针对 R 中的数字向量绘制字符向量 [英] Plot a character vector against a numeric vector in R

问题描述

我在 R 中有以下数据框:

I have the following data frame in R:

>AcceptData
  Mean.Rank Sentence.Type
1       2.5       An+Sp+a
2       2.6      An+Nsp+a
3       2.1       An+Sp-a
4       3.1      An+Nsp-a
5       2.4       In+Sp+a
6       1.7      In+Nsp+a
7       3.1       In+Sp-a
8       3.0      In+Nsp-a

我想绘制的图，x 轴为 Sentence.Type 列，每个单元格的实际名称作为 x 轴上的一个点.我希望 y 轴以 0.5 的步长从 1 到 4

Which I want to plot, with the Sentence.Type column in the x axis, with the actual name of each cell as a point in the x axis. I want the y axis to go from 1 to 4 in steps of .5

到目前为止，我无法绘制此图，无论是使用 plot() 还是 hist().我不断收到不同类型的错误，主要是因为 data.frame 中字符列的性质.

So far I haven't been able to plot this, neither with plot() not with hist(). I keep getting different types of errors, mainly because of the nature of the character column in the data.frame.

我知道这对大多数人来说应该很容易，但我仍然对 R 有点菜鸟，并且在下班后我无法正确理解情节.任何帮助深表感谢.

I know this should be easy for most, but I'm sort of noob with R still and after hours I can't get the plot right. Any help is much appreciated.

我遇到的一些错误:

> hist(AcceptData$Sentence.Type,AcceptData$Mean.Rank)
Error in hist.default(AcceptData$Sentence.Type, AcceptData$Mean.Rank) : 
  'x' must be numeric

或者:(这不会产生错误，但绝对不是我想要的图表.它的所有 x 值都挤在 x 轴的左侧)

Or: (this doesn't give an error, but definitely not the graph I want. It has all the x values cramped to the left of the x axis)

  plot(AcceptData$Sentence.Type,AcceptData$Mean.Rank,lty=5,lwd=2,xlim=c(1,16),ylim=c(1,4),xla b="Sentence Type",ylab="Mean Ranking",main="Mean Acceptability Ranking per Sentence")

推荐答案

默认的 plot 函数有一个方法可以让你在 x 轴上绘制因子，但是要使用它，你必须将您的文本数据转换为 factor:

The default plot function has a method that allows you to plot factors on the x-axis, but to use this, you have to convert your text data to a factor:

这是一个例子:

x <- letters[1:5]
y <- runif(5, 0, 5)

plot(factor(x), y)

以及您的示例数据:

AcceptData <- read.table(text="
Mean.Rank Sentence.Type
1       2.5       An+Sp+a
2       2.6      An+Nsp+a
3       2.1       An+Sp-a
4       3.1      An+Nsp-a
5       2.4       In+Sp+a
6       1.7      In+Nsp+a
7       3.1       In+Sp-a
8       3.0      In+Nsp-a", stringsAsFactors=FALSE)

plot(Mean.Rank~factor(Sentence.Type), AcceptData, las=2, 
     xlab="", main="Mean Acceptability Ranking per Sentence")

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