英文:
`get_all_outcomes()`: error when using cuminc
问题
我正在尝试在医学数据库中进行竞争风险分析。
安装所需的包后，我运行以下代码：
cuminc(Surv(HAPLO$SEGUIMIENTO, HAPLO$SEGFINAL) ~ 1, data = HAPLO)
但我一直收到以下错误：
Error in `get_all_outcomes()`:! The following outcomes were not found in `data`: 'HAPLO'.Run `rlang::last_trace()` to see where the error occurred.
如果我运行 rlang::last_trace()
我得到：
Backtrace:▆1. ├─tidycmprsk::cuminc(...)2. └─tidycmprsk:::cuminc.formula(...)3. ├─hardhat::mold(formula, data, blueprint = hardhat::default_formula_blueprint(intercept = TRUE))4. └─hardhat:::mold.formula(formula, data, blueprint = hardhat::default_formula_blueprint(intercept = TRUE))5. ├─hardhat::run_mold(blueprint, data = data)6. └─hardhat:::run_mold.default_formula_blueprint(blueprint, data = data)7. └─hardhat:::mold_formula_default_process(...)8. └─hardhat:::mold_formula_default_process_outcomes(...)9. └─hardhat:::get_all_outcomes(formula, data)
我不知道该怎么办。
提前感谢。
RAFA
英文:
I am trying to do a competing risk analysis in a medical database.
After installing packages I read were needed I run the following code:
cuminc(Surv(HAPLO$SEGUIMIENTO, HAPLO$SEGFINAL) ~ 1, data = HAPLO
And I keep getting this error
Error in get_all_outcomes():
! The following outcomes were not found in data: 'HAPLO'.
Run rlang::last_trace() to see where the error occurred.
If i run rlang::last_trace()
I get
Backtrace:
▆
- ├─tidycmprsk::cuminc(...)
- └─tidycmprsk:::cuminc.formula(...)
- ├─hardhat::mold(formula, data, blueprint = hardhat::default_formula_blueprint(intercept = TRUE))
- └─hardhat:::mold.formula(formula, data, blueprint = hardhat::default_formula_blueprint(intercept = TRUE))
-
├─hardhat::run_mold(blueprint, data = data)
-
└─hardhat:::run_mold.default_formula_blueprint(blueprint, data = data)
-
└─hardhat:::mold_formula_default_process(...)
-
└─hardhat:::mold_formula_default_process_outcomes(...)
-
└─hardhat:::get_all_outcomes(formula, data)
I dont know what to do.
Thanks in advance.
RAFA
答案1
得分: 1
Using the formula interface to a function, you supply just the column names, for example, SEGUIMIENTO, rather than HAPLO$SEGUIMIENTO.
So for data frame HAPLO, this should work:
library(tidycmprsk)cuminc(Surv(SEGUIMIENTO, SEGFINAL) ~ 1, data = HAPLO)
英文:
Using the formula interface to a function, you supply just the column names e.g. SEGUIMIENTO, rather than HAPLO$SEGUIMIENTO.
So for data frame HAPLO this should work:
library(tidycmprsk)cuminc(Surv(SEGUIMIENTO, SEGFINAL) ~ 1, data = HAPLO)
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