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感谢您的阅读。我发现我无法从现有数据中绘制线图,如下所示:

a=structure(list(ID = structure(1:3, .Names = c("V2", "V3", "V4"
), .Label = c(" day1", " day2", " day3"), class = "factor"), 
    Protein1 = structure(c(3L, 1L, 2L), .Names = c("V2", 
    "V3", "V4"), .Label = c("-0.651129553", "-1.613977035", "-1.915631511"
    ), class = "factor"), Protein2 = structure(c(3L, 
    1L, 2L), .Names = c("V2", "V3", "V4"), .Label = c("-1.438858662", 
    "-2.16361761", "-2.427593862"), class = "factor")), .Names = c("ID", 
"Protein1", "Protein2"), row.names = c("V2", 
"V3", "V4"), class = "data.frame")

我需要的是绘制如下图:

在此处输入图像描述

我尝试了以下代码,但结果不正确;

qplot(ID, Protein1, data=a, colour=ID, geom="line")

还:

a1<-melt(a, id.vars="ID")
ggplot(a1,aes(ID,value))+ geom_line()+geom_point()

非常感谢您的照顾。

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1 回答 1

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首先,您必须修改 data.frame 的结构:Protein1&Protein2应该是数字而不是因子。

a$Protein1 = as.numeric(as.character(a$Protein1))
a$Protein2 = as.numeric(as.character(a$Protein2))

如果只想绘制“Protein1”,则不需要先使用 melt。

ggplot(a, aes(x = ID, y = Protein1)) + geom_point() + geom_line(aes(group = 1)) + ylim(-3,3)

group = 1允许与geom_line(): source连接点


现在,如果你想在同一个情节上看到Protein1& ,你可以使用:Protein2melt

a1<-melt(a, id.vars="ID")
ggplot(a1, aes(x = ID, y = value, group = variable, color = variable)) + geom_point() + geom_line() + ylim(-3,3)

在此处输入图像描述

于 2016-04-18T07:13:45.920 回答