Question

我有从互联网收集的数据。日期列格式类似于“ 1个月前”，“ 2年前”，“ 4天前”，我需要更改此格式。

我尝试过：

googleR$`Date/Time` <- as.Date.character(googleR$`Date/Time`, 
                                         format = format(googleR$`Date/Time`), 
                                         tryFormats = c("%d/%m/%Y"))

但是它只给了我今天所有记录的日期

我也尝试了许多其他事情，但是出现了相同的错误

“字符串不是标准的明确格式”

这是我要转换的数据：

format.factor(googleR$`Date/Time`)
 [1] "7 months ago " "2 months ago " "a week ago   " "2 years ago  " "2 years ago  " "5 months ago " "10 months ago"
 [8] "2 years ago  " "4 years ago  " "2 years ago  " "2 years ago  " "11 months ago" "3 years ago  " "3 years ago  "
[15] "2 years ago  " "2 years ago  " "10 months ago" "10 months ago" "a year ago   " "a year ago   " "2 years ago  "
[22] "2 years ago  " "2 years ago  " "2 years ago  " "2 years ago  " "2 years ago  " "3 years ago  " "4 years ago  "
[29] "4 years ago  " "a week ago   " "a week ago   " "2 weeks ago  " "a month ago  " "2 months ago " "5 months ago "
[36] "7 months ago " "7 months ago " "8 months ago " "10 months ago" "10 months ago" "a year ago   " "a year ago   "
[43] "a year ago   " "a year ago   " "a year ago   " "a year ago   " "a year ago   " "2 years ago  " "2 years ago  "
[50] "2 years ago  " "4 years ago  " "6 years ago  "

Answer 1

您可以使用sub从每个元素中删除“ ago”，然后按如下所示使用lubridate的{{1}}

add_with_rollback

结果

library(lubridate)
add_with_rollback(Sys.time(), - as.period(sub("\\s+ago", "", x)))

数据

"2019-02-28 18:13:18 CET"  "2017-03-31 18:13:18 CEST" "2019-03-27 18:13:18 CET"

Answer 2

您可能首先需要使用正则表达式找出您拥有的时间间隔类型，然后再从那里开始。我喜欢这样的包装袋。例如，如果IN是您的输入字符串，OUT是您想要的输出，您可以说

 if(str_detect(IN, "day")){OUT <- as.numeric(str_extract(IN, "^[0-9]*"))}

现在，您已经有多少天了，您可以执行

之类的操作

 Sys.Date() - OUT

获取日期。然后，您可以在数月和数年内完成基本相同的操作。不可避免地，这将是近似的，例如，并非所有月份的天数都相同，但是听起来您的输入数据并不是绝对精确。

我需要将日期格式从“ xx时间前”更改为：“一年前，一周前”更改为“％Y /％m /％d”

2 个答案: