一天中时间的直方图或密度图

本文介绍了一天中时间的直方图或密度图的处理方法，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

我有data.frame，其中的每一行都是带有开始和结束时间戳记的情节。

I have the data.frame in which every row is an episode with a start and an end timestamp.

test.DF<-dput(head(test.DF, n=50)) structure(list(start = structure(c(1189494920, 1189495400, 1189496120, 1189496840, 1189497440, 1189498040, 1189498640, 1189501760, 1189503560, 1190453600, 1247458520, 1247480840, 1247482880, 1247483840, 1247485040, 1247486600, 1247487320, 1247488040, 1247488760, 1247490920, 1247491280, 1247492480, 1247493680, 1247502440, 1247503160, 1247503520, 1247548040, 1247549360, 1247550680, 1247552600, 1247553920, 1247557400, 1247558000, 1247558480, 1247559440, 1247560400, 1247563760, 1247564960, 1247566640, 1247567120, 1194935549, 1194936029, 1195722629, 1195724309, 1199691029, 1199692349, 1202560229, 1208063669, 1208322989, 1188188112), class = c("POSIXct", "POSIXt"), tzone = ""), end = structure(c(1189495280, 1189495520, 1189496360, 1189497080, 1189497560, 1189498160, 1189498760, 1189501880, 1189503920, 1190453720, 1247458640, 1247480960, 1247483480, 1247484080, 1247485640, 1247486840, 1247487560, 1247488640, 1247490440, 1247491160, 1247491520, 1247492600, 1247493920, 1247502680, 1247503400, 1247504120, 1247549240, 1247550560, 1247551280, 1247552720, 1247554400, 1247557880, 1247558240, 1247559080, 1247559560, 1247560760, 1247563880, 1247565080, 1247566760, 1247567240, 1194935669, 1194936269, 1195722749, 1195724429, 1199691269, 1199692469, 1202560349, 1208063789, 1208323109, 1188204792 ), class = c("POSIXct", "POSIXt"), tzone = "")), .Names = c("start", "end"), row.names = c(NA, 50L), class = "data.frame")

我想看看这些情节在24小时周期内的分布情况。这可以是直方图，也可以是密度图，x轴为24H日循环。这可能吗？我想忽略情节的日期。

I would like to see the distribution of these episodes within a 24 hour cycle. That is either a histogram or a density plot, with the 24H day cycle in the x axis. Is this possible? I would like to ignore the dates of the episodes.

推荐答案

通过转换为 POSIXlt format，您可以轻松提取时间中的小时：

By converting to a POSIXltformat, you can easily extract the hour of the time:

par(mar=c(6,4,1,1)) Hour <- as.POSIXlt(test.DF$start)$hour hist(Hour, breaks=seq(0, 23), main="Start time (hour)")

fun <- function(start.time, end.time){ seq.POSIXt( as.POSIXlt( paste0("2000-01-01 ", as.POSIXlt(start.time)$hour, ":", as.POSIXlt(start.time)$min) ), as.POSIXlt( paste0("2000-01-01 ", as.POSIXlt(end.time)$hour, ":", as.POSIXlt(end.time)$min) ), by="min" ) } HM <- vector(mode="list", dim(test.DF)[1]) for(i in seq(HM)){ HM[[i]] <- fun(test.DF$start[i], test.DF$end[i]) } HM2 <- as.POSIXlt(unlist(HM), origin="1970-01-01") Hour <- HM2$hour hist(Hour, breaks=seq(0, 23)) HourMinute <- HM2$hour + HM2$min/60 hist(HourMinute, breaks=seq(0, 23, by=1/60))