Question

我已经运行了一个逻辑回归，我给它命名了摘要。 “得分”因此，summary(score)给了我以下

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-1.3616  -0.9806  -0.7876   1.2563   1.9246  

                       Estimate Std. Error    z value    Pr(>|z|)
(Intercept)        -4.188286233 1.94605597 -2.1521921 0.031382230 *
Overall            -0.013407201 0.06158168 -0.2177141 0.827651866
RTN                -0.052959314 0.05015013 -1.0560154 0.290961160
Recorded            0.162863294 0.07290053  2.2340482 0.025479900 *
PV                 -0.086743611 0.02950620 -2.9398438 0.003283778 **
Expire             -0.035046322 0.04577103 -0.7656878 0.443862068
Trial               0.007220173 0.03294419  0.2191637 0.826522498
Fitness             0.056135418 0.03114687  1.8022810 0.071501212 .

---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 757.25  on 572  degrees of freedom
Residual deviance: 725.66  on 565  degrees of freedom
AIC: 741.66

Number of Fisher Scoring iterations: 4

我希望获得的是获取那些在*值旁边有**，***或Pr(>|z|)的变量的变量名称和系数。换句话说，我希望上述变量和系数具有Pr(>|z|)＆lt; 0.05。

理想情况下，我想将它们放在数据框中。不幸的是，我尝试过的以下代码无效。

variable_try <-
  summary(score)$coefficients[if(summary(score)$coefficients[, 4] <= .05, 
                                 summary(score)$coefficients[, 1]),]

Error: unexpected ',' in "variable_try <-
summary(score)$coefficients[if(summary(score)$coefficients[,4] < .05,"

Answer 1

这个怎么样：

data.frame(summary(score)$coef[summary(score)$coef[,4] <= .05, 4])

仅从逻辑模型中提取p值重要的系数

1 个答案: