16

我有一个带有多个时间序列向量的 data.frame 与 date:time 向量。我想绘制所有相关向量，垂直堆叠在具有相同 X 轴但唯一 Y 轴的单独图表上。与此类似的图表：在此处输入图像描述

我的数据如下所示：

 dt <- structure(list(DEPTH = c(156, 156.5, 157.4, 158.15, 158.8, 159.2, 
159.75, 160.35, 160.85, 161.1, 161.6, 162.05, 162.5, 162.65, 
163.15, 163.45, 163.55, 163.8, 163.65, 163.75, 163.8, 163.8, 
163.75, 164.45, 164.8, 165.35, 165.65, 165.75, 166.1, 166.75, 
167, 167.2, 167.65, 168, 168.8, 169.3, 169.7, 170.2, 170.65, 
170.9, 171.45, 171.65, 172, 172.1, 172.25, 173, 173.4, 173.9, 
174.2, 174.6, 175, 175.25, 175.45, 175.9, 176.25, 176.7, 177, 
177.15, 177.5, 178, 178.5, 179.05, 179.2, 180.7, 181.05, 181.25, 
181.5, 181.7, 182.1, 182.3, 182.35, 182.75, 183.1, 183.65, 184.3, 
184.6, 185.1, 185.15, 185.3, 185.15, 185.25, 185.3, 185.15), 
    Smooth.Vert.Speed = c(-0.550000000000011, -0.5, -0.900000000000006, 
    -0.75, -0.650000000000006, -0.399999999999977, -0.550000000000011, 
    -0.599999999999994, -0.5, -0.25, -0.5, -0.450000000000017, 
    -0.449999999999989, -0.150000000000006, -0.5, -0.299999999999983, 
    -0.100000000000023, -0.25, 0.150000000000006, -0.0999999999999943, 
    -0.0500000000000114, 0, 0.0500000000000114, -0.699999999999989, 
    -0.350000000000023, -0.549999999999983, -0.300000000000011, 
    -0.0999999999999943, -0.349999999999994, -0.650000000000006, 
    -0.25, -0.199999999999989, -0.450000000000017, -0.349999999999994, 
    -0.800000000000011, -0.5, -0.399999999999977, -0.5, -0.450000000000017, 
    -0.25, -0.549999999999983, -0.200000000000017, -0.349999999999994, 
    -0.0999999999999943, -0.150000000000006, -0.75, -0.400000000000006, 
    -0.5, -0.299999999999983, -0.400000000000006, -0.400000000000006, 
    -0.25, -0.199999999999989, -0.450000000000017, -0.349999999999994, 
    -0.449999999999989, -0.300000000000011, -0.150000000000006, 
    -0.349999999999994, -0.5, -0.5, -0.550000000000011, -0.149999999999977, 
    -1.5, -0.350000000000023, -0.199999999999989, -0.25, -0.199999999999989, 
    -0.400000000000006, -0.200000000000017, -0.049999999999983, 
    -0.400000000000006, -0.349999999999994, -0.550000000000011, 
    -0.650000000000006, -0.299999999999983, -0.5, -0.0500000000000114, 
    -0.150000000000006, 0.150000000000006, -0.0999999999999943, 
    -0.0500000000000114, 0.150000000000006), DIVE_SURF = c("dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21", "dive21", "dive21", 
    "dive21", "dive21", "dive21", "dive21"), X = c(2050L, 2062L, 
    2026L, 2078L, 2058L, 2076L, 2050L, 2068L, 2060L, 2078L, 2058L, 
    2088L, 2080L, 2065L, 2088L, 2076L, 2084L, 2105L, 2084L, 2102L, 
    2123L, 2096L, 2074L, 2054L, 2090L, 2089L, 2080L, 2078L, 2068L, 
    2092L, 2084L, 2082L, 2094L, 2056L, 2062L, 2067L, 2082L, 2084L, 
    2091L, 2058L, 2076L, 2098L, 2104L, 2090L, 2058L, 2050L, 2080L, 
    2074L, 2074L, 2082L, 2070L, 2088L, 2062L, 2062L, 2082L, 2086L, 
    2070L, 2081L, 2092L, 2058L, 2060L, 2076L, 2094L, 2083L, 2072L, 
    2107L, 2104L, 2066L, 2110L, 2104L, 2072L, 2076L, 2065L, 2042L, 
    2066L, 2093L, 2080L, 2083L, 2108L, 2107L, 2086L, 2096L, 2126L
    ), Y = c(2036L, 2000L, 2049L, 1966L, 2042L, 2078L, 2072L, 
    2055L, 2036L, 2128L, 2044L, 2112L, 2066L, 2051L, 2102L, 2060L, 
    2054L, 2043L, 2034L, 2086L, 1980L, 2076L, 2003L, 2033L, 2107L, 
    1992L, 2028L, 2027L, 2024L, 2005L, 2050L, 2010L, 1944L, 2010L, 
    2046L, 2020L, 2088L, 2086L, 2034L, 2066L, 2060L, 2152L, 2044L, 
    2078L, 2040L, 2067L, 2080L, 2072L, 2073L, 2028L, 2066L, 2082L, 
    2030L, 2042L, 1990L, 2076L, 2054L, 2064L, 2016L, 2048L, 2029L, 
    2008L, 2090L, 2038L, 2026L, 2096L, 2002L, 2025L, 2001L, 2098L, 
    2061L, 2022L, 2054L, 2064L, 2043L, 2090L, 2042L, 2086L, 2073L, 
    2066L, 2040L, 2081L, 2087L), Z = c(2488L, 2484L, 2490L, 2486L, 
    2488L, 2492L, 2498L, 2490L, 2492L, 2484L, 2491L, 2494L, 2497L, 
    2493L, 2488L, 2493L, 2494L, 2484L, 2486L, 2487L, 2478L, 2490L, 
    2478L, 2493L, 2490L, 2486L, 2488L, 2486L, 2488L, 2482L, 2488L, 
    2480L, 2480L, 2488L, 2490L, 2490L, 2490L, 2489L, 2492L, 2490L, 
    2486L, 2480L, 2488L, 2491L, 2486L, 2488L, 2488L, 2494L, 2490L, 
    2488L, 2492L, 2498L, 2484L, 2491L, 2480L, 2491L, 2497L, 2487L, 
    2482L, 2490L, 2490L, 2478L, 2488L, 2492L, 2492L, 2482L, 2484L, 
    2489L, 2482L, 2484L, 2485L, 2492L, 2488L, 2493L, 2487L, 2490L, 
    2492L, 2488L, 2490L, 2487L, 2484L, 2486L, 2478L)), .Names = c("DEPTH", 
"Smooth.Vert.Speed", "DIVE_SURF", "X", "Y", "Z"), row.names = 7222:7304, class = "data.frame")

我希望在具有共同 X 轴的单独图表上绘制 DEPTH、X、Y 和 Z。

4

5 回答 5

14

我同意@PaulHiemstra，ggplot2 是要走的路。

假设是Smooth.Vert.Speed您想要绘制的公共 x 轴变量DEPTH，X...YZ

library(ggplot2)
library(reshape2)

# Add time variable as per @BenBolker's suggestion
dt$time <- seq(nrow(dt))

# Use melt to reshape data so values and variables are in separate columns
dt.df <- melt(dt, measure.vars = c("DEPTH", "X", "Y", "Z"))

ggplot(dt.df, aes(x = time, y = value)) +
  geom_line(aes(color = variable)) +
  facet_grid(variable ~ ., scales = "free_y") +
  # Suppress the legend since color isn't actually providing any information
  opts(legend.position = "none")

针对一个常见的 x 变量绘制多个 y 变量

于 2012-08-03T13:09:49.307 回答
8

只是为了与众不同，让我提一个既不涉及 lattice 也不涉及 ggplot2 的解决方案——几年前我将其作为条目 65发布到 Romain 的 R Graph Gallery ，代码为 here。它只是将图表堆叠起来，使用par()设置来保持它们堆叠。

请注意，垂直尺寸因选择而异，它们也可以很容易地具有相同的高度。

在此处输入图像描述

于 2012-08-03T13:32:51.800 回答
6

如果你想成为老式的，你可以使用lattice. 与@aaronwolen 不同，我假设数据集中缺少一个time变量，所以我做了一个：

dt$time <- seq(nrow(dt))
library(reshape2)
mm <- melt(subset(dt,select=c(time,DEPTH,X,Y,Z)),id.var="time")
library(lattice)
xyplot(value~time|variable,data=mm,type="l",
       scales=list(y=list(relation="free")),
       layout=c(1,4))

在此处输入图像描述

于 2012-08-03T13:19:05.697 回答
5

实际上，我已经找到了另一种使用 zoo 库的有趣方法：

library(zoo)
z <- with(dt, zoo(cbind(DEPTH, X, Y, Z),as.POSIXct(time))) 
plot.zoo(z,  ylab=c("Depth (m)", "Pitch Angle (degrees)", "Swaying Acceleration (m/s^2)", "Heaving Acceleration (m/s^2)"), col=c("black", "blue", "darkred", "darkgreen"), 
     xlab = c("Time"), lwd=2, ylim=list((rev(range(dt$DEPTH))), c(-90,90), c(-10,10), c(-10,10)))

因此，在动物园图内，您可以将新轴标签创建为列表形式，并且所有图都可以有不同的颜色。

于 2012-08-03T16:19:44.653 回答
0

请阅读这个例子：

生成示例数据：

dt = read_table("Time        A       B       C       D
10:12:54    2376.2  1.462   3.462   48
10:12:55    2410    1.462   3.462   48
10:12:56    2400    1.462   3.462   48
10:12:57    2409    1.462   3.462   48.6
10:12:58    2400    1.462   3.462   48.6
10:12:59    2385.1  1.462   3.462   46.6
10:13:00    2400    1.462   3.462   46.6
10:13:01    2410    1.462   3.462   46.6
10:13:02    2400    1.462   3.462   46.6
10:13:03    2106    1.463   3.463   46.6
10:13:04    2406    1.463   3.463   44.8
10:13:05    2376.2  1.463   3.463   44.8
10:13:06    2406    1.463   3.463   44.8
10:13:07    2400    1.463   3.463   44.8")
dt$Time=as.POSIXct(dt$Time)

如果你想快速绘制它，试试这个：

library(foqat)
geom_ts_batch(dt, panelgap=4)

在此处输入图像描述

如果您想以更大的自由度绘制它，请尝试以下操作：

library(foqat)
library(patchwork)
blankx=theme(axis.title.x=element_blank(),axis.text.x=element_blank(),axis.ticks.x=element_blank())
p2=geom_ts(dt, yl=2, llist=2, lcc="blue", yllab="A")+blankx
p3=geom_ts(dt, yl=3, llist=3, lcc="red", yllab="B")+blankx
p4=geom_ts(dt, yl=4, llist=4, lcc="green", yllab="C")+blankx
p5=geom_ts(dt, yl=5, llist=5, lcc="grey", yllab="D", xlab="Time")
p2/p3/p4/p5

在此处输入图像描述

于 2022-02-01T07:20:11.927 回答