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我在 SPSS 中构建了每小时数据。开始时间 7:00 和结束时间 7:59。当我在 R 中阅读这些时,我得到这样的数字:57600 和 61140。如何让 R 将这些值识别为小时?感谢您的帮助*强文本*

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我猜你输入的那些数字R是以秒为单位的相应小时数,例如 02:51:03 = 2*60*60 + 51*60 + 3 = 10263 秒。与我的猜测不符的是 7:00 不是 57600 秒,而是 25200 秒(上午)或 68400 秒(下午)。57600 秒对应于 16:00(下午 4:00)。

如果这些值实际上是秒,那么您可以使用以下函数将它们转换为时间toHour

    toHour <- function(x, seconds = F)  # x is a value of seconds
    {
     hours <- x %/% 3600
     secs. <- x %% 3600
     mins <- secs. %/% 60
     secs <- secs. %% 60

     ## add a 0 to single-digit hours / minutes / seconds to make it look nicer  
     mytime <- c(hours, mins, secs)
     for(i in 1:length(mytime))  
      {
       if(length(unlist(strsplit(as.character(mytime[i]), split = ""))) == 1)
        {
         mytime[i] <- paste(0, mytime[i], sep = "", collapse = "")
        }
      }
     ##

    if(seconds == T)
      {
       return(paste(mytime[1], mytime[2], mytime[3], sep = ":", collapse = ""))
      }
     if(seconds == F)
      {
       return(paste(mytime[1], mytime[2], sep = ":", collapse = ""))
      }
    }

    toHour(10263, seconds = T) # check function
    #[1] "02:51:03" #checked
    my_seconds <- seq(57600,61140, 60) # I guess you have no seconds in your time and, so, "my_seconds" is the values you get in R
    sapply(my_seconds, toHour) # turn all values (seconds) to hours:minutes
    #[1] "16:00" "16:01" "16:02" "16:03" "16:04" ...
    sapply(my_seconds, toHour, seconds = T) # turn all values (seconds) to hours:minutes:seconds
    #[1] "16:00:00" "16:01:00" "16:02:00" "16:03:00" ...

如果我误解了这个问题或我的帖子无关紧要,我很抱歉,但由于我的声誉低,我无法在评论中询问详细信息。

编辑:toHour返回字符值。下一步是:

    library(chron)
    chron(times. = sapply(my_seconds, toHour, seconds = T))
    #[1] 16:00:00 16:01:00 16:02:00 16:03:00 16:04:00 ...
于 2013-10-05T23:39:11.253 回答