我想用 stargazer 创建一个分位数回归结果表。结果应该是这样的
所以到目前为止我所做的如下:首先拟合数据并为标准错误调用一些摘要函数。提取系数值每个分位数的相应 p_val 并将所有内容收集在一个小标题中。看起来像这样:
tau = c(0.1, 0.3, 0.5, 0.7, 0.9)
sum_male_qr <- tibble(tau = tau,
coef = c(0.02007477, 0.06914977, -0.07867272, -0.11111504, -0.15044577),
se = c(0.2572446, 0.1167441, 0.1099517, 0.1213363, 0.2465198),
p_val = c(0.9378020, 0.5536708, 0.4743298, 0.3598485, 0.5417125)) %>% as.matrix
#for sake of simplicity we just expect results for both genders to be the same
sum_female_qr <- tibble(tau = tau,
coef = c(0.02007477, 0.06914977, -0.07867272, -0.11111504, -0.15044577),
se = c(0.2572446, 0.1167441, 0.1099517, 0.1213363, 0.2465198),
p_val = c(0.9378020, 0.5536708, 0.4743298, 0.3598485, 0.5417125)) %>% as.matrix
我对观星者很陌生,但认为必须有办法做到这一点。任何帮助,将不胜感激。我已经完成了这个小插曲,但没有想出比以下更好的东西:
stargazer(column.labels = c("$\tau = 0.1$", "$\tau = 0.3$", "$\tau = 0.5$", "$\tau = 0.7$", "$\tau = 0.9$" ),
align = T,
coef = list(Males = exo_sum_male[,1],Females = exo_sum_female[,1]),
se = list(Males = exo_sum_male[,2],Females = exo_sum_female[,2]),
p = list(Males = exo_sum_male[,3],Females = exo_sum_female[,3]),
style = 'qje',
title = "Quantile Effects When Education is Treated as Exogenous"
)
这会产生以下错误:Error in if (substr(inside[i], 1, nchar("list(")) == "list(") { : missing value where TRUE/FALSE needed
编辑:我知道结果如何。可怕,当然。但请不要介意这一点。这是一个已知问题,不应成为此讨论的一部分。