Question

使用“级别”关于跨数据框列的折叠因子的快速问题。下面是基本数据帧的代码。

var keepIt = true;
var sample = [1,2,3,4].filter(function(){ 
              return keepIt; //It will be true always, 
                             //so all the items will be preserved. 
                             //If it is false, then all the elements will be ignored.
             });

我可以使用以下代码通过组合“满意”和“中立”轻松折叠级别。

Service<-c("AS","AS","PS","PS","RS","RS","ES","ES")
 Year<-c("2015","2016","2015","2016","2015","2016","2015","2016")
 Q1<-c("Dissatisfied","Satisfied","Satisfied","Neutral","Dissatisfied","Dissatisfied","Satisfied","Neutral")
Q2<-c("Dissatisfied","Neutral","Satisfied","Dissatisfied","Dissatisfied","Neutral","Satisfied","Satisfied")
Example<-data.frame(Service,Year,Q1,Q2)

但是，对于列Q1和Q2，如何在类似的列上同时执行此操作？我知道使用lapply或apply有一种简单的方法，但我似乎无法使它工作。

Answer 1

我认为你的手术不可行。我以为我记得这种崩溃因子水平被严重弃用以至于会引发错误，但由于它适用于一列，它应该（并且确实）适用于多列。

 Example[3:4] <- lapply(Example[3:4], 'levels<-', c("Dissatisfied","Satisfied","Satisfied") )
 Example
  Service Year           Q1           Q2
1      AS 2015 Dissatisfied Dissatisfied
2      AS 2016    Satisfied    Satisfied
3      PS 2015    Satisfied    Satisfied
4      PS 2016    Satisfied Dissatisfied
5      RS 2015 Dissatisfied Dissatisfied
6      RS 2016 Dissatisfied    Satisfied
7      ES 2015    Satisfied    Satisfied
8      ES 2016    Satisfied    Satisfied

Answer 2

我们可以使用lapply。创建要进行子集化的列索引。这里是以＆＃34; Q＆＃34;开头的列名。后跟数字（^Q\\d+）。使用grep获取索引，然后在这些列上使用lapply，并使用recode中的library(car)更改levels。

nm1 <- grep("^Q\\d+", names(Example))
Example[nm1] <-  lapply(Example[nm1], recode, 
         'c("Satisified", "Neutral")= "Satisfied"')

因为我们只是改变了＃34;中立＆＃34; ＆＃34;满意＆＃34;，而'＃34;满意＆＃34;应保持原样，我们也可以这样做

Example[nm1] <- lapply(Example[nm1], recode, '"Neutral"= "Satisfied"')

使用“levels”语句跨数据框的多个列折叠因子

2 个答案: