删除在选定列上满足多个条件的行

Question

我有一个大约1000列的数据框架。我对14个满意度评价变量感兴趣。

我需要删除任何14个评级变量中的任何一个包含“项目跳过”或NA的行。

是否有一种方法可以删除所有的行，其中NA或“项目跳过”出现在我感兴趣的满意度等级变量向量中，目前在向量'cols‘中。在下面的例子中，“cols”包含服务、效率和风味，但不包括经验和质量。

cols = c("Service","Efficiency","Flavour")
dat<-data.frame(Number = 1:6, University = c("A","B","C","D","E","F"), 
                Service=c("Satisfied","Item skipped",NA, "Not satisfied", "Neither","Item skipped" ), 
                Efficiency =c("Neither", "Neither", "Item skipped","Satisfied", NA, NA),
                Flavour =c("Satisfied", NA, "Item skipped",
                                     "Neither", NA, NA), Quality =c("Not satisfied", "Neither", NA,"Satisfied", NA, NA),
                Experience =c("Satisfied", NA, NA,
                                     "Not satisfied", NA, NA),Age =rep(c(18:19), times =3))

Answer 1

在基本R中，我们可以使用rowSums删除cols中存在“项跳过”或NA的行

cols = c("Service", "Efficiency", "Flavour")

dat[rowSums(dat[cols] == "Item skipped" | is.na(dat[cols])) == 0, ]

#  Number University       Service Efficiency   Flavour       Quality    Experience Age
#1      1          A     Satisfied    Neither Satisfied Not satisfied     Satisfied  18
#4      4          D Not satisfied  Satisfied   Neither     Satisfied Not satisfied  19

@amrrs建议的使用apply的另一种方法

dat[!apply(dat[cols], 1, function(x) any(x == 'Item skipped' | is.na(x))), ]

Answer 2

编辑：：使用我们可以使用的更新数据(假设NA总是与"Item_Skipped“一起出现-情况似乎是这样)：

 dat %>% 

     filter(!is.na(Experience))  

        Number University       Service Efficiency   Flavour       Quality    Experience Age
    1       1          A     Satisfied    Neither Satisfied Not satisfied     Satisfied  18
    2       4          D Not satisfied  Satisfied   Neither     Satisfied Not satisfied  19

原件：：

我们可以使用(数据见下文注)：

dat %>% 
  filter_at(vars(contains("rating")),all_vars(.!="Item Skipped"))

OR::

dat %>% 
  filter_all(all_vars(.!="Item Skipped"))

输出：

 Number University Service_rating Efficiency_rating Flavour_rating Age
1      1          A      Satisfied           Neither      Satisfied  18
2      4          D  Not satisfied         Satisfied        Neither  19

注意事项

dat<-data.frame(Number = 1:6, University = c("A","B","C","D","E","F"), 
                Service_rating=c("Satisfied","Item skipped",NA, "Not satisfied", "Neither","Item skipped" ), 
                Efficiency_rating =c("Neither", "Neither", "Item skipped","Satisfied", NA, NA),
                Flavour_rating =c("Satisfied", NA, "Item skipped",
                                  "Neither", NA, NA), Age =rep(c(18:19), times =3))