Question

sHour <- seq(from = as.POSIXct("2018-01-05 0:00", tz = "UTC"),
     to = as.POSIXct("2018-01-13 23:00", tz = "UTC"),
     by = "hour")
dayWeek <- strftime(sHour, format = "%u")

My_holidays <- as.POSIXct(c("2018-01-06", "2018-01-09"), tz = "UTC")

星期一是1 星期日是7 如何在dayWeek中将My_holidays添加为7？

所需的输出

 > dayWeek
"5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1"

对于也以My_holidays为单位的sHour中的每个日期（小时），我想将变量7的值从1,2,3等更改。所以每个假期和星期天将是7

当前输出

> dayWeek
"5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "5" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "6" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "7" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1"

Answer 1

像这样吗？

library( dplyr )

df <- data.table( days = sHour ) %>%
  mutate( holiday = ifelse( as.Date(days) %in% as.Date(My_holidays), "yes", "no"))

#                    days holiday
# ...
# 93  2018-01-08 20:00:00      no
# 94  2018-01-08 21:00:00      no
# 95  2018-01-08 22:00:00      no
# 96  2018-01-08 23:00:00      no
# 97  2018-01-09 00:00:00     yes
# 98  2018-01-09 01:00:00     yes
# 99  2018-01-09 02:00:00     yes
# 100 2018-01-09 03:00:00     yes
# 101 2018-01-09 04:00:00     yes
# 102 2018-01-09 05:00:00     yes
# 103 2018-01-09 06:00:00     yes
# 104 2018-01-09 07:00:00     yes
# 105 2018-01-09 08:00:00     yes
# 106 2018-01-09 09:00:00     yes
# 107 2018-01-09 10:00:00     yes
# 108 2018-01-09 11:00:00     yes
# 109 2018-01-09 12:00:00     yes
# 110 2018-01-09 13:00:00     yes
# 111 2018-01-09 14:00:00     yes
# 112 2018-01-09 15:00:00     yes
# 113 2018-01-09 16:00:00     yes
# 114 2018-01-09 17:00:00     yes
# 115 2018-01-09 18:00:00     yes
# 116 2018-01-09 19:00:00     yes
# 117 2018-01-09 20:00:00     yes
# 118 2018-01-09 21:00:00     yes
# 119 2018-01-09 22:00:00     yes
# 120 2018-01-09 23:00:00     yes
# 121 2018-01-10 00:00:00      no
# 122 2018-01-10 01:00:00      no
# ...

Answer 2

这本质上是一个join操作，几个R软件包可以提供帮助。我碰巧很喜欢data.table。

创建表格

R> dt <- data.table(myhour=seq(from = as.POSIXct("2018-01-05 0:00", tz = "UTC"),
+                              to = as.POSIXct("2018-01-13 23:00", tz = "UTC"), 
+                              by = "hour"))
R> dt[, weekday := wday(myhour)]
R> dt[]
                  myhour weekday
  1: 2018-01-05 00:00:00       6
  2: 2018-01-05 01:00:00       6
  3: 2018-01-05 02:00:00       6
  4: 2018-01-05 03:00:00       6
  5: 2018-01-05 04:00:00       6
 ---                            
212: 2018-01-13 19:00:00       7
213: 2018-01-13 20:00:00       7
214: 2018-01-13 21:00:00       7
215: 2018-01-13 22:00:00       7
216: 2018-01-13 23:00:00       7
R>

这基本上就是您所拥有的，但是作为data.table使用其辅助功能wday()可以在Date或POSIXt上使用。

创建假期

R> myholidays <- as.Date(c("2018-01-06", "2018-01-09")) # no UTC conversion

大致相同，但我们将它们保留为Date，因为它们是日期对象。

合并

R> dt[, isholiday:=FALSE][as.Date(myhour) %in% myholidays, isholiday:=TRUE][]
                  myhour weekday isholiday
  1: 2018-01-05 00:00:00       6     FALSE
  2: 2018-01-05 01:00:00       6     FALSE
  3: 2018-01-05 02:00:00       6     FALSE
  4: 2018-01-05 03:00:00       6     FALSE
  5: 2018-01-05 04:00:00       6     FALSE
 ---                                      
212: 2018-01-13 19:00:00       7     FALSE
213: 2018-01-13 20:00:00       7     FALSE
214: 2018-01-13 21:00:00       7     FALSE
215: 2018-01-13 22:00:00       7     FALSE
216: 2018-01-13 23:00:00       7     FALSE

这是关键，我们结合了三个data.table操作符。首先，我们创建 isholiday作为布尔值，默认为FALSE。我们将其设置为假日向量中的日期-从POSIXct到Date的转换是即时进行的。最后，我们打印结果。

检查

R> head(dt[ isholiday==TRUE, ])
                myhour weekday isholiday
1: 2018-01-06 00:00:00       7      TRUE
2: 2018-01-06 01:00:00       7      TRUE
3: 2018-01-06 02:00:00       7      TRUE
4: 2018-01-06 03:00:00       7      TRUE
5: 2018-01-06 04:00:00       7      TRUE
6: 2018-01-06 05:00:00       7      TRUE
R> summary(dt)
     myhour                       weekday     isholiday      
 Min.   :2018-01-05 00:00:00   Min.   :1.00   Mode :logical  
 1st Qu.:2018-01-07 05:45:00   1st Qu.:3.00   FALSE:168      
 Median :2018-01-09 11:30:00   Median :5.00   TRUE :48       
 Mean   :2018-01-09 11:30:00   Mean   :4.56                  
 3rd Qu.:2018-01-11 17:15:00   3rd Qu.:6.00                  
 Max.   :2018-01-13 23:00:00   Max.   :7.00                  
R>

检查无痛。我们希望两天的每小时数据达到48个条目。

如何在R的strftime中添加假期

2 个答案:

创建表格

创建假期

合并

检查