我想知道是否有可能只在Call:
函数中显示R
公式。
require(graphics)
## Annette Dobson (1990) "An Introduction to Generalized Linear Models".
## Page 9: Plant Weight Data.
ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group <- gl(2, 10, 20, labels = c("Ctl","Trt"))
weight <- c(ctl, trt)
lm.D9 <- lm(weight ~ group)
summary(lm.D9)
输出
Call:
lm(formula = weight ~ group)
Residuals:
Min 1Q Median 3Q Max
-1.0710 -0.4938 0.0685 0.2462 1.3690
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.0320 0.2202 22.850 9.55e-15 ***
groupTrt -0.3710 0.3114 -1.191 0.249
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.6964 on 18 degrees of freedom
Multiple R-squared: 0.07308, Adjusted R-squared: 0.02158
F-statistic: 1.419 on 1 and 18 DF, p-value: 0.249
答案 0 :(得分:4)
你可以尝试
ss <- capture.output(summary(lm.D9))
cat(ss[-(1:3)],sep="\n")
(更一般地说,您可以尝试使用grep()
&#34; [Cc] all&#34;在输出中做一些奇特的事情,并排除该行和后续行......)否则,您将&#39; d必须破解stats:::print.summary.lm
,例如
pp <- stats:::print.summary.lm
body(pp) <- body(pp)[-2]
pp(summary(lm.D9))
print.summary.lm <- pp
summary(lm.D9)
(请注意,这是相当脆弱的;它取决于Call:
打印是函数体的第一行,并且该行中没有任何内容对函数的其余部分是必需的。)