如何将日期转换为最近的周末（星期六）

Question

我有一个数据框，其日期格式为“%d-%m-%Y”，并且有周数。日期是工作日，我希望在另一栏中显示该周的星期六。

我最初使用 Chron 包中的函数检查日期是工作日还是周末，但这是一个布尔验证。我已将日期变量格式化为日期格式并提取每个日期的周数。

df = data.frame(date=c("2014-08-20", "2014-08-25", "2014-10-08")) 
df$date=as.Date(df$date,format="%Y-%m-%d")
df$week=week(ymd(df$date))

预期结果应该是：

date        week    EOW  
2014-08-20   34   2014-08-23

2014-08-25   34   2014-08-30

2014-10-08   41   2014-10-11

Answer 1

基础 R 选项。首先创建所有日期的列表，然后使用

match

weekdays

并从 6 中减去它（因为我们想要星期六），以获得我们需要在原始

date

列中添加多少天。

all_days <- c("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday")

#As @nicola mentioned this is locale dependent
#If your locale is not English you need weekdays in your current locale
#which you can manually write as shown above or do any one of the following

#all_days <- weekdays(seq(as.Date("2019-01-14"),by="day",length.out=7))
#OR
#all_days <- format(seq(as.Date("2019-01-14"),by="day",length.out=7), "%A")

df$EOW <- df$date + 6 - match(weekdays(df$date), all_days)

df
#        date week        EOW
#1 2014-08-20   34 2014-08-23
#2 2014-08-25   34 2014-08-30
#3 2014-10-08   41 2014-10-11

---

或者

lubridate

有一个函数

ceiling_date

，当与

unit = "week"

一起使用时，它会返回下一个“星期日”，所以我们从中减去 1 天来得到“星期六”。

library(lubridate)
df$EOW <- ceiling_date(df$date, unit = "week") - 1

Answer 2

另一种使用方式

library(data.table)
df <- data.table(date=c("2014-08-20", "2014-08-25", "2014-10-08")) 
df$date=as.Date(df$date,format="%Y-%m-%d")
df$week=week(ymd(df$date))

## if the locale is not English, please use the local values for days 
days <- data.frame(DOW = c("Monday", "Tuesday", "Wednesday", "Thursday","Friday", "Saturday", "Sunday"))
days$day <- seq(1,7,1)

df <- df[,DOW:= weekdays(date)]
df <- merge(df, days, all.x = T, by = "DOW")

df <- df[, EOW := date + (6 - day)]
df

         DOW       date week day        EOW
1:    Monday 2014-08-25   34   1 2014-08-30
2: Wednesday 2014-08-20   34   3 2014-08-23
3: Wednesday 2014-10-08   41   3 2014-10-11

Answer 3

使用

data.table

连接语法：

library(data.table)

# Create a saturdays dataset
saturdays2014 <- data.table(date = seq(as.Date("2014-01-01"), as.Date("2014-12-31"), by = 1))
Sys.setlocale("LC_ALL","English")
saturdays2014 <- saturdays2014[weekdays(date) == "Saturday"]

# convert df to data.table and date to a Date variable
setDT(df)[, date := as.Date(date)]

# Join
df[saturdays2014, on = "date", roll = 6, EOW := i.date]
df    
#          date        EOW
# 1: 2014-08-20 2014-08-23
# 2: 2014-08-25 2014-08-30
# 3: 2014-10-08 2014-10-11

Answer 4

另一种方法。这是周六的。如果您想要该周的星期四，只需在计算中使用“5”而不是“7”，等等..

library(lubridate)
library(dplyr)

df <- data.table(date=as.Date(c("2019-01-04","2019-01-07", "2019-01-15", "2019-01-26","2019-01-27")))

df %>% mutate(cw = wday(date)) %>%
  mutate(nSaturday = date+(7-cw))

        date cw  nSaturday
1 2019-01-04  6 2019-01-05
2 2019-01-07  2 2019-01-12
3 2019-01-15  3 2019-01-19
4 2019-01-26  7 2019-01-26
5 2019-01-27  1 2019-02-02