将秒转换为分钟

Question

我是 R 新手，请帮助我理解从秒到分钟的转换。

\n

I'm doing a travel time analysis, and I get the result in seconds, which is not convenient for descriptive analysis and subsequent visualizations. There are a lot of calculations in different time sections, here is a small piece:

\n

trip_stats <- cyclistic_df %>%\n  group_by(member_casual) %>% \n  summarise(average_ride_length = round((mean(ride_length), 2),  # average ride length (total ride time / trips)\n            median_ride_length = round(median(ride_length), 2),  # median ride length\n            min_ride_length = round(min(ride_length), 2),  # minimum ride length\n            max_ride_length = round(max(ride_length), 2))  # maximum ride length\nhead(trip_stats)\n

\n

Conclusion:

\n

# A tibble: 2 \xc3\x97 5\n  member_casual average_ride_length median_ride_length min_ride_length max_ride_length\n  <chr>         <drtn>              <drtn>             <drtn>          <drtn>         \n1 casual        22.72 secs          785 secs           1 secs          1922127 secs   \n2 member        12.19 secs          525 secs           1 secs            89872 secs\n

\n

Or here's another example:

\n

# Average ride length (ride_length):\nride_lengt_avg <- round(mean(cyclistic_df$ride_length), 2)\nprint(ride_lengt_avg)\n

\n

Conclusion:

\n

Time difference of 977.28 secs\n

\n

I tried different variants with as_hms, format "%M:%S", minutes(), but unfortunately it still outputs the results of calculations in seconds. For example, I have created a number of columns, among which there is a column with only hour:

\n

# Format time as HH:MM:SS:\ncyclistic_df$time <- format(as.Date(cyclistic_df$date), "%H:%M:%S")\n\n# Create new column for time:\ncyclistic_df$time <- as_hms((cyclistic_df$started_at))\n\n# Create new column for hour:\ncyclistic_df$hour <- hour(cyclistic_df$time)\n

\n

Conclusion:

\n

0      1      2      3      4      5      6      7      8      9     10     11     12     13     14     15     16     17\n  casual  32053  20813  12246   6763   4515   8707  23390  40346  55066  56190  72233  93437 109905 114531 121187 134839 153614 170534\n  member  25582  15645   8736   5268   6097  26049  81660 151664 180585 120035 109540 129717 149403 147804 148800 183330 249039 297724\n        \n             18     19     20     21     22     23\n  casual 148452 111916  81388  69348  61476  44808\n  member 233303 165938 115161  88668  65505  41332\n

\n

Unfortunately by analogy I can't represent the output in minutes instead of seconds.

\n

Dividing by 60 also doesn't help, and is only misleading because "secs" is retained.

\n

I have added the following columns:

\n

# Default format is yyyy-mm-dd, use start date:\ncyclistic_df$date <- as.Date(cyclistic_df$started_at)  # the default format is yyyy-mm-dd\n\n# Create column for year:\ncyclistic_df$year <- format(as.Date(cyclistic_df$date), "%Y")\n\n# Create column for month:\ncyclistic_df$month <- format(as.Date(cyclistic_df$date), "%m")\n\n# Create column for day:\ncyclistic_df$day <- format(as.Date(cyclistic_df$date), "%d")\n\n# Calculate the day of the week:\ncyclistic_df$day_of_week <- wday(cyclistic_df$started_at)\n\n# Create column for day of week:\ncyclistic_df$day_of_week <- format(as.Date(cyclistic_df$date), "%A")  # wday(cyclistic_df$started_at, label = T, abbr = T)\n\n# Format time as HH:MM:SS:\ncyclistic_df$time <- format(as.Date(cyclistic_df$date), "%H:%M:%S")\n\n# Create new column for time:\ncyclistic_df$time <- as_hms((cyclistic_df$started_at))\n\n# Create new column for hour:\ncyclistic_df$hour <- hour(cyclistic_df$time)\n\n# Calculate & Create ride length column by subtracting ended_at time from started_at time and converted it to minutes:\ncyclistic_df$ride_length <- as_hms(difftime(cyclistic_df$ended_at, cyclistic_df$started_at))\n

\n

如果我理解正确的话，那么可以一次性使用数字类型创建年、月、日、小时、骑行长度列，这样以后就不必每次都将它们转换为数字。\n最好不要使用格式（，“%__”），对吧？

\n

我将非常感谢您的帮助！

\n

Answer 1

as.numeric接受units=：

set.seed(42)
d <- (Sys.time()) - (Sys.time() - runif(10, 0, 99))
d
# Time differences in secs
#  [1] 90.56578 92.77045 28.32780 82.21430 63.53279 51.39048 72.92223 13.33198 65.04222 69.80140
as.numeric(d, units = "mins")
#  [1] 1.5094297 1.5461741 0.4721299 1.3702383 1.0588798 0.8565080 1.2153704 0.2221996 1.0840370 1.1633566

您可以使用类似的内容将其应用于所有这些列（假设dplyr）：

trip_stats <- trip_stats %>%
  mutate(across(ends_with("_length"), ~ as.numeric(., units = "mins")))

或者更一般地说

trip_stats <- trip_stats %>%
  mutate(across(where(~ inherits(., "difftime")), ~ as.numeric(., units = "mins")))

如果以 R 为基础，您可以执行以下操作：

isdifftm <- sapply(trip_stats, inherits, "difftime")
trip_stats[,isdifftm] <- lapply(trip_stats[,isdifftm,drop=FALSE], as.numeric, units = "mins")

编辑

也许这就是完整的？

trip_stats <- cyclistic_df %>%
  group_by(member_casual) %>% 
  summarize(across(disp, .names = "{.fn}", list(
    average_ride_length = mean,
    median_ride_length = median,
    min_ride_length = min,
    max_ride_length = max))) %>%
  mutate(across(ends_with("_length"), ~ round(as.numeric(., units = "mins"), 2)))