从R中的时间减去一个数字

Question

I have a data frame with two time columns containing times ranging from 0:00 - 23:59. This data was supposed to be collected in a 0:00-12:00 format, but there was an error made in the Redcap survey allowing the respondents to choose in military times.

I need to subtract 12:00 from all values that are greater than 12:00 so that I get the range from military to standard, yet I cannot find an obvious way to go about this.

Here is an approximation of the for loop I was trying - I just am unaware of how to subtract a number from hours in the column once I have transformed it with lubridate which is why I have not tried that.

Additionally, I tried difftime but am not sure how to use it when I am not subtracting two columns.

    if (is.na(df$var1)) {
    next
  }
    else if (x > "12:00") {
      x<-as.numeric(x)-12
    }
    print(x)
  }

Thanks

Answer 1

This quite simple approach requires conversion of your time data stored as characters to POSIXlt using strptime() , enabling to choose another format based on your individual specifications using format() :

x <- c("04:12", "11:38", "15:39", "23:59")

strptime(x, format="%H:%M") |> format("%I:%M")
#> [1] "04:12" "11:38" "03:39" "11:59"