如何定义散点图的固定纵横比 [英] How to define fixed aspect-ratio for scatter-plot

问题描述

我正在绘制在来自两个种群的每个个体中测量的两种同位素的相关系数(值 = 0.0:1.0).我希望我的散点图有一个固定的纵横比，这样无论图形设备如何，x 轴和 y 轴的大小都完全相同.建议?

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?

这是我在 R 中的第一个情节，对改进我的代码的任何评论表示赞赏?最后，是否值得投资学习基本的绘图技术，还是应该直接跳到 ggplot2 或lattice?

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?

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)

推荐答案

使用 asp=1 作为绘图参数将被低级 plot.window 调用解释，并且应该给你一个统一的纵横比.使用 ylim 和 xlim 的调用有可能与纵横比规范冲突，并且 asp 应该占上风".顺便说一句，这是一个非常令人印象深刻的第一个 R 图.以及出色的问题构建.高分.

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 the asp should "prevail". That's a very impressive first R graph, by the away. And an excellent question construction. High marks.

一个不和谐的地方是你对结构 xlim=c(0:1.0) 的使用.因为 xlim 需要一个二元素向量，所以我期望 xlim=c(0,1).如果您更改为一组不同的限制，那么击键次数会减少，并且将来出错的可能性也会降低，因为如果您尝试使用0:2.5"，:"运算符会给您带来意想不到的结果.

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".

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