ggplot中的一组行的图例

Question

我正在尝试绘制一组线条，并为图例中的这些线条指定一个标签。这是我正在使用的数据（df2）的一个例子。

structure(list(
true = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), 
.Label = c("model1", "model2"), class = "factor"), 
test = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), 
.Label = c("false.model1", "false.model2", "false.model3", "true model"), class = "factor"), 
est.interval = c(0, 5, 5, 7, 7, 10, 10, 11, 11, 0, 2, 2, 3, 3, 0, 2, 2, 4, 4, 0, 2, 2, 3, 3, 4, 4, 0, 7, 7, 10, 10, 13, 13, 0, 4, 4, 5, 5, 0, 5, 5, 6, 6, 7, 7, 9, 9, 0, 2, 2, 3, 3, 4, 4), 
sens = c(1, 1, 0.75, 0.75, 0.5, 0.5, 0.25, 0.25, 0, 1, 1, 0.25, 0.25, 0, 1, 1, 0.5, 0.5, 0, 1, 1, 0.5, 0.5, 0.25, 0.25, 0, 1, 1, 0.75, 0.75, 0.25, 0.25, 0, 1, 1, 0.5, 0.5, 0, 1, 1, 0.75, 0.75, 0.5, 0.5, 0.25, 0.25, 0, 1, 1, 0.5, 0.5, 0.25, 0.25, 0)), 
.Names = c("true", "test", "est.interval", "sens"), 
class = "data.frame")

在这里，我使用ggplot绘制曲线。

ggplot(df2, aes(x=est.interval, y=sens, color=test)) +
    ylim(0,1) +
    geom_line(size = 0.6) +
    facet_wrap(~true, nrow=1) +
    scale_colour_manual(values=c(rep('#CCCCCC', 3), "#000000"), name="Estimation Model") + 
    guides(color=guide_legend(keywidth = 3, keyheight = 1))

我想要三个灰色＆＃34; false.model＆＃34;曲线在情节上分别可见，但在图例中我只想要一个＆＃34;假模型＆＃34;在一条灰线旁边进入。有关如何做到这一点的任何想法？

Answer 1

我的dput出了问题。它似乎缺少rownames，使其成为无效的data.frame。这是一个纠正的转储

df2 <- structure(list(true = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("model1", 
"model2"), class = "factor"), test = structure(c(1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 
4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 
2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L), .Label = c("false.model1", "false.model2", "false.model3", 
"true model"), class = "factor"), est.interval = c(0, 5, 5, 7, 
7, 10, 10, 11, 11, 0, 2, 2, 3, 3, 0, 2, 2, 4, 4, 0, 2, 2, 3, 
3, 4, 4, 0, 7, 7, 10, 10, 13, 13, 0, 4, 4, 5, 5, 0, 5, 5, 6, 
6, 7, 7, 9, 9, 0, 2, 2, 3, 3, 4, 4), sens = c(1, 1, 0.75, 0.75, 
0.5, 0.5, 0.25, 0.25, 0, 1, 1, 0.25, 0.25, 0, 1, 1, 0.5, 0.5, 
0, 1, 1, 0.5, 0.5, 0.25, 0.25, 0, 1, 1, 0.75, 0.75, 0.25, 0.25, 
0, 1, 1, 0.5, 0.5, 0, 1, 1, 0.75, 0.75, 0.5, 0.5, 0.25, 0.25, 
0, 1, 1, 0.5, 0.5, 0.25, 0.25, 0)), .Names = c("true", "test", 
"est.interval", "sens"), row.names = c(NA, -54L), class = "data.frame")

最简单的方法是创建一个可用于着色的新因子。例如

df2$testcol<-factor(ifelse(df2$test=="true model",1,2), 
   levels=1:2,
   labels=c("True Model","False Model"))

现在所有的假模型都会共享一个共同的价值。然后就可以了

ggplot(df2, aes(x=est.interval, y=sens, group=test, colour=testcol)) +
    ylim(0,1) +
    geom_line(size = 0.6) +
    facet_wrap(~true, nrow=1) +
    scale_colour_manual(values=c("True Model"='#CC0000', 
        "False Model"="#999999"), name="Estimation Model") + 
    guides(color=guide_legend(keywidth = 3, keyheight = 1))

获取