Question

这是我的数据框的示例：

    I  UserID | Day_of_week | hour |   min 
   #1    1           1          0     0     
   #2    1           1          0     30     
   #3    1           1          1     0     
   #4    1           1          1     30   
   #5    1           1          2     0      
   #6    1           1          2     30   
   ...   ...         ...       ...    ...  
   #10    1          2          0     0     
   #11    1          2          0     30    
   #12    1          2          1     0     
   #13    1          2          1     30    
   ...   ...         ...       ...    ...

Day_of_week列从1到7，其中：

2 =星期一，3 =星期二...... 6 =星期五

和7和1分别是周六和周日。

我想在Day_of_Week的列中更改星期的名称。像这样：

       I  UserID | Day_of_week | hour |   min 
       #1    1       Sunday        0     0     
       #2    1       Sunday        0     30     
       #3    1       Sunday        1     0      
       ...   ...         ...      ...    ...  
       #10    1      Monday        0     0     
       #11    1      Monday        0     30    
       #12    1      Monday        1     0     
       #13    1      Monday        1     30    
       ...   ...         ...      ...    ...

Answer 1

只需使用矢量。

MIMEText()

Answer 2

 library(expss)

 #Creating a sample dataset like yours
 Day_of_week <- c(1,7,1,7,1,3,2,2,7,2,2,4,3,5,3)
 userid <- c(1,3,1,1,1,3,2,2,4,4,3,3,32,3,3)
 TableExample <- as.data.frame(cbind(userid, Day_of_week))


 #Creating a week day reference table
 DayID <- c(1,2,3,4,5,6,7)
 DayName <- c("Sunday", "Monday", "Tuesday", "Wed", "Thur", "Friday", "Sat")
 Weektable <- as.data.frame(cbind(DayID,DayName))

 #Vlookup formula to look for x in the referencetable
 FindWeekFunc <- function(x){
   vlookup(x,Weektable,2)}

 #maply to find teh result
 TableExample$Name_of_Day <- mapply(FindWeekFunc, Day_of_week)

Answer 3

我认为您可以将列转换为因子变量。

您可以尝试下面的内容

df$day_of_week <- factor(df$day_of_week,
                          labels = c("Sunday", "Monday","Tuesday","Wednesday", "Thursday", "Friday", "Saturday"),
                          levels = c(1,2,3,4,5,6,7))

我希望这应该有用。

Answer 4

创建一个带数字的函数

library(dplyr)

day_of_week <- function(num){

  ifelse(num == 1, "Sunday", 
         ifelse(num ==2, "Monday", 
                ifelse(num == 3, "Tuesday", 
                       ifelse(num == 4, "Wednesday", 
                              ifelse(num == 5, "Thursday", 
                                     ifelse(num == 6, "Friday", "Saturday"))))))

  }

然后使用dplyr来改变您的数据框

dat <- 
  data.frame(x = rnorm(10), 
             day = sample(1:7, 10, TRUE)) %>%
  mutate(day_of_week = day %>% day_of_week())

> dat
               x day day_of_week
1  -0.4235498649   4   Wednesday
2   1.0366219455   2      Monday
3   0.4280368042   7    Saturday
4  -1.0335602213   5    Thursday
5  -0.6016099023   5    Thursday
6   1.4246527069   2      Monday
7  -0.3303051311   6      Friday
8  -0.0009648844   4   Wednesday
9   1.0168062274   6      Friday
10  0.2896002781   7    Saturday

Answer 5

如果您通过将数字添加到星期六的任何特定日期将数字转换为日期，则可以使用base::weekdays功能：

df <- data.frame(I = c(1L, 2L, 3L, 4L, 5L, 6L, 10L, 11L, 12L, 13L), 
                 UserID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), 
                 Day_of_week = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), 
                 hour = c(0L, 0L, 1L, 1L, 2L, 2L, 0L, 0L, 1L, 1L), 
                 min = c(0L, 30L, 0L, 30L, 0L, 30L, 0L, 30L, 0L, 30L))

df$weekday <- weekdays(as.Date('1970-01-01') + 2 + df$Day_of_week)

df
#>     I UserID Day_of_week hour min weekday
#> 1   1      1           1    0   0  Sunday
#> 2   2      1           1    0  30  Sunday
#> 3   3      1           1    1   0  Sunday
#> 4   4      1           1    1  30  Sunday
#> 5   5      1           1    2   0  Sunday
#> 6   6      1           1    2  30  Sunday
#> 7  10      1           2    0   0  Monday
#> 8  11      1           2    0  30  Monday
#> 9  12      1           2    1   0  Monday
#> 10 13      1           2    1  30  Monday

替换列中的值

5 个答案: