ggplot多项式拟合与日期作为解释变量
问题描述:
我有这样的数据,以便在不同的日子(POSIXct格式)给一个主题得分。ggplot多项式拟合与日期作为解释变量
head(test)
Date Subject score
1 2012-08-10 Black6 0
2 2012-08-11 Black6 0
3 2012-08-12 Black6 0
4 2012-08-13 Black6 0
5 2012-08-14 Black6 0
6 2012-08-15 Black6 0
拟合黄土曲线很简单。
ggplot(test,aes(Date,score))+geom_smooth()+geom_point()
我希望做的是安装一个3阶多项式线。如果我在下面的类型我得到一个错误:
ggplot(test,aes(Date,score))+stat_smooth(method = "lm", formula = score ~ poly(Date, 3), size = 1)+geom_point()
Error in eval(expr, envir, enclos) : object 'score' not found
我得到同样的错误,如果我指定的日期为数字stat_smooth内()。有没有办法在ggplot中做到这一点?
这里的数据:
test<-structure(list(Date = structure(c(1344556800, 1344643200, 1344729600,
1344816000, 1344902400, 1344988800, 1345075200, 1345161600, 1345248000,
1345334400, 1345420800, 1345507200, 1345593600, 1345680000, 1345766400,
1345852800, 1345939200, 1346025600, 1346112000, 1346198400, 1346284800,
1346371200, 1346457600, 1346544000, 1346630400, 1346716800, 1346803200,
1346889600, 1346976000, 1347062400, 1347148800, 1347235200, 1347321600,
1347408000, 1347494400, 1347580800, 1347667200, 1347753600, 1347840000,
1347926400, 1348012800, 1348099200, 1348185600, 1348272000, 1348358400,
1348444800, 1348531200, 1348617600, 1348704000, 1348790400, 1348876800,
1348963200, 1349049600, 1349136000, 1349222400, 1349308800, 1349395200
), class = c("POSIXct", "POSIXt"), tzone = "UTC"), Subject = 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, 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, 1L, 1L, 1L, 1L, 1L), .Label = "Black6", class = "factor"),
score = c(0, 0, 0, 0, 0, 0, 0, 0.25, 0.25, 0.25, 0.25, 0.25,
0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 1, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 1.5, 1.5, 1.25, 1.25, 1.25, 1,
1, 1, 1, 1, 1, 1, 1, 1.5, 1.5, 1.25, 1, 1, 1, 0.5, 0.5, 0.5,
0.25, 0.25)), .Names = c("Date", "Subject", "score"), row.names = c(NA,
57L), class = "data.frame")
答
在stat_smooth
的公式参数必须在美学y
和x
,不映射到那些美学原始变量的术语来指定。
ggplot(test,aes(Date,score)) +
stat_smooth(method = "lm", formula = y ~ poly(x, 3), size = 1) +
geom_point()
答
即使问题是关于ggplot2
,我给他使用lattice
和latticeExtra
包类似的解决方案。
library(lattice)
library(latticeExtra)
xyplot(score ~ Date, test,par.settings = ggplot2like()) +
layer(panel.smoother(y ~ poly(x, 3), method = "lm"), style = 2)
大非常感谢! – iantist 2013-02-08 20:19:40