在函数中使用带引号的变量并将其裸露 [英] Taking quoted variable in a function and making it bare

问题描述

我有一个函数，它从R中的estimatr包中调用lm_robust函数.我希望能够指定一个变量，以在该变量上集群标准错误，但是lm_robust函数仅允许在lm_robust的集群中使用裸露的(未加引号的)变量名.选项，而我的函数需要输入的内容是用引号引起来的变量名.

I have a function that calls the lm_robust function from the estimatr package in R. I want to be able to specify a variable on which to cluster standard errors, but the lm_robust function only allows bare (unquoted) variable names in lm_robust's cluster option while my function needs the input to be a quoted variable name.

如何获取输入到函数中的变量(例如"cl")并将其转换为未加引号的变量(例如cl)?

How do I take a variable that is input into a function (such as "cl") and turn it into a unquoted variable (such as cl)?

推荐答案

在 do.call 中使用 as.name :

library(estimatr)
example(lm_robust)
## ... snip ...

clname <- "clusterID"
do.call("lm_robust", list(y ~ x + z, data = quote(dat), weights = quote(w),
   clusters = as.name(clname)))

给予:

              Estimate Std. Error    t value     Pr(>|t|)  CI Lower  CI Upper
(Intercept)  3.4261621  0.2009692 17.0481986 1.332761e-05  2.908643  3.943681
x           -0.6734741  0.1351184 -4.9843254 4.300590e-03 -1.022076 -0.324872
z            0.5850340  0.9436175  0.6199907 5.566933e-01 -1.689652  2.859720
                  DF
(Intercept) 4.970885
x           4.940162
z           6.396615

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