在R中對字符向量進行排序

Question

我有這個向量，我需要按降序排序。 首先是最新的txt：

d<-c("/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_04_2015.txt","/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_11_2015.txt","/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_18_2015.txt","/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_08-01_25_2015.txt","/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-11_25-01_20_2015.txt")

當我這樣做時：

d <- sort(d)

d[1]
# "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_04_2015.txt"

必須是這樣的：

"/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_08-01_25_2015.txt"

我應該能夠通過文本“ 11_25-01_20_2015”中的此條目對此進行排序，其中11是小時， 25分鍾， 01個月， 20天和2015年，即hour_minute-month_day_year。

我該怎么辦？

Answer 1

如果字符串的末尾是一致的（bla-bla-bla- time-date.txt ），則可以使用substring提取時間。 然后將時間轉換為as.POSIXct並按order使用它們

time <- substring(d, first = nchar(d)-19)
d[order(as.POSIXct(time, format = "%H_%M-%m_%d_%Y.txt"), decreasing = TRUE)]
# [1] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_08-01_25_2015.txt"
# [2] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-11_25-01_20_2015.txt"
# [3] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_18_2015.txt"
# [4] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_11_2015.txt"
# [5] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_04_2015.txt"

Answer 2

首先，您應該提取時間並將其設置為明智的格式：

times <- as.POSIXct(sub("^.+Report-([0-9]+)_([0-9]+)-([0-9]+)_([0-9]+)_([0-9]+)\\.txt$","\\5-\\3-\\4 \\1:\\2",d))
times
[1] "2015-01-04 02:06:00 GMT" "2015-01-11 02:06:00 GMT"
[3] "2015-01-18 02:06:00 GMT" "2015-01-25 02:08:00 GMT"
[5] "2015-01-20 11:25:00 GMT"

然后，您可以使用這些命令訂購原始數據：

d[order(times, decreasing=TRUE)][1]
[1] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_08-01_25_2015.txt"

Answer 3

嘗試這個：

# trim everythin before the string 'Report-'
dateSting <- gsub('^.*Report-','',d )
# trim the '.txt' from the end.
dateSting <- gsub('\\.txt$','',dateSting )
#convert the date string to a date-time object
dateTime  <-  as.POSIXct(dateSting,'%H_%M-%m_%d_%Y')
# sort on date time 
d <- d[order(dateTime)]

Answer 4

您可能會提取日期，將其轉換為POSIXct類，然后使用which.max獲取最新日期。

library(stringi)
indx <- as.POSIXct(stri_extract_first_regex(d, "(?<=Report-).*(?=\\.txt)"), format = "%H_%M-%m_%d_%Y")
d[which.max(indx)]
# [1] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_08-01_25_2015.txt"

或者您可以按降序排列

d[order(indx, decreasing = TRUE)]
# [1] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_08-01_25_2015.txt"
# [2] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-11_25-01_20_2015.txt"
# [3] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_18_2015.txt"
# [4] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_11_2015.txt"
# [5] "/SiteScope/accounts/login59/htdocs/Reports-1722992141/Report-02_06-01_04_2015.txt"