How to convert time (not date-time) variable into numeric? - r

Question

I have a data of swimming times that I would like to be able to plot over time. I was wondering if there was a quick way to change these variables from character to numeric?
I started by trying to convert the times to a POSIX date-time format, but that proved to not be helpful, especially because I would like to do some ARIMA predictions on the data.
Here is my data
times <- c("47.45","47.69",
"47.69","47.82",
"47.84","47.92",
"47.96","48.13",
"48.16","48.16",
"48.16","48.31",
"49.01","49.27",
"49.33","49.40",
"49.48","49.51",
"52.85","52.89",
"53.14","54.31",
"54.63","56.91",
"1:18.39","1:20.26",
"1:38.30")
dates <- c("2017-02-24 MST",
"2017-02-24 MST",
"2016-02-26 MST",
"2018-02-23 MST",
"2015-12-04 MST",
"2015-03-06 MST",
"2015-03-06 MST",
"2016-12-02 MST",
"2016-02-26 MST",
"2017-11-17 MST",
"2016-12-02 MST",
"2017-11-17 MST",
"2014-11-22 MST",
"2017-01-13 MST",
"2017-01-21 MST",
"2015-10-17 MDT",
"2017-01-27 MST",
"2016-01-29 MST",
"2017-10-20 MDT",
"2016-11-05 MDT",
"2015-11-07 MST",
"2015-10-30 MDT",
"2014-11-22 MST",
"2016-11-11 MST",
"2014-02-28 MST",
"2014-02-28 MST",
"2014-02-28 MST",)
df <- cbind(as.data.frame(dates),as.data.frame(times))
I hope to get a column for time, probably in seconds, so the first 24 obs would stay the same, but the last 3 obs would change to 78.39,80.26, and 98.30

Answer 1

One way is to pre-pend those times that don't have minutes with "00:".
Then you can use lubridate::ms to do the time conversion.
library(dplyr)
library(lubridate)
data.frame(times = times,
stringsAsFactors = FALSE) %>%
mutate(times2 = ifelse(grepl(":", times), times, paste0("00:", times)),
seconds = as.numeric(ms(times2)))
Result:
times times2 seconds
1 47.45 00:47.45 47.45
2 47.69 00:47.69 47.69
3 47.69 00:47.69 47.69
4 47.82 00:47.82 47.82
5 47.84 00:47.84 47.84
6 47.92 00:47.92 47.92
7 47.96 00:47.96 47.96
8 48.13 00:48.13 48.13
9 48.16 00:48.16 48.16
10 48.16 00:48.16 48.16
11 48.16 00:48.16 48.16
12 48.31 00:48.31 48.31
13 49.01 00:49.01 49.01
14 49.27 00:49.27 49.27
15 49.33 00:49.33 49.33
16 49.40 00:49.40 49.40
17 49.48 00:49.48 49.48
18 49.51 00:49.51 49.51
19 52.85 00:52.85 52.85
20 52.89 00:52.89 52.89
21 53.14 00:53.14 53.14
22 54.31 00:54.31 54.31
23 54.63 00:54.63 54.63
24 56.91 00:56.91 56.91
25 1:18.39 1:18.39 78.39
26 1:20.26 1:20.26 80.26
27 1:38.30 1:38.30 98.30

Answer 2

as.difftime, and a quick regex to add the minutes when they are not present, should handle it:
as.difftime(sub("(^\\d{1,2}\\.)", "0:\\1", times), format="%M:%OS")
#Time differences in secs
# [1] 47.45 47.69 47.69 47.82 47.84 47.92 47.96 48.13 48.16 48.16 48.16 48.31
#[13] 49.01 49.27 49.33 49.40 49.48 49.51 52.85 52.89 53.14 54.31 54.63 56.91
#[25] 78.39 80.26 98.30

Answer 3

You can use separate in the Tidyverse tidyr package to split the strings into minutes and seconds:
library(tidyr)
library(dplyr)
separate(tibble(times = times), times, sep = ":",
into = c("min", "sec"), fill = "left", convert = T) %>%
mutate(min = ifelse(is.na(min), 0, min),
seconds = 60 * min + sec)
# A tibble: 27 x 3
min sec seconds
<dbl> <dbl> <dbl>
1 0 47.4 47.4
2 0 47.7 47.7
3 0 47.7 47.7
4 0 47.8 47.8
5 0 47.8 47.8
6 0 47.9 47.9
7 0 48.0 48.0
8 0 48.1 48.1
9 0 48.2 48.2
10 0 48.2 48.2
# ... with 17 more rows
The new column seconds is the number of seconds, multiplying the number of minutes by 60.