用一个范围内的随机数替换数据帧中的NA

Question

我有以下名为cars的数据框

Brand      year     mpg        reputation      Luxury
Honda      2010     30            8.5            0.5
Honda      2011     28            8.5            0.6
Dodge      2010     20            6.5            0.6
Dodge      2011     23            7.0            0.7
Mercedes   2010     22            9.5            NA
Mercedes   2011     25            9.0            NA

我想用0.9 and 1.0之间随机生成的实数替换NA

我正在尝试以下，但它正在用数字0.9替换NA

cars[is.na(cars)] <-  sample(0.9:1, sum(is.na(cars)),replace=TRUE)

数据表看起来像这样：

Brand      year     mpg        reputation      Luxury
Honda      2010     30            8.5            0.5
Honda      2011     28            8.5            0.6
Dodge      2010     20            6.5            0.6
Dodge      2011     23            7.0            0.7
Mercedes   2010     22            9.5           *0.91*
Mercedes   2011     25            9.0           *0.97*

数据结构代码：

cars <- structure(list(Brand = c("Honda","Honda", "Dodge", "Dodge","Mercedes","Mercedes"), 
   year = c(2010L, 2011L,2010L, 2011L, 2010L, 2011L), 
   mpg = c(30L, 28L, 20L, 23L, 22L, 25L), reputation = c(8.5, 8.5, 6.5, 7L, 9.5, 9.5), Luxury = c(5L, 5.5, 6L, 6.5)), 
  class = "data.frame", row.names = c(NA, -4L))      

Answer 1

使用runif而不是sample：

cars[is.na(cars)] <-  runif(sum(is.na(cars)), min = 0.9, max = 1)

Answer 2

那是因为0.9:1只给你一个0.9的数字。尝试，

0.9:1
#[1] 0.9

因此，它将这些数字替换为0.9。

假设您需要序列为

vals <- seq(0.9, 1, 0.01)
vals
#[1] 0.90 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1.00

现在，我们可以sample这个序列

df[is.na(df)] <- sample(vals, sum(is.na(df)), replace = TRUE)

df
#     Brand year mpg reputation Luxury
#1    Honda 2010  30        8.5   5.00
#2    Honda 2011  28        8.5   5.50
#3    Dodge 2010  20        6.5   6.00
#4    Dodge 2011  23        7.0   6.50
#5 Mercedes 2010  22        9.5   0.91
#6 Mercedes 2011  25        9.0   0.92

数据

df <- structure(list(Brand = structure(c(2L, 2L, 1L, 1L, 3L, 3L), 
.Label = c("Dodge", 
"Honda", "Mercedes"), class = "factor"), year = c(2010L, 2011L, 
2010L, 2011L, 2010L, 2011L), mpg = c(30L, 28L, 20L, 23L, 22L, 
25L), reputation = c(8.5, 8.5, 6.5, 7, 9.5, 9), Luxury = c(5, 
5.5, 6, 6.5, NA, NA)), class = "data.frame", row.names = c(NA, -6L))