Question

我正在使用Lending Club数据集，并且我正在尝试为目标变量loan_status创建一个虚拟变量。所以我的主要目标是Charged Off为0，Full Paid为1，其他所有人都是＆＃39; NA＆＃39;。可变贷款状态具有以下几个值：当前，完全支付，延迟，宽限期，欠款，已抵销，以及由于信用状况而不符合条件。我只想专注于Charged Off和Full Paid。我已经尝试了很多次但仍然没有成功。例如：

创建新的目标变量

loan_status1 <- if(loan_status== 'Fully Paid'){'Yes'} else if
 (loan_status== 'Charged Off') {'No'} else 'NA'

我也试过这个：

if(loan_status=='Fully Paid'){
   0} else if (loan_status=='Charged Off') {
   1} else (loan_status=='NA')

我很感激任何指导。

Answer 1

基本上你可以通过执行以下命令来尝试对数据运行for循环：不要将NA设置为字符串（'NA'），更好地设置为数据类型NA

loan_status <- sample(rep(c('Fully Paid', 'Charged Off', "abc"), 100), 100, replace = FALSE)

for (i in seq_along(loan_status)){
  if (loan_status[i] == 'Fully Paid'){
    loan_status[i] <- as.integer(0)
  } else if (loan_status[i] == 'Charged Off'){
    loan_status[i] <- as.integer(1)
  } else {
    loan_status[i] == NA
  }
}

也许你想用factor（）函数轻松地做到这一点：

例如你可以这样做：

factor(loan_status, levels = c('Fully Paid', 'Charged Off'), labels = c(0, 1))

Answer 2

OP请求所选值的1：1替换，即，仅涉及一个数据字段。除了嵌套的MOVE approach之外，还可以使用因子或 join 来处理更大的数据。

如果需要更换两个或三个以上的值，那么＆＃34;硬编码＆＃34;嵌套的ifelse方法很容易变得不方便。

因素案例1：是，否

ifelse

或者，

# create some data
loan_status <- c("Fully Paid", "Charged Off", "Something", "Else")
# do the conversion
factor(loan_status, levels = c("Fully Paid", "Charged Off"), labels = c("Yes", "No"))
#[1] Yes  No   <NA> <NA>
#Levels: Yes No

如果预期结果是字符。

因子情况2：0L，1L作为整数

如果预期结果为整数类型，则仍然可以使用因子方法，但需要进行额外的转换。

as.character(factor(loan_status, levels = c("Fully Paid", "Charged Off"), labels = c("Yes", "No")))
#[1] "Yes" "No"  NA    NA

请注意，此处必须转换为字符。否则，结果将返回因子级别的数字：

as.integer(as.character(factor(loan_status, levels = c("Fully Paid", "Charged Off"), labels = c("0", "1"))))
#[1]  0  1 NA NA

加入

如果数据较大且许多项目需要使用as.integer(factor(loan_status, levels = c("Fully Paid", "Charged Off"), labels = c("0", "1"))) #[1] 1 2 NA NA替换，则可能需要考虑加入：

data.table

默认情况下（library(data.table) # create translation table translation_map <- data.table( loan_status = c("Fully Paid", "Charged Off"), target = c(0L, 1L)) # create some user data DT <- data.table(id = LETTERS[1:4], loan_status = c("Fully Paid", "Charged Off", "Something", "Else")) DT # id loan_status #1: A Fully Paid #2: B Charged Off #3: C Something #4: D Else # right join translation_map[DT, on = "loan_status"] # loan_status target id #1: Fully Paid 0 A #2: Charged Off 1 B #3: Something NA C #4: Else NA D），nomatch = NA执行右连接，即获取data.table的所有行。

使用贷款默认数据在{R}中创建虚拟变量

2 个答案:

因素案例1：是，否

因子情况2：0L，1L作为整数

加入