Question

我想控制包含超过一百个级别的因子变量，而不将该控件的结果输出到摘要表。请注意，我也有兴趣复制Stata命令的速度，而不仅仅是对输出的外观改变。

在Stata我可以像这样使用“吸收”：

use http://www.stata-press.com/data/r14/abdata.dta, clear
. xtreg n w k i.year, fe

Fixed-effects (within) regression               Number of obs     =      1,031
Group variable: id                              Number of groups  =        140

R-sq:                                           Obs per group:
     within  = 0.6277                                         min =          7
     between = 0.8473                                         avg =        7.4
     overall = 0.8346                                         max =          9

                                                F(10,881)         =     148.56
corr(u_i, Xb)  = 0.5666                         Prob > F          =     0.0000

------------------------------------------------------------------------------
           n |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
           w |  -.2731482   .0551503    -4.95   0.000    -.3813896   -.1649068
           k |   .5648036   .0212211    26.62   0.000     .5231537    .6064535
             |
        year |
       1977  |  -.0347963   .0188134    -1.85   0.065    -.0717206    .0021281
       1978  |  -.0583286   .0190916    -3.06   0.002    -.0957989   -.0208583
       1979  |   -.070047   .0190414    -3.68   0.000    -.1074187   -.0326752
       1980  |  -.0889378   .0189788    -4.69   0.000    -.1261867   -.0516889
       1981  |  -.1401502   .0186309    -7.52   0.000    -.1767163   -.1035841
       1982  |  -.1603768   .0188132    -8.52   0.000    -.1973008   -.1234528
       1983  |  -.1621103   .0222902    -7.27   0.000    -.2058585   -.1183621
       1984  |  -.1258136   .0282391    -4.46   0.000    -.1812373   -.0703899
             |
       _cons |   2.255419   .1772614    12.72   0.000     1.907515    2.603323
-------------+----------------------------------------------------------------
     sigma_u |  .64723143
     sigma_e |  .12836859
         rho |  .96215208   (fraction of variance due to u_i)
------------------------------------------------------------------------------
F test that all u_i=0: F(139, 881) = 126.32                  Prob > F = 0.0000

使用吸收会消除固定效果

. reghdfe n w k, absorb(id year)
(converged in 7 iterations)

HDFE Linear regression                            Number of obs   =      1,031
Absorbing 2 HDFE groups                           F(   2,    881) =     362.67
                                                  Prob > F        =     0.0000
                                                  R-squared       =     0.9922
                                                  Adj R-squared   =     0.9908
                                                  Within R-sq.    =     0.4516
                                                  Root MSE        =     0.1284

------------------------------------------------------------------------------
           n |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
           w |  -.2731482   .0551503    -4.95   0.000    -.3813896   -.1649068
           k |   .5648036   .0212211    26.62   0.000     .5231537    .6064535
-------------+----------------------------------------------------------------
    Absorbed |       F(147, 881) =    120.660   0.000             (Joint test)
------------------------------------------------------------------------------

Absorbed degrees of freedom:
---------------------------------------------------------------+
 Absorbed FE |  Num. Coefs.  =   Categories  -   Redundant     |
-------------+-------------------------------------------------|
          id |          140             140              0     |
        year |            8               9              1     |
---------------------------------------------------------------+

Answer 1

我不知道有这样做的内置方法，但broom::tidy加上一些基于因子名称的过滤将做你想做的事情：

示例数据：

set.seed(101)
dd <- data.frame(y=rnorm(1000),
                 f=factor(sample(1:50,size=1000,replace=TRUE)),
                 x=rnorm(1000))

m <- lm(y~f+x,data=dd)

一种方式（取决于grepl()，它是基础R [我更熟悉]，这是混合和匹配范式一点点）

library(broom)
library(dplyr)
tidy(m) %>%
    filter(!grepl("^f[0-9]+",term))
##          term    estimate  std.error statistic   p.value
## 1 (Intercept) -0.22643955 0.18852186 -1.201131 0.2299999
## 2           x -0.03330846 0.03101449 -1.073964 0.2831116

或者您可以使用stringr::str_detect

执行此操作

library(stringr)
tidy(m) %>%
    filter(!str_detect(term,"^f[0-9]+"))

我使用的特定正则表达式基于因子的名称加上级别的名称。在您的情况下，如果您感到幸运，它将是"^year[0-9]+"，或仅"^year"

Answer 2

我能找到的最佳替代方案是lfe包，它实现了具有高维固定效果或/和工具变量的模型。

您可以在垂直条之后指定固定效果，如下所示：

felm(n ~ w _ k | year, df)

年度系数将在最终规范中被吸收。这种方法的问题在于它现在允许您预测观察结果。

R等价于Stata的Absorb

2 个答案: