对StepAIC的“系数”表进行排序

Question

有什么方法可以对StepAIC的“系数”部分的输出进行排序（通过估计值或t值）？

> summary(sta9); 

Call: 
lm(formula = rating ~ `b/d` + `v/w` + `korte klank` + `lange klank` + 
    `-eer, -oor, -eur` + `-aai, -ooi, -oei` + `-eeuw, -ieuw, -uw` + 
    `-ng` + `-nk` + schwa + `eind d` + `eind b` + open + gesloten + 
    `-ee` + `mv klinker 's` + tussenletter + hoofdletter + trema + 
    regelverkleinwoorden + `-ch, -cht` + `ei/ij` + `ou, au, ouw, auw` + 
    `-sch, -schr` + `i/ie` + `be-, ge-, ver-` + e + `-lijk` + 
    `-heid` + `-teit` + `-tie` + `-ge` + verklein + `-er, -el, -en, -em` + 
    leenwoord + `-air` + `c/k, c/s` + q + x + y + `-th` + accent + 
    ch + length + freq - 1, data = as.data.frame(scale(items16[c(7, 
    21:78, 85:86)]))) 

Residuals: 
    Min  1Q Median  3Q  Max 
-1.8271 -0.3729 -0.0522 0.3585 1.9980 

Coefficients: 
         Estimate Std. Error t value Pr(>|t|)  
`b/d`     0.026588 0.009423 2.822 0.004805 ** 
`v/w`     0.032920 0.009350 3.521 0.000435 *** 
`korte klank`  -0.029310 0.013324 -2.200 0.027885 * 
`lange klank`  -0.024760 0.010828 -2.287 0.022271 * 
`-eer, -oor, -eur` -0.020068 0.009321 -2.153 0.031378 * 
`-aai, -ooi, -oei` 0.015499 0.009259 1.674 0.094217 . 
`-eeuw, -ieuw, -uw` 0.019706 0.010097 1.952 0.051045 . 
`-ng`    -0.045828 0.009522 -4.813 1.55e-06 *** 
`-nk`    -0.022971 0.009299 -2.470 0.013543 * 
schwa    -0.014353 0.009340 -1.537 0.124461  
`eind d`    0.033766 0.010208 3.308 0.000950 *** 
`eind b`    0.041695 0.009045 4.610 4.18e-06 *** 
open     0.042792 0.010284 4.161 3.24e-05 *** 
gesloten    0.088218 0.009863 8.944 < 2e-16 *** 
`-ee`     0.014626 0.009043 1.617 0.105879  
`mv klinker 's`  0.012957 0.009101 1.424 0.154628  
tussenletter   -0.052541 0.009850 -5.334 1.02e-07 *** 
hoofdletter   0.097015 0.009283 10.451 < 2e-16 *** 
trema     0.141985 0.009932 14.296 < 2e-16 *** 
regelverkleinwoorden 0.031238 0.010874 2.873 0.004093 ** 
`-ch, -cht`   -0.044193 0.009947 -4.443 9.15e-06 *** 
`ei/ij`    0.041017 0.011251 3.646 0.000271 *** 
`ou, au, ouw, auw` 0.021683 0.010342 2.097 0.036092 * 
`-sch, -schr`  -0.055558 0.009299 -5.974 2.54e-09 *** 
`i/ie`    0.043021 0.010286 4.182 2.95e-05 *** 
`be-, ge-, ver-`  -0.022401 0.009356 -2.394 0.016705 * 
e      0.051668 0..184 2.93e-05 *** 
`-lijk`    -0.032245 0.009964 -3.236 0.001222 ** 
`-heid`    -0.075044 0.010324 -7.269 4.42e-13 *** 
`-teit`    -0.018272 0.009826 -1.860 0.063034 . 
`-tie`    -0.013415 0.009448 -1.420 0.155715  
`-ge`    -0.019328 0.009757 -1.981 0.047673 * 
verklein    -0.077275 0.012170 -6.350 2.42e-10 *** 
`-er, -el, -en, -em` -0.046764 0.009867 -4.739 2.23e-06 *** 
leenwoord    0.117488 0.009910 11.856 < 2e-16 *** 
`-air`    0.036930 0.009062 4.075 4.69e-05 *** 
`c/k, c/s`   0.091685 0.009851 9.307 < 2e-16 *** 
q      0.052842 0.009360 5.645 1.77e-08 *** 
x      0.045299 0.009039 5.011 5.67e-07 *** 
y      0.063955 0.009375 6.822 1.05e-11 *** 
`-th`     0.038773 0.009380 4.134 3.65e-05 *** 
accent    0.072422 0.009434 7.677 2.09e-14 *** 
ch     0.029126 0.009621 3.027 0.002484 ** 
length    0.681405 0.013451 50.658 < 2e-16 *** 
freq     -0.025164 0.009157 -2.748 0.006026 ** 
--- 
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 0.5466 on 3597 degrees of freedom 
    (71 observations deleted due to missingness) 
Multiple R-squared: 0.6963, Adjusted R-squared: 0.6925 
F-statistic: 183.3 on 45 and 3597 DF, p-value: < 2.2e-16 

我已经发现这

sort(sta9$coefficients, decreasing = T); 

的作品，但只给了我“估计”列...

Answer 1

这里是一个线性回归的例子，你想要做什么：

set.seed(1) 
y = rnorm(10);x1 = rnorm(10);x2 = rnorm(10) 
m = lm(y ~ x1+ x2) 

s = summary(m) 
sc = as.data.frame(s$coefficients) 

sc[with(sc, order(-Estimate)), ] 

这给：

R> sc[with(sc, order(-Estimate)), ] 
      Estimate Std. Error t value Pr(>|t|) 
x1   0.06398  0.2405 0.2661 0.79785 
(Intercept) 0.03234  0.2128 0.1519 0.88352 
x2   -0.62799  0.2691 -2.3332 0.05236 

你的主要问题是，从sta9而不是由summary命令生成的表中使用的系数的载体，即比较：

sta9$coefficients 

与

summary(sta9)$coefficients