如何将函数应用于 R 中数据框中的特定列集以替换 NA

How to Apply functions to specific set of columns in data frame in R to replace NAs

我有一个数据集,我想在其中以不同方式替换不同列中的 NA。以下是虚拟数据集和复制它的代码。

test <- data.frame(ID = c(1:5),
               FirstName = c(NA,"Sid",NA,"Harsh","CJ"),
               LastName = c("Snow",NA,"Lapata","Khan",NA),
               BillNum = c(6:10),
               Phone = c(1213,3123,3123,NA,NA),
               Married = c("Yes","Yes",NA,"NO","Yes"),
               ZIP = c(1111,2222,333,444,555),
               Gender = c("M",NA,"F",NA,"M"),
               Address = c("A","B",NA,"C","D"))
> test
  ID FirstName LastName BillNum Phone Married  ZIP Gender Address
1  1      <NA>     Snow       6  1213     Yes 1111      M       A
2  2       Sid     <NA>       7  3123     Yes 2222   <NA>       B
3  3      <NA>   Lapata       8  3123    <NA>  333      F    <NA>
4  4     Harsh     Khan       9    NA      NO  444   <NA>       C
5  5        CJ     <NA>      10    NA     Yes  555      M       D

在某些专栏中,我想指出一个值是否由客户提供,而不保留提供的值,如下所示。

Availability_Indicator <- function(x){
  x <- ifelse(is.na(x),"NotAvialable","Available")
  return(x)
}
test$FirstName <- Availability_Indicator(test$FirstName)
test$LastName <- Availability_Indicator(test$LastName)
test$Phone <- Availability_Indicator(test$Phone)
test$Address <- Availability_Indicator(test$Address)

我得到以下数据

> test
ID    FirstName     LastName BillNum        Phone Married  ZIP Gender
 1 NotAvialable    Available       6    Available     Yes 1111      M
 2    Available NotAvialable       7    Available     Yes 2222   <NA> 
 3 NotAvialable    Available       8    Available    <NA>  333      F
 4    Available    Available       9 NotAvialable      NO  444   <NA>
 5    Available NotAvialable      10 NotAvialable     Yes  555      M

Address
Available
Available
NotAvialable
Available
Available

在 married 和 gender 变量中,我不想丢失列的值,只需将 NA 替换如下。

NotAvailable_Indicator <- function(x){
  x[is.na(x)]<-"NotAvailable"
  return(x)
}
test$Married <- NotAvailable_Indicator(test$Married)
test$Gender <- NotAvailable_Indicator(test$Gender)

得到如下数据集

ID    FirstName     LastName BillNum        Phone      Married  ZIP       Gender      Address
 1 NotAvialable    Available       6    Available          Yes 1111            M    Available
 2    Available NotAvialable       7    Available          Yes 2222 NotAvailable    Available
 3 NotAvialable    Available       8    Available NotAvailable  333            F NotAvialable
 4    Available    Available       9 NotAvialable           NO  444 NotAvailable    Available
 5    Available NotAvialable      10 NotAvialable          Yes  555            M    Available

我的问题是我不想分别为每一列重复函数调用,因为我有大约 200 列。我无法使用应用函数,因为我必须对数据进行子集化,然后使用 lapply 应用函数,然后再次 cbind 到更改列顺序的原始数据。有什么方法可以提供列名和函数名,并且我可以在 return 中将修改后的列与其他列(未更改)一起作为数据集,或者在没有 [=30= 的情况下就地修改列]ing 任何东西(比如 python 中的 DataFrame.fillna 有参数 inplace=logical)

我们可以使用 tidyverse 来做到这一点

library(dplyr)
#specify the columns of interest 
#if there are any patterns, we can use `matches` or `grep`
nm1 <- names(test)[c(2, 3, 5, 9)]
nm2 <- names(test)[c(6, 8)]


#use `mutate_at` by specifying the arguments 'vars' and 'funs'
test %>% 
    mutate_at(vars(one_of(nm1)), funs(Availability_Indicator)) %>%
    mutate_at(vars(one_of(nm2)), funs(NotAvailable_Indicator))
#ID    FirstName     LastName BillNum        Phone      Married  ZIP       Gender      Address
#1  1 NotAvialable    Available       6    Available          Yes 1111            M    Available
#2  2    Available NotAvialable       7    Available          Yes 2222 NotAvailable    Available
#3  3 NotAvialable    Available       8    Available NotAvailable  333            F NotAvialable
#4  4    Available    Available       9 NotAvialable           NO  444 NotAvailable    Available
#5  5    Available NotAvialable      10 NotAvialable          Yes  555            M    Available

一个base R选项是使用lapply遍历列,应用函数并更新数据集列

test[nm1] <- lapply(test[nm1], Availability_Indicator)
test[nm2] <- lapply(test[nm2], NotAvailable_Indicator)

数据

factor class 列相比,更改 character 的值更容易。因此,在 'data.frame' 调用中使用 stringsAsFActors=FALSE,非数字列将是 character class

test <- data.frame(ID = c(1:5),
           FirstName = c(NA,"Sid",NA,"Harsh","CJ"),
           LastName = c("Snow",NA,"Lapata","Khan",NA),
           BillNum = c(6:10),
           Phone = c(1213,3123,3123,NA,NA),
           Married = c("Yes","Yes",NA,"NO","Yes"),
           ZIP = c(1111,2222,333,444,555),
           Gender = c("M",NA,"F",NA,"M"),
           Address = c("A","B",NA,"C","D"), stringsAsFactors=FALSE)