r - 以日期作为解释变量的ggplot多项式拟合

Question

我有这样的数据，可以在不同的日子给受试者打分(采用 POSIXct 格式)。

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()

我想做的是拟合三阶多项式线。如果我输入以下内容，则会出现错误:

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

如果我在 stat_smooth() 中将 Date 指定为数字，我会得到同样的错误。有什么办法可以在 ggplot 中做到这一点吗？

这是数据:

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")

Answer 1

stat_smooth 中的公式参数必须根据美学 y 和 x 指定，而不是映射到那些的原始变量美学。

ggplot(test,aes(Date,score)) + 
  stat_smooth(method = "lm", formula = y ~ poly(x, 3), size = 1) + 
  geom_point()