如何绘制仅一天重叠的不同日期(基于小时:分钟)中R中的数据? [英] How to plot data in R from different days overlayed in only one day (based on hours:minutes)?
问题描述
我在使用R时遇到此问题:
I am having this problem with R:
我有一个名为"teste"的数据集，其中有一个日期"列(以POSIXct格式，格式为％Y-％m-％d％H:％M:％S")，每个列都有读数5个月内10分钟.
I have a dataset called "teste" that has a column for 'Date' (it's in POSIXct, format = "%Y-%m-%d %H:%M:%S"), which has readings every 10 minutes during a 5 months period.
我需要在不同的时间同时比较变量.例如，在数据集中的每个星期六绘制图形，并进行叠加.我已经有了用于设置data.frame的代码，并且只有星期六.
I need to make a comparison of the variables at the same time but in different days. For example plot every Saturday in the Dataset, overlayed. I already have the code for subsetting the data.frame and have only the Saturdays.
以下是数据示例:
DATE ID VAR1 VAR2 VAR3
1 2016-09-19 00:07:47 79 19 0 OPN
2 2016-09-19 00:17:47 79 18 1 OPN
3 2016-09-19 00:27:47 79 16 3 OPN
4 2016-09-19 00:37:47 79 15 4 OPN
5 2016-09-19 00:47:47 79 16 3 OPN
6 2016-09-19 00:57:47 79 16 3 OPN
这是来自数据的输出:
structure(list(FECHA = structure(c(1474236467，1474237067，1474237667，1474841253，1474841853，1474842453)，class = c("POSIXct"，"POSIXt")，tzone =")，ID = c(79L，79L，79L，79L，79L，79L)，SLOTS = c(19L，18L，16L，14L，15L，15L)，BIKES = c(0L，1L，3L，8L，7L，7L)，状态=结构(c(2L，2L，2L，2L，2L，2L)，.Label = c("CLS"，"OPN")，class ="factor"))，.Names = c("FECHA"，"ID"，"SLOTS"，"BIKES"，"STATUS")，row.names = c(1L，2L，3L，1004L，1005L，1006L)，class ="data.frame")
structure(list(FECHA = structure(c(1474236467, 1474237067, 1474237667, 1474841253, 1474841853, 1474842453), class = c("POSIXct", "POSIXt" ), tzone = ""), ID = c(79L, 79L, 79L, 79L, 79L, 79L), SLOTS = c(19L, 18L, 16L, 14L, 15L, 15L), BIKES = c(0L, 1L, 3L, 8L, 7L, 7L), STATUS = structure(c(2L, 2L, 2L, 2L, 2L, 2L), .Label = c("CLS", "OPN"), class = "factor")), .Names = c("FECHA", "ID", "SLOTS", "BIKES", "STATUS"), row.names = c(1L, 2L, 3L, 1004L, 1005L, 1006L ), class = "data.frame")
我尝试做:(使用的包裹lubridate)
I tried doing: (used package lubridate)
plot(as.POSIXct(paste(hour(teste$FECHA),":",minute(teste$FECHA),sep = ""), format = "%H:%M"),teste$BIKES)
它可以工作，但是使用粘贴"绝对不是实现此目的的最佳方法.可能有一种更简单，更优雅的方法，对吗?如果是，怎么办?
It works but using 'paste' is definetely not the best way for doing this. There probably is a easier and more elegant way, right? If yes, how?
如果我用type ="lines"绘制它，我会遇到一个问题，因为它不知道一天的最后一个读数不应该连接"到下一个读数的第一个读数，从而得出以下结果:(请参见从24小时到0小时跨越所有图表的线)
And I have a problem if I plot it with type = "lines", because it doesnt know that the last reading of a day shouldnt be 'connected' to the first of the next one, giving this result: (see the lines crossing all the graph from 24 hours to 0 hours)
我虽然想一次绘制一天，使用绘图然后使用线条函数，但是问题是我不知道每天有多少读数.每天应该有6 * 24 = 144个读数，但有些读数是143、142(由于获取数据时出现问题).
I though about ploting one day at a time, using the plot and then using lines functions, but the problem is that I dont know how many readings are there in each day. Each day should have 6*24=144 readings, but some have 143, 142 (because of problems while getting data).
感谢您的帮助.
推荐答案
您可以尝试
# add factor to differentiate between the different days
d$day <- factor(strftime(d$FECHA, format="%Y-%m-%d"))
# add hours, set one day
d$time <- as.POSIXct(strftime(d$FECHA, format="%H:%M:%S"), format="%H:%M:%S")
# color per day
ggplot(d, aes(x=time, y=BIKES, col=day)) +
geom_point() +
geom_line() +
scale_x_datetime(limits = as.POSIXct(c("2017-04-19 00:00:00", "2017-04-19 24:00:00")), date_labels = "%H:%M")
# or facet_wrap/grid
ggplot(d, aes(x=time, y=BIKES)) + geom_point() + geom_line() +
facet_grid(~day)+
scale_x_datetime(limits = as.POSIXct(c("2017-04-19 00:00:00", "2017-04-19 24:00:00")), date_labels = "%H:%M")
# or use the wrap function instead of the grid function
facet_wrap(~day, scales = "free_y")
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