2017-06-27 30 views
1

我想要一個函數,其中的參數可以帶一個變量名(它是數據集的一部分,但不作爲對象存儲在環境中)並將該變量名稱插入到模型公式中。R - 如何在函數中使用列名作爲參數並插入到模型公式中

例如:

# Some data with a couple of variables 
my_df <- data.frame(y = rbinom(10, 1,0.5), var1 = runif(10), var2 = runif(10)) 

# A function that fits a model using predictor specified in the arguments 
my_fun <- function(var_name, df){ 
    glm(y ~ var_name, data = df, family = "binomial") 
} 

當我嘗試使用我收到以下錯誤消息

my_fun(var1, my_df) 
Error in eval(expr, envir, enclos) : object 'var1' not found 

# What I want the function to do 
glm(y ~ var1, data = my_df, family = "binomial") 

是否有一種方式來獲得這種功能工作的功能?

+0

裏面my_fun,您可以用'GLM(重新制定(VAR_NAME, 「Y」),數據= DF, family =「binomial」)'然後使用'my_fun(「var1」,df)'來運行迴歸。 – lmo

回答

4

您可以分析不帶引號var_namesubstitute

my_fun <- function(var_name, df){ 
    glm.formula <- substitute(y ~ x, list(x = substitute(var_name))) 
    glm(glm.formula, data = df, family = "binomial") 
} 

一個例子:

my_fun(var1, my_df) 

# Call: glm(formula = glm.formula, family = "binomial", data = df) 
# 
# Coefficients: 
# (Intercept)   var1 
#  -1.226  3.108 
# 
# Degrees of Freedom: 9 Total (i.e. Null); 8 Residual 
# Null Deviance:  13.46 
# Residual Deviance: 11.35 AIC: 15.35 

glm(y ~ var1, data = my_df, family = "binomial") 

# Call: glm(formula = y ~ var1, family = "binomial", data = my_df) 
# 
# Coefficients: 
# (Intercept)   var1 
#  -1.226  3.108 
# 
# Degrees of Freedom: 9 Total (i.e. Null); 8 Residual 
# Null Deviance:  13.46 
# Residual Deviance: 11.35 AIC: 15.35 
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