从 POSIXct 中提取小时和秒以在 R 中绘图 [英] extract hours and seconds from POSIXct for plotting purposes in R

问题描述

假设我有以下 data.frame foo

           start.time duration
1 2012-02-06 15:47:00      1
2 2012-02-06 15:02:00      2
3 2012-02-22 10:08:00      3
4 2012-02-22 09:32:00      4
5 2012-03-21 13:47:00      5

然后 class(foo$start.time) 返回

[1] "POSIXct" "POSIXt" 

我想创建一个 foo$duration 与 foo$start.time 的图.在我的场景中，我只对一天中的时间感兴趣，而不是一年中的实际日期.如何从 POSIXct 类向量中提取一天中的时间为 hours:seconds?

I'd like to create a plot of foo$duration v. foo$start.time. In my scenario, I'm only interested in the time of day rather than the actual day of the year. How does one go about extracting the time of day as hours:seconds from POSIXct class of vector?

推荐答案

这是一个很好的问题，突出了在 R 中处理日期的一些困难.lubridate 包非常方便，所以下面我介绍两种方法，一种使用碱(如@RJ- 所建议的)，另一种使用 lubridate.

This is a good question, and highlights some of the difficulty in dealing with dates in R. The lubridate package is very handy, so below I present two approaches, one using base (as suggested by @RJ-) and the other using lubridate.

重新创建原始帖子中的(前两行)数据框:

Recreate the (first two rows of) the dataframe in the original post:

foo <- data.frame(start.time = c("2012-02-06 15:47:00", 
                                 "2012-02-06 15:02:00",
                                 "2012-02-22 10:08:00"),
                  duration   = c(1,2,3))

转换为 POSIXct 和 POSIXt 类(两种方法)

Convert to POSIXct and POSIXt class (two ways to do this)

# using base::strptime
t.str <- strptime(foo$start.time, "%Y-%m-%d %H:%M:%S")

# using lubridate::ymd_hms
library(lubridate)
t.lub <- ymd_hms(foo$start.time)

现在，将时间提取为十进制小时

Now, extract time as decimal hours

# using base::format
h.str <- as.numeric(format(t.str, "%H")) +
               as.numeric(format(t.str, "%M"))/60

# using lubridate::hour and lubridate::minute
h.lub <- hour(t.lub) + minute(t.lub)/60

证明这些方法是相同的:

Demonstrate that these approaches are equal:

identical(h.str, h.lub)

然后选择上述方法之一将十进制小时分配给foo$hr:

Then choose one of above approaches to assign decimal hour to foo$hr:

foo$hr <- h.str

# If you prefer, the choice can be made at random:
foo$hr <- if(runif(1) > 0.5){ h.str } else { h.lub }

然后使用 ggplot2 包进行绘图:

then plot using the ggplot2 package:

library(ggplot2)
qplot(foo$hr, foo$duration) + 
             scale_x_datetime(labels = "%S:00")

这篇关于从 POSIXct 中提取小时和秒以在 R 中绘图的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！