如何绘制 R [xts] 中的某些列？

Question

我正在使用以下方法绘制它：

plot(fifty_twoweekmovavg)
pdf("52_week_moving_average_chartNSW.pdf",onefile=TRUE)
addLegend("topleft",lty = 1,cex=1.2)
dev.off()

如何绘制它以便只包含几个变量？例如。绘制新南威尔士州价格和煤炭价格与时间的关系图，而不是绘制每个变量与时间的关系图？

谢谢

可重现的例子：

NSW1.Price Black.Coal Gas Hydro Liquid.Fuel 2011-01-01 30.89336 32.33668 41.63653 69.82661 108.06855 2011-01-08 30.98103 32.24805 41.33295 69.44308 104.36587 2011-01-15 30.73076 32.11497 40.76273 69.59129 97.30812 2011-01-22 30.76028 30.50381 36.56215 62.50329 61.78828 2011-01-29 29.76733 34.65090 43.94289 93.20954 113.42410

Edit2，我是如何创建数据的：

mydata=read.csv(file="nem_tech_dataTAS.csv") library(xts) library(zoo) date <- seq(from=as.POSIXct("2010-01-01 00:30", format = "%Y-%m-%d %H:%M"), length.out = nrow(mydata), by = "30 min") mydata_dated <- xts(mydata, date) fifty_twoweekmovavg=rollapply(mydata_dated,17520,mean,by = 336,na.pad = FALSE)

Edit3，图例格式：

当前图例：

所需图例：

Answer 1

如果您想绘制两条线来显示随时间变化的天然气和水电值，您首先需要创建一个时间序列。创建一个获取日期的列，并使用 as.Date 将其转换为日期格式。在你上面的例子中，你会写：

fifty_twoweekmovavg$date=as.Date(rownames(fifty_twoweekmovavg))

这会得到你的 x 轴值。

现在，为了同时获得 Gas 和 Hydro 值，您必须确保 y 轴适合它们，因为 Gas 和 Hydro 的值不相交。

一种方法是：

extents=range(c(fifty_twoweekmovavg$Gas,fifty_twoweekmovavg$Hydro))

设置日期和范围后，您最终可以继续绘制线条：

plot(fifty_twoweekmovavg$date,fifty_twoweekmovavg$Gas,type='l',ylim=extents)
lines(fifty_twoweekmovavg$date,fifty_twoweekmovavg$Hydro,col='red')

Answer 2

下面的代码可以渲染带有图例的多行图。

df2 = data.frame(matrix(data=c(
30.89336,   32.33668, 41.63653,  69.82661,   108.06855,
30.98103,   32.24805, 41.33295,  69.44308,   104.36587,
30.73076,   32.11497, 40.76273,  69.59129,    97.30812,
30.76028,   30.50381, 36.56215,  62.50329,    61.78828,
29.76733,   34.65090, 43.94289,  93.20954,   113.42410
), ncol = 5, byrow = TRUE ))
colnames(df2) = c("NSW1.Price", "Black.Coal", "Gas", "Hydro", "Liquid.Fuel")
df1 = data.frame("time" = as.Date(
      c("2011-01-01",   
        "2011-01-08",   
        "2011-01-15",   
        "2011-01-22",   
        "2011-01-29"),"%Y-%m-%d"))
df = cbind(df1, df2)
plot(0, cex=0, xlim=range(df$time), 
     ylim=c(0,max(c(df$NSW1.Price, df$Black.Coal))))
lines(df$time, df$NSW1.Price, col="cyan", lty = 1)
lines(df$time, df$Black.Coal, col="black", lty=2)
legend("bottomleft", legend = c("NSW1.Price", "Black.Coal"), 
       col = c("cyan", "black"), lty = c(1,2))

Answer 3

如果已加载包 xts，则可以使用 plot 命令绘制 xts 对象。有关如何绘制 xts 对象的更多详细信息，请使用 ?plot.xts.

至select 只有 2 列您可以在 xts 对象内使用 grep。

library(xts)

plot(fifty_twoweekmovavg[, grep("NSW1|Coal", names(fifty_twoweekmovavg))],
     legend(grep("NSW1|Coal", names(fifty_twoweekmovavg))),
     main = "52_week_moving_average",
     legend.loc = "topleft")

编辑： 操纵图例，这使它更容易，并且会产生相同的情节，但图例用线条而不是正方形：

plot(fifty_twoweekmovavg[, grep("NSW1|Coal", names(fifty_twoweekmovavg))],
     main = "52_week_moving_average")

# on = 1 is for main plot. lty is for showing a line in the legend.
# see ?addLegend and ?legend
addLegend("topleft", on = 1, lty=1)

如何绘制 R [xts] 中的某些列？

How to plot certain columns in R [xts]?

graphing

r

zoo

xts