如何在r中將因子水平轉換為整數

Question

我在 R 中有以下數據框

  ID      Season      Year       Weekday
  1       Winter      2017       Monday
  2       Winter      2018       Tuesday
  3       Summer      2017       Monday
  4       Summer      2018       Wednsday

我想將這些因子級別轉換為整數，以下是我想要的數據框

  ID      Season      Year       Weekday
  1       1           1          1
  2       1           2          2
  3       2           1          1
  4       2           2          3

  Winter = 1,Summer =2
  2017 = 1 , 2018 = 2
  Monday = 1,Tuesday = 2,Wednesday = 3

目前，我正在為以上 3 做ifelse

  otest_xgb$Weekday <- as.integer(ifelse(otest_xgb$Weekday == "Monday",1,
                                   ifelse(otest_xgb$Weekday == "Tuesday",2,
                                          ifelse(otest_xgb$Weekday == "Wednesday",3,
                                                 ifelse(otest_xgb$Weekday == "Thursday",4,5)))))

有什么辦法可以避免寫長的ifelse嗎？

Answer 1

m=dat
> m[]=lapply(dat,function(x)as.integer(factor(x,unique(x))))
> m
  ID Season Year Weekday
1  1      1    1       1
2  2      1    2       2
3  3      2    1       1
4  4      2    2       3

Answer 2

我們可以將match與unique元素一起使用

library(dplyr)
dat %>%
      mutate_all(funs(match(., unique(.))))
#   ID Season Year Weekday
#1  1      1    1       1
#2  2      1    2       2
#3  3      2    1       1
#4  4      2    2       3

Answer 3

有序和名義因子變量需要分別處理。 直接將因子列轉換為整數或數字將提供字典意義上的值。

這里Weekday在概念上是有序的， Year是整數， Season通常是名義上的。 然而，這又是主觀的，取決於所需的分析類型。

例如。 當您直接從因子轉換為整數變量時。 在Weekday列中， Wednesday將獲得比星期六和星期二更高的值：

 dat[] <- lapply(dat, function(x)as.integer(factor(x)))
 dat 

#  ID Season Year Weekday
#1  1      2    1       1
#2  2      2    2       3
#3  3      1    1       2   (Saturday)
#4  4      1    2       4   (Wednesday): assigned value greater than that ofSaturday        

因此，您只能將“ Season和“ Year列的因子直接轉換為整數。 可能會注意到，對於year列，它可以正常工作，因為詞典意義與其序數意義相匹配。

dat[c('Season', 'Year')] <- lapply(dat[c('Season', 'Year')], 
                                   function(x) as.integer(factor(x)))

Weekday需要從具有所需級別順序的有序因子變量轉換而來。 如果進行一般聚合可能無害，但在實施統計模型時會極大地影響結果。

dat$Weekday <- as.integer(factor(dat$Weekday, 
                          levels = c("Monday", "Tuesday", "Wednesday", "Thursday", 
                                     "Friday", "Saturday", "Sunday"), ordered = TRUE))

dat
#  ID Season Year Weekday
#1  1      2    1       1
#2  2      2    2       2
#3  3      1    1       6  (Saturday)
#4  4      1    2       3  (Wednesday): assigned value less than that of Saturday

使用的數據：

dat <- read.table(text="  ID      Season      Year       Weekday
1       Winter      2017       Monday
2       Winter      2018       Tuesday
3       Summer      2017       Saturday
4       Summer      2018       Wednesday", header = TRUE)

Answer 4

您可以簡單地使用as.numeric()將因子轉換為數字。 每個值將更改為該因子級別表示的相應整數：

library(dplyr)

### Change factor levels to the levels you specified
otest_xgb$Season  <- factor(otest_xgb$Season , levels = c("Winter", "Summer"))
otest_xgb$Year    <- factor(otest_xgb$Year   , levels = c(2017, 2018))
otest_xgb$Weekday <- factor(otest_xgb$Weekday, levels = c("Monday", "Tuesday", "Wednesday"))

otest_xgb %>% 
  dplyr::mutate_at(c("Season", "Year", "Weekday"), as.numeric)


# ID Season Year Weekday
# 1  1      1    1       1
# 2  2      1    2       2
# 3  3      2    1       1
# 4  4      2    2      NA

Answer 5

將季節、年份和工作日轉換為因子后，請使用此代碼更改為虛擬指標變量

contrasts(factor(dat$season) 
contrasts(factor(dat$year)
contrasts(factor(dat$weekday)