本文介绍了在R中的分类日期的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧! 问题描述 这是我的数据集: > str(temp.df)'data.frame':74602 obs。的2个变量: $ time:POSIXct,格式:2011-04-09 03:53:202011-04-09 03:53:152011-04-09 03:53:07 2011-04-09 03:52:39... $ value:num 1 1 1 1 1 1 1 1 1 1 ... >头(temp.df $ time,n = 10) [1]2011-04-09 03:53:20 EDT2011-04-09 03:53:15 EDT2011-04- 09 03:53:07 EDT2011-04-09 03:52:39 EDT [5]2011-04-09 03:52:29 EDT2011-04-09 03:51 :56 EDT2011-04-09 03:51:54 EDT2011-04-09 03:51:46 EDT [9]2011-04-09 03:51:44 EDT 2011-04-09 03:51:26 EDT 为了方便... > dput(head(temp.df $ time,n = 10))结构(c(1302335600,1302335595,1302335587,1302335559,1302335549, 1302335516,1302335514,1302335506,1302335504,1302335486),class = c(POSIXct,POSIXt),tzone =) 我正在寻找: 如何找到最小和最大日期/时间之间的时间? 使用1小时时间段创建数据摘要的最佳方式是什么? 任何帮助,您可以提供将不胜感激解决方案使用正确的时间序列包动物园和/或 xts 。这个例子是直接从 aggregate.zoo()的帮助页面,它每10分钟聚合POSIXct秒数据 tt x< - zoo(tt,structure(tt,class = c(POSIXt ,POSIXct))) aggregate(x,time(x) - as.numeric(time(x))%% 600,mean) to.period()函数。 org / package = xtsrel =nofollow noreferrer> xts 也是一个肯定的赢家。在这里有关于SO和r-sig财务列表的无数例子。 I struggle with dates and times in R, but I am hoping this is a fairly basic task.Here is my dataset:> str(temp.df)'data.frame': 74602 obs. of 2 variables: $ time : POSIXct, format: "2011-04-09 03:53:20" "2011-04-09 03:53:15" "2011-04-09 03:53:07" "2011-04-09 03:52:39" ... $ value: num 1 1 1 1 1 1 1 1 1 1 ...> head(temp.df$time, n=10) [1] "2011-04-09 03:53:20 EDT" "2011-04-09 03:53:15 EDT" "2011-04-09 03:53:07 EDT" "2011-04-09 03:52:39 EDT" [5] "2011-04-09 03:52:29 EDT" "2011-04-09 03:51:56 EDT" "2011-04-09 03:51:54 EDT" "2011-04-09 03:51:46 EDT" [9] "2011-04-09 03:51:44 EDT" "2011-04-09 03:51:26 EDT"and for convenience...> dput(head(temp.df$time, n=10))structure(c(1302335600, 1302335595, 1302335587, 1302335559, 1302335549, 1302335516, 1302335514, 1302335506, 1302335504, 1302335486), class = c("POSIXct", "POSIXt"), tzone = "")What I am looking to do:How can I find how many hours are between the min and max date/time?What's the best way to create summaries of my data using 1-hour time buckets?Any help you can provide will be greatly appreciated 解决方案 Use the proper time series packages zoo and/or xts. This example is straight from the help pages of aggregate.zoo() which aggregates POSIXct seconds data every 10 minutes tt <- seq(10, 2000, 10) x <- zoo(tt, structure(tt, class = c("POSIXt", "POSIXct"))) aggregate(x, time(x) - as.numeric(time(x)) %% 600, mean)The to.period() function in xts is also a sure winner. There are countless examples here on SO and on the r-sig-finance list. 这篇关于在R中的分类日期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! 10-26 15:52