Question

我想绘制从 VGAM 包中的vglm()函数派生的多项逻辑回归的预测值。

我使用 VGAM 非常重要，因为我试图复制在Stata中进行的同事分析，这是我使用此功能/包实现的。

数据的一个子集：

structure(list(
caretime3 = c(0, 2, 2, 0, 0, 2, 1, 1, 0, 2, 2, 0, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 1, 0, 1, 1, 2, 2, 0, 1), 
pmt05allz = c(0.1315678358078, 2.57276844978333, -0.86949759721756, -0.844452261924744, -0.48639452457428, 1.87834203243256, -0.988184869289398, -1.02298593521118, 0.570109307765961, 1.00886857509613, -0.972711682319641, -0.713021039962769, -0.70054304599762, 1.02071666717529, -0.571928858757019, -0.786627769470215, -0.628270447254181, 1.76193022727966, 0.75188934803009, 1.22556257247925, -0.205045282840729, -0.163282126188278, -0.149484217166901, -0.710245132446289, -0.631508588790894, -0.372817307710648, -0.0988877564668655, -0.28418955206871, -0.386095404624939, -1.8762229681015), 
arz = c(0.283046782016754, 0.283046782016754, -0.00598874036222696, -0.00598874036222696, 0.572082281112671, 0.283046782016754, 0.283046782016754, -0.295024245977402, -0.295024245977402, -0.584059774875641, 1.43918883800507, 0.861117839813232, -0.00598874036222696,-0.584059774875641, 0.283046782016754, -1.16213083267212, -0.584059774875641, -0.295024245977402, 1.1501532793045, -0.00598874036222696, -1.74020183086395,4.90761518478394, 1.43918883800507, -0.873095273971558, -0.295024245977402, 0.283046782016754, 1.1501532793045, 0.861117839813232, -0.295024245977402, 1.1501532793045), 
arlevel = structure(c(2L, 2L, 2L, 2L, 3L, 2L, 2L, 1L, 1L, 1L, 3L, 3L, 2L, 1L, 2L, 1L, 1L, 1L, 3L, 2L, 1L, 3L, 3L, 1L, 1L, 2L, 3L, 3L, 1L, 3L), .Label = c("short", "medium", "long"), class = "factor")), .Names = c("caretime3", "pmt05allz", "arz", "arlevel"), row.names = c(1566L, 1142L, 1637L, 574L, 507L, 1500L, 1393L, 1609L, 877L, 753L, 895L, 1608L, 1827L, 1342L, 1435L, 451L, 1606L, 368L, 848L, 1829L, 395L, 81L, 1021L, 87L, 1388L, 1765L, 491L, 29L, 5L, 1020L), class = "data.frame")

模型如下：

ctime.ml2 <-vglm(caretime3~ pmt05allz*arlevel, 
                 family = multinomial(refLevel = 1), data = CAG.sort)

结果如下：

Call:
vglm(formula = caretime3 ~ pmt05allz * arz, 
     family = multinomial(refLevel = 1), data = CAG.sort)

Pearson residuals:
                      Min      1Q  Median    3Q   Max
log(mu[,2]/mu[,1]) -1.771 -0.7532 -0.3770 1.089 2.177
log(mu[,3]/mu[,1]) -1.572 -0.8929 -0.3578 1.288 1.890

Coefficients:
                 Estimate Std. Error z value Pr(>|z|)  
(Intercept):1    0.24763    0.16787   1.475   0.1402  
(Intercept):2    0.12888    0.17101   0.754   0.4511  
pmt05allz:1     -0.28920    0.16643  -1.738   0.0823 .
pmt05allz:2     -0.13245    0.15691  -0.844   0.3986  
arz:1            0.40889    0.18664   2.191   0.0285 *
arz:2           -0.08447    0.19705  -0.429   0.6681  
pmt05allz:arz:1  0.56149    0.24221   2.318   0.0204 *
pmt05allz:arz:2  0.39024    0.22904   1.704   0.0884 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Number of linear predictors:  2 

Names of linear predictors: log(mu[,2]/mu[,1]), log(mu[,3]/mu[,1])

Dispersion Parameter for multinomial family:   1

Residual deviance: 499.5317 on 466 degrees of freedom

Log-likelihood: -249.7659 on 466 degrees of freedom

Number of iterations: 4

使用predict(m1, newdata)函数为我提供了两列。

log(mu[,2]/mu[,1]) log(mu[,3]/mu[,1])
1          1.837926621       1.6387672851
2          1.784309766       1.5924054498
3          1.730692911       1.5460436146
4          1.677076056       1.4996817793
5          1.623459202       1.4533199440

Q1。这两列是相对于参考水平（reflevel = 1）的两个水平中每一个的线性预测，对吗？

相反，使用predict(m1, newdata = newdata, type = "response")为我提供了三列（0,1和2）。

             0          1          2
1   0.08043554 0.50541645 0.41414801
2   0.08423871 0.50168094 0.41408035
3   0.08820341 0.49786976 0.41392683
4   0.09233480 0.49398103 0.41368418
5   0.09663804 0.49001289 0.41334907
...

Q2。这三列是什么？哪些匹配上面的比较（对比度2和3到1级）？

Q3。我是否也可以获得响应变量中预测值的标准误差（95％CI），然后我可以绘制？如果是这样，怎么样？

摘要：通过多项逻辑回归，我尝试从Stata生成类似的内容：

实际上看起来更像是这样：

基本上，我希望预测响应变量（caretime3）由x2（pmt05allz）范围内的预测变量x（arz）之一，但最终用于可视化分组通过arz（arlevel）的三分位数。

用95％CI绘制VGLM多项Logistic回归

0 个答案: