警告信息:在`...`:无效因子级别,NA生成

Question

我不明白为什么我收到这条警告信息.

> fixed <- data.frame("Type" = character(3), "Amount" = numeric(3))
> fixed[1, ] <- c("lunch", 100)
Warning message:
In `[<-.factor`(`*tmp*`, iseq, value = "lunch") :
  invalid factor level, NA generated
> fixed
  Type Amount
1 <NA>    100
2           0
3           0

Answer 1

警告消息是因为您的"类型"变量是一个因素而"午餐"不是定义的级别.stringsAsFactors = FALSE在创建数据框时使用标志强制"类型"为字符.

> fixed <- data.frame("Type" = character(3), "Amount" = numeric(3))
> str(fixed)
'data.frame':   3 obs. of  2 variables:
 $ Type  : Factor w/ 1 level "": NA 1 1
 $ Amount: chr  "100" "0" "0"
> 
> fixed <- data.frame("Type" = character(3), "Amount" = numeric(3),stringsAsFactors=FALSE)
> fixed[1, ] <- c("lunch", 100)
> str(fixed)
'data.frame':   3 obs. of  2 variables:
 $ Type  : chr  "lunch" "" ""
 $ Amount: chr  "100" "0" "0"

Answer 2

如果您直接从CSV文件中阅读,请执行此操作.

myDataFrame <- read.csv("path/to/file.csv", header = TRUE, stringsAsFactors = FALSE)

Answer 3

这是一种灵活的方法,它可以在所有情况下使用,特别是:

1. 以只影响一列,或
2. 在dataframe已经从先前施加的操作(例如获得不立即打开一个文件,或者创建一个新的数据帧).

首先,使用函数对字符串进行非因式分解as.character,然后使用(或简单)函数重新分解:as.factorfactor

fixed <- data.frame("Type" = character(3), "Amount" = numeric(3))

# Un-factorize (as.numeric can be use for numeric values)
#              (as.vector  can be use for objects - not tested)
fixed$Type <- as.character(fixed$Type)
fixed[1, ] <- c("lunch", 100)

# Re-factorize with the as.factor function or simple factor(fixed$Type)
fixed$Type <- as.factor(fixed$Type)

Answer 4

解决此问题的最简单方法是为列添加新因子.使用级别功能确定您拥有的因子数量,然后添加新因子.

    > levels(data$Fireplace.Qu)
    [1] "Ex" "Fa" "Gd" "Po" "TA"
    > levels(data$Fireplace.Qu) = c("Ex", "Fa", "Gd", "Po", "TA", "None")
    [1] "Ex"   "Fa"   "Gd"   "Po"   " TA"  "None"