3

在下面的 R 会话中，summary(model)公式显示为model_str。如何让它显示为mpg ~ cyl + hp同时仍然能够通过字符串设置模型公式？

> data(mtcars)
> names(mtcars)
 [1] "mpg"  "cyl"  "disp" "hp"   "drat" "wt"   "qsec" "vs"   "am"   "gear" "carb"
> model_str <- 'mpg ~ cyl + hp'
> model <- lm(model_str, data=mtcars)
> summary(model)

Call:
lm(formula = model_str, data = mtcars)

Residuals:
    Min      1Q  Median      3Q     Max 
-4.4948 -2.4901 -0.1828  1.9777  7.2934 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 36.90833    2.19080  16.847  < 2e-16 ***
cyl         -2.26469    0.57589  -3.933  0.00048 ***
hp          -0.01912    0.01500  -1.275  0.21253    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.173 on 29 degrees of freedom
Multiple R-squared:  0.7407,    Adjusted R-squared:  0.7228 
F-statistic: 41.42 on 2 and 29 DF,  p-value: 3.162e-09
4

4 回答 4

6

使用do.call以便model_str在发送到之前进行评估，lm但引用mtcars以便它不是（否则会有一个巨大的输出显示实际值mtcars）。

do.call("lm", list(as.formula(model_str), data = quote(mtcars)))

给予：

Call:
lm(formula = mpg ~ cyl + hp, data = mtcars)

Coefficients:
(Intercept)          cyl           hp  
   36.90833     -2.26469     -0.01912
于 2020-10-22T17:55:20.577 回答
1

这有点 hack 并且可能很脆弱，但是修改模型formula元素的call元素是有效的：

model$call$formula <- formula(model_str)
summary(model)
## Call:
## lm(formula = mpg ~ cyl + hp, data = mtcars)
于 2020-10-22T17:47:10.017 回答
1

您可以在一行中直接构建和评估调用：

eval(as.call(list(quote(lm), formula = model_str, data = quote(mtcars))))
#> 
#> Call:
#> lm(formula = "mpg ~ cyl + hp", data = mtcars)
#> 
#> Coefficients:
#> (Intercept)          cyl           hp  
#>    36.90833     -2.26469     -0.01912
于 2020-10-22T17:51:31.690 回答
0

使用str2lang，然后do.call。

fo <- str2lang("mpg ~ hp + am")
do.call("lm", list(fo, quote(mtcars)))
# 
# Call:
#   lm(formula = mpg ~ hp + am, data = mtcars)
# 
# Coefficients:
# (Intercept)           hp           am  
#    26.58491     -0.05889      5.27709
于 2020-10-22T17:52:59.490 回答