Question

我运行glm并获得结果正常。现在我想得到95％显着的那些预测因子的名称，即p值小于或等于显着性水平5e-2。我跑：

fit <- glm(data=dfa, formula=response~.)
sig <- summary(fit)$coefficients[,4]

 (Intercept)       close0       close1       close2       close3       close4      closema        open0 
0.000000e+00 3.147425e-19 7.210909e-04 1.046019e-02 4.117580e-03 2.778701e-01 2.829958e-05 0.000000e+00 
       open1        open2        open3        open4       openma         low0         low1         low2 
8.627202e-30 1.138499e-02 1.112236e-03 7.422145e-03 3.967735e-03 3.036329e-42 3.033847e-05 3.237155e-01 
        low3         low4        lowma        high0        high1        high2        high3        high4 
8.198750e-01 6.647138e-02 4.350488e-05 6.177130e-58 2.625192e-02 4.143373e-01 3.964651e-01 3.694272e-01 
      highma      volume0      volume1      volume2      volume3      volume4     volumema 
1.416310e-05 8.027502e-02 1.975302e-01 1.630341e-09 8.979313e-03 1.274195e-06 8.246661e-01

> str(sig)
  Named num [1:31] 0.00 3.15e-19 7.21e-04 1.05e-02 4.12e-03 ...
  - attr(*, "names")= chr [1:31] "(Intercept)" "close0" "close1" "close2" ...

那么“命名数”类型是什么？

我想有一个像这样的列名数组，因为那些预测变量的p值低于显着性水平5e-2，即

best <- c('close0', 'close1', 'close2', 'close3', 'closema', ... etc)

注意close4不存在......如何以矢量化方式提取这些列名？

更新：我找到了如何在循环中完成它

fit <- glm(data=dfa, formula=response~.)
summary(fit)
sig <- summary(fit)$coefficients[,4]
best <- NULL
columnLabels <- names(sig)
for (columnLabel in columnLabels) {
    if (as.numeric(sig[columnLabel]) <= 5e-2) {
        if (is.null(best)) {
            best <- columnLabel
        } else {
            best <- c(best, columnLabel)
        }
    }
}

Answer 1

names(sig)[sig <= 0.05]正是您要找的。 names(sig)返回所有名称，sig <= 0.05有助于提取所需的子集。

Answer 2

utils::data(anorexia, package="MASS")

anorex.1 <- glm(Postwt ~ Prewt + Treat + offset(Prewt),
                family = gaussian, data = anorexia)
summary(anorex.1)

ll<-summary(anorex.1)$coefficients

ll<-as.data.frame(ll)

ll
              Estimate Std. Error   t value     Pr(>|t|)
(Intercept) 49.7711090 13.3909581  3.716770 0.0004101067
Prewt       -0.5655388  0.1611824 -3.508689 0.0008034250
TreatCont   -4.0970655  1.8934926 -2.163761 0.0339993147
TreatFT      4.5630627  2.1333359  2.138933 0.0360350847

rownames(ll[ll[,4]<0.005,])

[1] "(Intercept)" "Prewt"

如何矢量化提取重要的预测变量？

2 个答案: