,15), Species = iris$Species, resid1 = runif(150)) > head(df1) time

我使用以下方法创建了一个plm对象：

require(plm)
plm1 <- plm(Sepal.Length ~ Petal.Length + Petal.Width, data = iris, index = "Species")

如果我有这样的东西就好了：

> df1 <- data.frame(time = rep(1:10,15), Species = iris$Species, resid1 = runif(150))
> head(df1)
  time Species    resid1
1    1  setosa 0.7038776
2    2  setosa 0.2164597
3    3  setosa 0.1988884
4    4  setosa 0.9311872
5    5  setosa 0.7087211
6    6  setosa 0.9914357

>df1头（df1）
物种残留时间1
1 setosa 0.7038776
2 setosa 0.2164597
3 3 setosa 0.1988884
4刚毛0.9311872
5刚毛0.7087211
6 setosa 0.9914357

我可以使用ddply或聚合来找到每个物种的RSquare

有什么建议吗？

可能是这样的，这样做可以奏效

library(plm)
plm1 <- plm(Sepal.Length ~ Petal.Length + Petal.Width, data = iris, index = "Species")
res <- residuals(plm1)
df <- cbind(as.vector(res), attr(res, "index"))
names(df) <- c("resid", "species", "time")
str(df)
## 'data.frame':    150 obs. of  3 variables:
##  $ resid  : num  0.1499 -0.0501 -0.1595 -0.4407 0.0499 ...
##  $ species: Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
##  $ time   : Factor w/ 50 levels "1","2","3","4",..: 1 2 3 4 5 6 7 8 9 10 ...

库（plm）
plm1这是一个老问题，但我想指出一些容易漏掉并可能导致严重错误的问题。答案是正确的，但我想我应该澄清为什么需要这样一种变通方法，因为它可能并不明显
在阅读过程中，我学到了以下几点：如前所述，plm不一定按照给定给函数的顺序保存数据。这意味着，如果您不小心，只需在plm对象上使用residuals（）
函数，然后将其连接到您的数据，就会导致错误的残差分组到错误的数据行！作为一个例子，考虑如下：
require(plm)
data("Gasoline") # The Gasoline dataset from the plm package

plm1 <- plm(lgaspcar ~ lincomep + lrpmg + lcarpcap, data=Gasoline, method = "within", index = c("country", "year"))

coef(plm1)
  lincomep      lrpmg   lcarpcap 
 0.6622497 -0.3217025 -0.6404829 

head(residuals(plm1))
          1           2           3           4           5           6 
-0.18814207 -0.19642727 -0.14874420 -0.12476346 -0.12114060 -0.08684045 

这意味着，如果我们使用汽油2作为我们正在使用的数据集，那么在Gasoline2
和residuals（plm2）
上使用类似cbind（）
的函数将导致错误的残差连接到观测值
head(cbind(Gasoline, residuals(plm1)))
  country year lgaspcar  lincomep      lrpmg  lcarpcap residuals(plm1)
1 AUSTRIA 1960 4.173244 -6.474277 -0.3345476 -9.766840     -0.18814207
2 AUSTRIA 1961 4.100989 -6.426006 -0.3513276 -9.608622     -0.19642727
3 AUSTRIA 1962 4.073177 -6.407308 -0.3795177 -9.457257     -0.14874420
4 AUSTRIA 1963 4.059509 -6.370679 -0.4142514 -9.343155     -0.12476346
5 AUSTRIA 1964 4.037689 -6.322247 -0.4453354 -9.237739     -0.12114060
6 AUSTRIA 1965 4.033983 -6.294668 -0.4970607 -9.123903     -0.08684045

head(cbind(Gasoline2, residuals(plm2)))
     country year lgaspcar  lincomep      lrpmg  lcarpcap residuals(plm2)
258   SWEDEN 1970 3.989372 -7.732610 -2.7335921 -8.164506     -0.18814207
7    AUSTRIA 1966 4.047537 -6.252545 -0.4668377 -9.019822     -0.19642727
64   DENMARK 1966 4.233643 -5.851866 -0.3961885 -8.681541     -0.14874420
73   DENMARK 1975 4.033015 -5.612967 -0.3939543 -8.274632     -0.12476346
268 SWITZERL 1961 4.441330 -6.111640 -0.8655847 -9.158229     -0.12114060
186    JAPAN 1974 4.007964 -5.852553 -0.1909064 -8.846520     -0.08684045

如上所述，在汽油2示例中，残差分配到错误的行
发生了什么事？正如前面提到的，plm
并没有保持观测的顺序。使用dickoa在前面的回答中指出的attr（）
函数，我们可以看到plm
按国家和年份重新组织数据
head( attr(residuals(plm2), "index") )
  country year
1 AUSTRIA 1960
2 AUSTRIA 1961
3 AUSTRIA 1962
4 AUSTRIA 1963
5 AUSTRIA 1964
6 AUSTRIA 1965

这就是原始汽油数据的结构，这就是残差以相同顺序呈现的原因
因此，我们可以使用attr（残差（plm2），“指数”）
为我们提供残差及其相应的国家和年份指标，以便将残差添加到原始数据中。如前所述，plyr
包对此非常有帮助
require(plyr)
resids2 <- data.frame(residual = residuals(plm2), attr(residuals(plm2), "index"))
Gasoline2$year <- factor(Gasoline2$year) # Needed since resids2$year is a factor, and Gasoline2$years was an integer. plyr does not accept them to be of different types.
Gasoline2 <- join(Gasoline2, resids2, by = c("country", "year"))

head(Gasoline2)
   country year lgaspcar  lincomep      lrpmg  lcarpcap    residual
1   SWEDEN 1970 3.989372 -7.732610 -2.7335921 -8.164506 -0.02468148
2  AUSTRIA 1966 4.047537 -6.252545 -0.4668377 -9.019822 -0.02479759
3  DENMARK 1966 4.233643 -5.851866 -0.3961885 -8.681541  0.03175032
4  DENMARK 1975 4.033015 -5.612967 -0.3939543 -8.274632 -0.06575219
5 SWITZERL 1961 4.441330 -6.111640 -0.8655847 -9.158229 -0.05789130
6    JAPAN 1974 4.007964 -5.852553 -0.1909064 -8.846520 -0.21957156

require（plyr）
这能满足你的需要吗<代码>iris$residuals@RichardHerron：行与索引匹配吗？是的，除非你缺少观察结果。我使用complete.cases来确保我的数据中没有遗漏的观察值。由于plm能够自动使用不平衡面板，如果残差有na，我们是否也可以将其合并到原始数据帧中？
head(cbind(Gasoline, residuals(plm1)))
  country year lgaspcar  lincomep      lrpmg  lcarpcap residuals(plm1)
1 AUSTRIA 1960 4.173244 -6.474277 -0.3345476 -9.766840     -0.18814207
2 AUSTRIA 1961 4.100989 -6.426006 -0.3513276 -9.608622     -0.19642727
3 AUSTRIA 1962 4.073177 -6.407308 -0.3795177 -9.457257     -0.14874420
4 AUSTRIA 1963 4.059509 -6.370679 -0.4142514 -9.343155     -0.12476346
5 AUSTRIA 1964 4.037689 -6.322247 -0.4453354 -9.237739     -0.12114060
6 AUSTRIA 1965 4.033983 -6.294668 -0.4970607 -9.123903     -0.08684045

head(cbind(Gasoline2, residuals(plm2)))
     country year lgaspcar  lincomep      lrpmg  lcarpcap residuals(plm2)
258   SWEDEN 1970 3.989372 -7.732610 -2.7335921 -8.164506     -0.18814207
7    AUSTRIA 1966 4.047537 -6.252545 -0.4668377 -9.019822     -0.19642727
64   DENMARK 1966 4.233643 -5.851866 -0.3961885 -8.681541     -0.14874420
73   DENMARK 1975 4.033015 -5.612967 -0.3939543 -8.274632     -0.12476346
268 SWITZERL 1961 4.441330 -6.111640 -0.8655847 -9.158229     -0.12114060
186    JAPAN 1974 4.007964 -5.852553 -0.1909064 -8.846520     -0.08684045

head( attr(residuals(plm2), "index") )
  country year
1 AUSTRIA 1960
2 AUSTRIA 1961
3 AUSTRIA 1962
4 AUSTRIA 1963
5 AUSTRIA 1964
6 AUSTRIA 1965

require(plyr)
resids2 <- data.frame(residual = residuals(plm2), attr(residuals(plm2), "index"))
Gasoline2$year <- factor(Gasoline2$year) # Needed since resids2$year is a factor, and Gasoline2$years was an integer. plyr does not accept them to be of different types.
Gasoline2 <- join(Gasoline2, resids2, by = c("country", "year"))

head(Gasoline2)
   country year lgaspcar  lincomep      lrpmg  lcarpcap    residual
1   SWEDEN 1970 3.989372 -7.732610 -2.7335921 -8.164506 -0.02468148
2  AUSTRIA 1966 4.047537 -6.252545 -0.4668377 -9.019822 -0.02479759
3  DENMARK 1966 4.233643 -5.851866 -0.3961885 -8.681541  0.03175032
4  DENMARK 1975 4.033015 -5.612967 -0.3939543 -8.274632 -0.06575219
5 SWITZERL 1961 4.441330 -6.111640 -0.8655847 -9.158229 -0.05789130
6    JAPAN 1974 4.007964 -5.852553 -0.1909064 -8.846520 -0.21957156