Question

我正在尝试使用knitr / xtable从报表中的coxph（）对象生成表。当我在模型中不包含pspline术语时，一切都按预期工作。在一个块中：

<<results = 'asis'>>=
require(survival, quietly = T)
require(xtable, quietly = T)
data(cancer,  package = "survival")
fit0 <- coxph(Surv(time, status) ~ meal.cal + ph.ecog + age, cancer)
# construct data frame for tables - no spline
fit0table <- data.frame(Variable = c("Calories Consumed", "ECOG Performance Score","Age"), RiskRatio = summary(fit0)$conf.int[,1], Lower = summary(fit0)$conf.int[,3],   Upper = summary(fit0)$conf.int[,4], Pval = summary(fit0)$coeff[,5])
# print latex table
print(xtable(fit0table, digits = 3), include.rownames = F)
@

但是当我包含一个惩罚的样条曲线术语时，summary（）对象的结构会发生变化，$conf.int和$coeff个插槽将不再可用。

> fit1 <- coxph(Surv(time, status) ~ meal.cal + ph.ecog + pspline(age, 3), cancer)
> str(summary(fit0))
List of 14
$ call        : language coxph(formula = Surv(time, status) ~ meal.cal + ph.ecog + age, data = cancer)
$ fail        : NULL 
$ na.action   :Class 'omit'  Named int [1:48] 3 5 12 13 14 16 23 25 33 44 ...
  .. ..- attr(*, "names")= chr [1:48] "3" "5" "12" "13" ...
$ n           : int 180
$ loglik      : num [1:2] -574 -567
$ nevent      : num 133
$ coefficients: num [1:3, 1:5] 3.84e-05 4.00e-01 1.10e-02 1.00 1.49 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : chr [1:3] "meal.cal" "ph.ecog" "age"
  .. ..$ : chr [1:5] "coef" "exp(coef)" "se(coef)" "z" ...
$ conf.int    : num [1:3, 1:4] 1 1.491 1.011 1 0.671 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : chr [1:3] "meal.cal" "ph.ecog" "age"
.. ..$ : chr [1:4] "exp(coef)" "exp(-coef)" "lower .95" "upper .95"
$ logtest     : Named num [1:3] 13.2142 3 0.0042
..- attr(*, "names")= chr [1:3] "test" "df" "pvalue"
$ sctest      : Named num [1:3] 13.46468 3 0.00373
..- attr(*, "names")= chr [1:3] "test" "df" "pvalue"
$ rsq         : Named num [1:2] 0.0708 0.9983
..- attr(*, "names")= chr [1:2] "rsq" "maxrsq"
$ waldtest    : Named num [1:3] 13.28 3 0.00407  
..- attr(*, "names")= chr [1:3] "test" "df" "pvalue"
$ used.robust : logi FALSE
$ concordance : Named num [1:2] 0.6061 0.0291
..- attr(*, "names")= chr [1:2] "concordance.concordant" "se.std(c-d)"
- attr(*, "class")= chr "summary.coxph"


> str(summary(fit1))
Call: 
coxph(formula = Surv(time, status) ~ meal.cal + ph.ecog + pspline(age, 
3), data = cancer)

n= 180, number of events= 133 
(48 observations deleted due to missingness)

                        coef     se(coef) se2      Chisq DF   p     
meal.cal                3.65e-05 0.000228 0.000228 0.03  1.00 0.8700
ph.ecog                 3.98e-01 0.131938 0.131738 9.10  1.00 0.0026
pspline(age, 3), linear 1.07e-02 0.010694 0.010694 1.00  1.00 0.3200
pspline(age, 3), nonlin                            2.90  2.07 0.2500

          exp(coef) exp(-coef) lower .95 upper .95
meal.cal       1.00     1.0000     1.000      1.00
ph.ecog        1.49     0.6717     1.150      1.93
ps(age)3       1.75     0.5717     0.473      6.47
ps(age)4       3.03     0.3302     0.365     25.14
ps(age)5       4.49     0.2228     0.395     50.96
ps(age)6       4.65     0.2150     0.405     53.43
ps(age)7       3.96     0.2526     0.363     43.12
ps(age)8       3.84     0.2604     0.360     41.01
ps(age)9       4.44     0.2250     0.413     47.84
ps(age)10      5.39     0.1855     0.486     59.82
ps(age)11      7.94     0.1260     0.599    105.23
ps(age)12     12.25     0.0816     0.537    279.91

Iterations: 4 outer, 12 Newton-Raphson
     Theta= 0.836 
Degrees of freedom for terms= 1.0 1.0 3.1 
Concordance= 0.616  (se = 0.029 )
Rsquare= 0.092   (max possible= 0.998 )
Likelihood ratio test= 17.5  on 5.06 df,   p=0.00389
Wald test            = 15.9  on 5.06 df,   p=0.0073
 NULL


> coefficients(fit1) # doesn't give p-values
meal.cal      ph.ecog     ps(age)3     ps(age)4     ps(age)5     ps(age)6         ps(age)7     ps(age)8 
3.647054e-05 3.980039e-01 5.590767e-01 1.108052e+00 1.501557e+00 1.537249e+00     1.375833e+00 1.345564e+00 
ps(age)9    ps(age)10    ps(age)11    ps(age)12 
1.491454e+00 1.684622e+00 2.071641e+00 2.505932e+00 

> confint(fit1) # getting closer
                  2.5 %       97.5 %
meal.cal  -0.0004104346 0.0004833757
ph.ecog    0.1394097867 0.6565980826
ps(age)3  -0.7493022459 1.8674555679
ps(age)4  -1.0084545140 3.2245588375
ps(age)5  -0.9278798219 3.9309933396
ps(age)6  -0.9038092211 3.9783071434
ps(age)7  -1.0122388810 3.7639051908
ps(age)8  -1.0226368192 3.7137644105
ps(age)9  -0.8849251510 3.8678337954
ps(age)10 -0.7221442743 4.0913878825
ps(age)11 -0.5129062130 4.6561876883
ps(age)12 -0.6226068259 5.6344701023

Answer 1

我认为没有一个数字（甚至两个或三个）对于描述适合包含在表中的惩罚样条函数拟合的置信区间是有意义的，我当然这样做（编辑：意为说不）认为confint产生的长间隔列表是有意义的。（没有confint.coxph.penal功能。）当7年前在R-help上提出了一个类似的问题（虽然请求图形显示）时，Terry Therneau发布了这段代码，用于显示他认为有意义的内容，我已修改为适合您的姓名并显示“年龄”的适合度和CI：

fit1 <- coxph(Surv(time, status) ~ meal.cal + ph.ecog + pspline(age, 3), na.omit(cancer) )
temp <- predict(fit1, type='terms', se=TRUE) 
matplot(na.omit(cancer)$age, exp(cbind( temp$fit[, 3], 
                                  temp$fit[,3] - 2* temp$se.fit[,3], 
                                  temp$fit[,3] + 2* temp$se.fit[,3])), 
        log='y', xlab="Age", ylab="Estimated Relative Risk", col=c('red',"blue","blue") )

enter image description here

BTW：summary(fit0)除了invisible()之外没有任何内容返回，因此您从str(summary(fit1))看到的只是cat来电后发送到控制台的输出孤独的小NULL。如果您怀疑我的用法，只需使用getAnywhere(summary.coxph.penal)审核代码。

将pspline项添加到coxph模型更改R中的汇总表

1 个答案: