使用 lapply() 函数执行逻辑回归时出错（选择了未定义的列）答案

【问题标题】：Error while using lapply() function to perform logistic regression (undefined columns selected)使用 lapply() 函数执行逻辑回归时出错（选择了未定义的列）
【发布时间】：2021-03-22 23:33:41
【问题描述】：

所以这里取到数据https://archive.ics.uci.edu/ml/machine-learning-databases/00497/divorce.rar

当我运行代码执行逻辑回归时，它显示错误。但它可以在其他 R 程序上完美运行。我有什么遗漏的吗？

set.seed(123)

divorce = read.csv("C://Users//User//Documents//Y2S3//Predictive Modelling//divorce//divorce.csv")

dim（离婚）

结果：[1] 170 1

总结（离婚）

结果： Atr1.Atr2.Atr3.Atr4.Atr5.Atr6.Atr7.Atr8.Atr9.Atr10.Atr11.Atr12.Atr13.Atr14.Atr15.Atr16.Atr17.Atr18.Atr19.Atr20.Atr21.Atr22.Atr23.Atr24.Atr25。 Atr26.Atr27.Atr28.Atr29.Atr30.Atr31.Atr32.Atr33.Atr34.Atr35.Atr36.Atr37.Atr38.Atr39.Atr40.Atr41.Atr42.Atr43.Atr44.Atr45.Atr46.Atr47.Atr48.Atr49.Atr50。 Atr51.Atr52.Atr53.Atr54.Class

长度：170

类：字符

模式：字符

colnames（离婚）

结果： [1]“Atr1.Atr2.Atr3.Atr4.Atr5.Atr6.Atr7.Atr8.Atr9.Atr10.Atr11.Atr12.Atr13.Atr14.Atr15.Atr16.Atr17.Atr18.Atr19.Atr20.Atr21.Atr22.Atr23。 Atr24.Atr25.Atr26.Atr27.Atr28.Atr29.Atr30.Atr31.Atr32.Atr33.Atr34.Atr35.Atr36.Atr37.Atr38.Atr39.Atr40.Atr41.Atr42.Atr43.Atr44.Atr45.Atr46.Atr47.Atr48。 Atr49.Atr50.Atr51.Atr52.Atr53.Atr54.Class"

申请（离婚，班级）

col_fac = c("Atr1","Atr2","Atr3","Atr4","Atr5","Atr6","Atr7","Atr8","Atr9","Atr10", +"Atr11","Atr12","Atr13","Atr14","Atr15","Atr16","Atr17","Atr18","Atr19","Atr20", +"Atr21","Atr22","Atr23","Atr24","Atr25","Atr26","Atr27","Atr28","Atr29","Atr30", +"Atr31","Atr32","Atr33","Atr34","Atr35","Atr36","Atr37","Atr38","Atr39","Atr40", +"Atr41","Atr42","Atr43","Atr44","Atr45","Atr46","Atr47","Atr48","Atr49","Atr50", +"Atr51","Atr52","Atr53","Atr54","类")

离婚[col_fac] = lapply(离婚[col_fac],因素)

结果：[.data.frame(divorce, col_fac) 中的错误：选择了未定义的列）

【问题讨论】：

欢迎来到 SO！如果您提供最小可重复的示例，您将最大限度地获得有用的答案。 This post 可能会有所帮助。具体来说，请不要提供您的代码或数据作为屏幕截图。请使用dput() 获取数据并粘贴您的代码。

标签： r dataframe

【解决方案1】：

唯一的问题是您读取的文件由“;”分隔并不是 ”，”。 sep = ";" 会解决这个问题。

# downloaded and extracted from https://archive.ics.uci.edu/ml/machine-learning-databases/00497/divorce.rar
divorce <- read.csv("./divorce.csv", sep = ";")
dim(divorce)
summary(divorce)
colnames(divorce)
sapply(divorce,class)
col_fac = c("Atr1","Atr2","Atr3","Atr4","Atr5","Atr6","Atr7","Atr8","Atr9","Atr10",
            "Atr11","Atr12","Atr13","Atr14","Atr15","Atr16","Atr17","Atr18","Atr19","Atr20",
            "Atr21","Atr22","Atr23","Atr24","Atr25","Atr26","Atr27","Atr28","Atr29","Atr30", 
            "Atr31","Atr32","Atr33","Atr34","Atr35","Atr36","Atr37","Atr38","Atr39","Atr40", 
            "Atr41","Atr42","Atr43","Atr44","Atr45","Atr46","Atr47","Atr48","Atr49","Atr50", 
            "Atr51","Atr52","Atr53","Atr54","Class")

divorce[col_fac] = lapply(divorce[col_fac],factor)

使用 dplyr 不易出错的版本

通过应用函数as.factoracross 这些变量where 函数is.numeric 返回TRUE，以下将mutate 您的数据集。请注意，across 和 where 中传递的函数没有通常的括号。

library(dplyr)
divorce <- read.csv("./divorce.csv", sep = ";") %>% 
  mutate(across(where(is.numeric), as.factor))
glimpse(divorce)

有关变异的详细信息，请在 R 控制台中输入 ?across。

【讨论】：