Question

最近我偶然发现了一个问题。不幸的是，我的日期变量尚未统一记录。

我得到了类似的数据框，如下图所示

myFile

我只需要提取日期即可。因此，我想基于“日期”创建一个新变量。

Date变量应仅包含日期。该小时与我的目的无关，可以忽略。

我的目标是获取以下数据框：

Variable1 <- c(10,20,30,40,50)
Variable2 <- c("a", "b", "c", "d", "d")
Date <- c("today 10:45", "yesterday 3:10", "28 october 2018 5:32", "28 october 2018 8:32", "27 october 2018 5:32")
df <- data.frame(Variable1, Variable2, Date)
df

最好Date的值也应采用正确的格式（日期）。

已经谢谢您了。

Answer 1

tolower(                                               # not strictly necessary, but for consistency
  gsub("yesterday", format(Sys.Date()-1, "%d %B %Y"),  # convert *day to dates
       gsub("today", format(Sys.Date(), "%d %B %Y"),
            gsub("\\s*[0-9:]*$", "",                   # remove the times
                 c("today 10:45", "yesterday 3:10", "28 october 2018 5:32", "28 october 2018 8:32", "27 october 2018 5:32")))))
# [1] "31 october 2018" "30 october 2018" "28 october 2018" "28 october 2018" "27 october 2018"

Answer 2

df$NewDate[grepl("today",df$Date)]<-Sys.Date() # Convert today to date
df$NewDate[grepl("yesterday",df$Date)]<-Sys.Date()-1  # covert yesterday to date
df$NewDate[is.na(df$NewDate)]<-df$Date[is.na(df$NewDate)] %>% as.Date(format="%d %b %Y")  # Convert explicit dates to date format
class(df$NewDate)<-"Date"  # Convert column to Date class

df
  Variable1 Variable2                 Date    NewDate
1        10         a          today 10:45 2018-10-31
2        20         b       yesterday 3:10 2018-10-30
3        30         c 28 october 2018 5:32 2018-10-28
4        40         d 28 october 2018 8:32 2018-10-28
5        50         d 27 october 2018 5:32 2018-10-27

Answer 3

使用索引的另一种解决方案。

Date <- c("today 10:45", "yesterday 3:10", "28 october 2018 5:32", "28 october 2018 8:32", "27 october 2018 5:32")

Date <- sub("today", Sys.Date(), Date)
Date <- sub("yesterday", Sys.Date() - 1, Date)
i <- grep("[[:alpha:]]", Date)
Date[i] <- format(as.POSIXct(Date[i], format = "%d %B %Y %H:%M"), format = "%d %B %Y")
Date[-i] <- format(as.POSIXct(Date[-i]), format = "%d %B %Y")

Date
#[1] "31 October 2018" "30 October 2018" "28 October 2018"
#[4] "28 October 2018" "27 October 2018"

然后我注意到solution by user r2evans，它将所有内容都转换为小写。因此，如有必要，以

结尾

Date <- tolower(Date)

在R中的数据框中修改日期

3 个答案: