如何为(基本R)散点图定义固定的方面比例
我正在绘制相关系数(值= 0.0:1.0)的两个同位素在每个人群中测得的两个同位素。我想为我的散点图有一个固定的方面比例,以使x-和y轴与图形设备的大小完全相同。建议?
这是我在R中的第一个情节,对我的代码的改进有任何评论吗?最后,是否值得投资学习基本的绘图技术,或者我应该直接跳到GGPLOT2或晶格?
我的情节脚本:
## Create dataset
WW_corr <-
structure(list(South_N15 = c(0.7976495, 0.1796725, 0.5338347,
0.4103769, 0.7447027, 0.5080296, 0.7566544, 0.7432026, 0.8927161
), South_C13 = c(0.76706752, 0.02320767, 0.88429902, 0.36648357,
0.73840937, 0.0523504, 0.52145159, 0.50707858, 0.51874445), North_N15 = c(0.7483608,
0.4294148, 0.9283554, 0.8831571, 0.5056481, 0.1945943, 0.8492716,
0.5759033, 0.7483608), North_C13 = c(0.08114805, 0.47268136,
0.94975596, 0.06023815, 0.33652839, 0.53055943, 0.30228833, 0.8864435,
0.08114805)), .Names = c("South_N15", "South_C13", "North_N15",
"North_C13"), row.names = c(NA, -9L), class = "data.frame")
opar <- par()
## Plot results
par(oma = c(1, 0, 0, 0), mar = c(4, 5, 2, 2))
plot(1,1,xlim=c(0:1.0), ylim=c(0:1.0), type="n", las=1, bty="n", main = NULL,
ylab=expression(paste("Correlation Coefficient (r) for ", delta ^{15},"N ",
"\u0028","\u2030","\u0029")),
xlab=expression(paste("Correlation Coefficient (r) for ", delta ^{13},"C ",
"\u0028","\u2030","\u0029")))
points(WW_corr$South_N15, WW_corr$South_C13, pch = 23, cex = 1.25,
bg ="antiquewhite4", col = "antiquewhite4")
points(WW_corr$North_N15, WW_corr$North_C13, pch = 15, cex = 1.25,
bg ="black")
axis(1, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
axis(2, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
abline(h=.86, v=.86, col = "gray60", lty = 2)
legend("topleft", c("North", "South"), pch = c(15, 23),
col = c("black", "antiquewhite4"), pt.bg = c("black", "antiquewhite4"),
horiz=TRUE, bty = "n")
par(opar)
I am plotting correlation coefficients (values = 0.0:1.0) for two isotopes measured in each individual from two populations. I would like to have a fixed aspect-ratio for my scatter-plot so that the x- and y-axis are exactly the same size no matter the graphics device. Suggestions?
This is my first plot in R, any comments on refinements to my code is appreciated? Finally, is it worth investing in learning the basic plotting techniques or should I jump right to ggplot2 or lattice?
My plot script:
## Create dataset
WW_corr <-
structure(list(South_N15 = c(0.7976495, 0.1796725, 0.5338347,
0.4103769, 0.7447027, 0.5080296, 0.7566544, 0.7432026, 0.8927161
), South_C13 = c(0.76706752, 0.02320767, 0.88429902, 0.36648357,
0.73840937, 0.0523504, 0.52145159, 0.50707858, 0.51874445), North_N15 = c(0.7483608,
0.4294148, 0.9283554, 0.8831571, 0.5056481, 0.1945943, 0.8492716,
0.5759033, 0.7483608), North_C13 = c(0.08114805, 0.47268136,
0.94975596, 0.06023815, 0.33652839, 0.53055943, 0.30228833, 0.8864435,
0.08114805)), .Names = c("South_N15", "South_C13", "North_N15",
"North_C13"), row.names = c(NA, -9L), class = "data.frame")
opar <- par()
## Plot results
par(oma = c(1, 0, 0, 0), mar = c(4, 5, 2, 2))
plot(1,1,xlim=c(0:1.0), ylim=c(0:1.0), type="n", las=1, bty="n", main = NULL,
ylab=expression(paste("Correlation Coefficient (r) for ", delta ^{15},"N ",
"\u0028","\u2030","\u0029")),
xlab=expression(paste("Correlation Coefficient (r) for ", delta ^{13},"C ",
"\u0028","\u2030","\u0029")))
points(WW_corr$South_N15, WW_corr$South_C13, pch = 23, cex = 1.25,
bg ="antiquewhite4", col = "antiquewhite4")
points(WW_corr$North_N15, WW_corr$North_C13, pch = 15, cex = 1.25,
bg ="black")
axis(1, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
axis(2, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
abline(h=.86, v=.86, col = "gray60", lty = 2)
legend("topleft", c("North", "South"), pch = c(15, 23),
col = c("black", "antiquewhite4"), pt.bg = c("black", "antiquewhite4"),
horiz=TRUE, bty = "n")
par(opar)
如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱,如果其他用户或者作者回复了您的评论,将不能在第一时间通知您!
发布评论
评论(2)
将情节类型设置为正方形,在您的情况下,它将完成工作(我认为),因为您的X和Y范围是相同的。相当隐藏的选项已记录在?par中。
sets the plot type to be square, which will do the job (I think) in your case because your x and y ranges are the same. Fairly well hidden option documented in ?par.
使用
asp = 1
作为绘图的参数,将由低级绘图来解释。窗口调用,应为您提供单一的长宽比。使用Ylim和Xlim的调用可能与纵横比分类相抵触,而ASP
应该“占上风”。这是一个令人印象深刻的第一r图。以及一个很好的问题构造。高分。一个令人震惊的说明是您使用构造
xlim = c(0:1.0)
。由于Xlim期望有两个元素向量,因此我会期望Xlim = C(0,1)。如果您更改为不同的限制,则击键较少,并且将来会遇到错误,因为“操作员”尝试使用“ 0:2.5”,将为您带来意外的结果。Using
asp=1
as a parameter to plot will get interpreted by the low-level plot.window call and should give you a unitary aspect ratio. There is the potential that a call using ylim and xlim could conflict with an aspect ratio scpecification and theasp
should "prevail". That's a very impressive first R graph, by the away. And an excellent question construction. High marks.The one jarring note was your use of the construction
xlim=c(0:1.0)
. Since xlim expects a two element vector, I would have expected xlim=c(0,1). Fewer keystrokes and less subject to error in the future if you changed to a different set of limits, since the ":" operator would give you unexpected results if you tried that with "0:2.5".