Question

我目前在差异模型方面存在差异，并希望以有效的方式衡量同一模型的不同指标。

例如，我有一个数据框，其中包含Miles Driven，Work Hoursed，State，Group，Time的列。

目前，我可以使用代码复制并粘贴每个指标的模型：

# Create DID models
model_miles <- lm(df$miles_driven ~ df$Group 
                    + df$Time 
                    + df$Group * df$Time, data = df)
model_hours <- lm(df$hours_worked ~ df$Group 
       + df$Time 
       + df$Group * df$Time, data = df)

# Select results using clustered standard errors. The purpose of this is to 
# avoid making distributional assumptions about the errors in the models. 
results_miles <- clubSandwich::coef_test(model_miles, 
                                        vcov = "CR2", 
                                        cluster = df$state, 
                                        test = "Satterthwaite")
results_hours <- clubSandwich::coef_test(model_hours, 
                           vcov = "CR2", 
                           cluster = df$state, 
                           test = "Satterthwaite")

 results <- data.table::rbindlist(list(results_miles, results_hours))
View(results)

我想以某种方式创建我的度量标准名称列表，并使用用户定义的函数循环遍历此列表，以便使此过程更快，更自动化，但我无法使其工作正确地：

#list of metrics
metrics <- c("miles_driven", "hours_worked")

udf <- function(metric, dataframe){
 # Create DID model
 model <- lm(dataframe$metric ~ df$Group 
            + dataframe$Time 
            + dataframe$Group * df$Time, data = dataframe)

 # Select results using clustered standard errors. The purpose of this 
 is to 
 # avoid making distributional assumptions about the errors in the 
 models. 
 results_miles <- clubSandwich::coef_test(model_miles, 
                                   vcov = "CR2", 
                                   cluster = dataframe$state, 
                                   test = "Satterthwaite")[4,]
 View(results)
}

lapply(metrics, udf)

任何见解都将受到赞赏。谢谢！

Answer 1

如果您使用[[代替$，则会有效。

例如：

dat <- data.frame(A=1:1, B=1:2)
x <- "A"
lm(dat[[x]] ~ dat$B)

您可以使用get和assign在循环中执行此类更高级的操作。

Answer 2

在这些情况下，我始终lm在lm中data一个公式。以下示例。如果您提供dataframe$varname library(glue) dostuff <- function(metric, dataframe){ model.form <- '{metric} ~ b + c + c*d' lm(as.formula(glue(model.form)), data = dataframe) } df <- data.frame(a = runif(10), b = runif(10), c = runif(10), d = runif(10)) dostuff('a', df)参数，则可以只提供变量名称，而不是显式传递glue('{metric} ~ b + c + c*d') paste(metric, '~ b + c + c*d')。

a = [1,2,3] 
# for loop begins
a[0] = a[0]    # a is now [1,2,3]
print(a[0])    # prints 1
a[0] = a[1]    # a is now [2,2,3]
print(a[0])    # prints 2
a[0] = a[2]    # a is now [3,2,3]
print(a[0])    # prints 3
# for loop ends
print(a)       # prints [3,2,3]

如果你想坚持基础R，下面的两个对象是相等的

    entityManager.getTransaction().begin();

    CriteriaQuery<Object> criteriaQuery = criteriaBuilder.createQuery();
    Root<Profile> fromRoot = criteriaQuery.from(Profile.class);
    criteriaQuery.select(fromRoot);

    criteriaQuery.where(criteriaBuilder.equal(fromRoot.get("userName"), username),
                        criteriaBuilder.equal(fromRoot.get("password"), password));

    List<Object> resultList = entityManager.createQuery(criteriaQuery).getResultList();
    Profile dbProfile = null;
    if (resultList.isEmpty()) {
        // Handle Error
    } else {
        dbProfile = (Profile) resultList.get(0);
    }

    entityManager.getTransaction().commit();

R - 循环用户定义模型上的度量列表

2 个答案: