问题

I want to do t-test（or chi^2 test） to estimate the difference of variables between grou=0 and grou=1. All variables in the dataset are imputed by MICE. Variables include AGE,SCORE,GENDER, HEART, etc.; AGE and SCORE are continuous variables, and GENDER and HEART are categorical variables.

If the t-test is done for only one variable at a time, I know the code is：

library(MICE)
data_im <- mice(data, m=5, seed=6666)
summary(pool(with(data_im, glm(AGE~grou))))

The output p-value is also the p-value of the t-test.

However, there are too many variables I need to evaluate, thus I would like to write a for loop or create a function to output the summary test results of multiple variables at once.

I have tried to write:

vars <- c("AGE","SCORE","GENDER","HEART")
afterMICE <- c()
for(i in 1:4){
  pool_fitMICE <- pool(with(data_im, glm(substitute(y ~ grou, list(y=as.name(vars[i])))))
}

**Error in eval(predvars, data, env) : object 'AGE' not found**

afterMICE <- rbind(afterMICE, c(vars[i], coef(summary(pool_fitMICE))[2,c(1,2,4)])

I know the reason for the error is that data_im is not a regular dataframe structure.

How to modify the code to achieve batch output of summary results from different variables？

#------------------------------------------------------------------------------

Edit:
mice() is a function to do multivariate imputation for missing data. https://www.rdocumentation.org/packages/mice/versions/3.15.0/topics/mice

Take the data("nhanes2") for example and we can see the structure of data_im.

library(MICE)
data("nhanes2")
vars <- c("bmi", "chl", "age", "hyp")
catvars <- c("age", "hyp")
data_im <- mice(nhanes2, m=5, seed=6666)
pool.fits <- pool(with(data_im, glm(age~hyp)))

But we need to batch pool the results of pool(with(data_im, glm(vars~hyp))) (Variables in vars take turns being a dependent variable in glm(), with hyp as the independent variable.)

library(MICE)
data_im<-mice(data, m=5,seed=6666)
summary(pool(with(data_im,glm(AGE~grou))))

vars <- c("AGE","SCORE","GENDER","HEART")
afterMICE <- c()
for(i in 1:4){
  pool_fitMICE <- pool(with(data_im,glm(substitute(y ~ grou,list(y=as.name(vars[i]))))))
}

**Error in eval(predvars, data, env) : object 'AGE' not found**

afterMICE <- rbind(afterMICE,C(vars[i],coef(summary(pool_fitMICE))[2,c(1,2,4)]))

library(MICE)
data("nhanes2")
vars=c("bmi","chl","age","hyp")
catvars=c("age","hyp")
data_im=mice(nhanes2,m=5,seed=6666)
pool.fits <- pool(with(data_im, glm(age~hyp)))

答案1

得分: 0

用于模型动态生成公式的，您可以查看`reformulate`。`reformulate`可以帮助您避免`paste`/`substitute`/`as.name`的麻烦。我认为我们也不需要使用`with()`。这可能会导致一些环境混淆。

vars <- c("AGE","SCORE","GENDER","HEART")

list_of_formulas <- lapply(vars, (x) reformulate(termlabels = 'grou', response = x))

查看创建的公式列表：

[[1]]
AGE ~ grou
<environment: 0x557191bff8b8>

[[2]]
SCORE ~ grou
<environment: 0x557191c07b58>

[[3]]
GENDER ~ grou
<environment: 0x557191cb30f8>

[[4]]
HEART ~ grou
<environment: 0x557191ccad40>

然后在您的for循环中使用此列表。

pool_fitMICE <- list()
for(i in 1:4){
pool_fitMICE[[i]] <- pool(glm(formula = list_of_formulas[[i]], data = data_im))
}

如果我们使用一个循环函数，还可以避免初始化输出列表或使用for循环的需要：

lapply(list_of_formulas, (x) pool(glm(formula = x, data = data_im))

<details>
<summary>英文:</summary>

for dynamic generation of formulas for models, you may look into `reformulate`. `reformulate` may save you from the `paste`/`substitute`/`as.name` hell. I believe we do not need `with()`too. It may generate some environment confusion.

vars <- c("AGE","SCORE","GENDER","HEART")

list_of_formulas <- lapply(vars, (x) reformulate(termlabels = 'grou', response = x))

see the created list of formulas:

[[1]]
AGE ~ grou
<environment: 0x557191bff8b8>

[[2]]
SCORE ~ grou
<environment: 0x557191c07b58>

[[3]]
GENDER ~ grou
<environment: 0x557191cb30f8>

[[4]]
HEART ~ grou
<environment: 0x557191ccad40>

Then use this list in your for loop.

pool_fitMICE <- list()
for(i in 1:4){
pool_fitMICE[[i]] <- pool(glm(formula = list_of_formulas[[i]], data = data_im))
}

We can also obviate the need to initiate the output list or using a for loop if we use a loop-function:

lapply(list_of_formulas, (x) pool(glm(formula = x, data = data_im))

答案2

得分: 0

不需要在这种情况下使用eval和substitute。您可以正常创建公式（例如，使用as.formula），当在with内部评估公式时，它将自动找到缺失数据中的变量。

library(mice)
set.seed(123)
vars <- c("age", "bmi", "chl")
imps <- mice(nhanes, printFlag = FALSE)
mods <- list()
for(i in seq_along(vars)) {
  mods[[i]] <- pool(with(imps, 
                     glm(as.formula(paste0(vars[[i]], "~ hyp"))))   
}
names(mods) <- vars
lapply(mods, summary)

编辑：将输出放入数据框中。

out <- data.frame(var = vars,
                  coef_int = 0,
                  coef_hyp = 0)
for(i in seq_along(vars)) {
  out[i, 2:3] <- mods[[i]]$pooled$estimate
}
out

这是您提供的代码的翻译部分。

library(mice)
set.seed(123)
vars <- c("age", "bmi", "chl")
imps <- mice(nhanes, printFlag = FALSE)
mods <- list()
for(i in seq_along(vars)) {
  mods[[i]] <- pool(with(imps, 
                         glm(as.formula(paste0(vars[[i]], "~ hyp")))))
}
names(mods) <- vars
lapply(mods, summary)
#> $age
#>          term  estimate std.error statistic       df   p.value
#> 1 (Intercept) 0.5216374 0.4873196  1.070422 17.71793 0.2987954
#> 2         hyp 1.0163241 0.3744249  2.714360 19.13894 0.0136971
#> 
#> $bmi
#>          term   estimate std.error   statistic       df      p.value
#> 1 (Intercept) 27.0037636  2.882558  9.36798555 16.45801 5.297937e-08
#> 2         hyp -0.1502389  2.191507 -0.06855506 18.35948 9.460849e-01
#> 
#> $chl
#>          term  estimate std.error statistic       df      p.value
#> 1 (Intercept) 157.88698  24.65957  6.402665 20.91373 2.445428e-06
#> 2         hyp  29.06627  19.62684  1.480945 19.90150 1.542804e-01

out <- data.frame(var = vars,
                  coef_int = 0,
                  coef_hyp = 0)
for(i in seq_along(vars)) {
  out[i, 2:3] <- mods[[i]]$pooled$estimate
}
out
#>   var    coef_int   coef_hyp
#> 1 age   0.5216374  1.0163241
#> 2 bmi  27.0037636 -0.1502389
#> 3 chl 157.8869758 29.0662740