如何将函数应用于R中数据帧中的特定列集以替换NAs
我有一个数据集,我想在其中以不同的方式替换不同列中的NAs。下面是虚拟数据集和复制它的代码如何将函数应用于R中数据帧中的特定列集以替换NAs,r,function,dataframe,dry,na,R,Function,Dataframe,Dry,Na,我有一个数据集,我想在其中以不同的方式替换不同列中的NAs。下面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
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
我的问题是我不想对每一列分别重复函数调用,因为我有大约200列。我不能使用apply函数,因为我必须先对数据进行子集处理,然后使用lappy将函数应用到原始数据中,然后使用cbind再次应用到原始数据中,这改变了列的顺序。是否有任何方法可以提供列和函数的名称,并将修改后的列与其他列(未更改)一起作为数据集返回,或者在不返回任何内容的情况下就地修改列(例如python中的DataFrame.fillna,其参数inplace=logical)我们可以使用
tidyverse
进行此操作
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
数据
与因子
类列相比,更改字符
的值更容易。因此,在'data.frame'调用中使用stringsAsFActors=FALSE
,非数字列将是character
类
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)
test我们可以很容易地做到这一点nm1
NotAvailable_Indicator <- function(x){
x[is.na(x)]<-"NotAvailable"
return(x)
}
test$Married <- NotAvailable_Indicator(test$Married)
test$Gender <- NotAvailable_Indicator(test$Gender)
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
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
test[nm1] <- lapply(test[nm1], Availability_Indicator)
test[nm2] <- lapply(test[nm2], NotAvailable_Indicator)
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)