summary.coxph.RdProduces a summary of a fitted coxph model
# S3 method for class 'coxph'
summary(object, conf.int=0.95, scale=1,...)the result of a coxph fit
level for computation of the confidence intervals. If set to FALSE no confidence intervals are printed
vector of scale factors for the coefficients, defaults to 1. The printed coefficients, se, and confidence intervals will be associated with one scale unit.
for future methods
An object of class summary.coxph, with components:
number of observations and number of events, respectively, in the fit
the log partial likelihood at the initial and final values
a matrix with one row for each coefficient, and columns containing the coefficient, the hazard ratio exp(coef), standard error, Wald statistic, and P value.
a matrix with one row for each coefficient, containing the confidence limits for exp(coef)
the overall likelihood ratio, score, and Wald test statistics for the model
the concordance statistic and its standard error
whether an asymptotic or robust variance was used
an approximate R^2 based on Nagelkerke (Biometrika 1991).
a message, if the underlying coxph call failed
a copy of the call
information on missing values
The pseudo r-squared of Nagelkerke is attractive because it is simple,
but further work has shown that it has poor properties and it is now
deprecated. The value is no longer printed by default, and will
eventually be removed from the object.
The royston function now includes it along with several other
measures of association.
fit <- coxph(Surv(time, status) ~ age + sex, lung)
summary(fit)
#> Call:
#> coxph(formula = Surv(time, status) ~ age + sex, data = lung)
#>
#> n= 228, number of events= 165
#>
#> coef exp(coef) se(coef) z Pr(>|z|)
#> age 0.017045 1.017191 0.009223 1.848 0.06459 .
#> sex -0.513219 0.598566 0.167458 -3.065 0.00218 **
#> ---
#> Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#>
#> exp(coef) exp(-coef) lower .95 upper .95
#> age 1.0172 0.9831 0.9990 1.0357
#> sex 0.5986 1.6707 0.4311 0.8311
#>
#> Concordance= 0.603 (se = 0.025 )
#> Likelihood ratio test= 14.12 on 2 df, p=9e-04
#> Wald test = 13.47 on 2 df, p=0.001
#> Score (logrank) test = 13.72 on 2 df, p=0.001
#>