如何在ggplot2中正确使用facet_grid？

Question

我正在尝试使用以下代码为每个配置文件生成一个图表，但我不断得到“至少一个图层必须包含用于构面的所有变量。”错误。我花了最后几个小时试图让它工作，但我不能。

我相信anwser必须简单，任何人都可以帮忙吗？

d = structure(list(category = structure(c(2L, 2L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 
3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L), .Label = c("4X4", 
"HATCH", "SEDAN"), class = "factor"), profile = structure(c(1L, 
1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 
1L), .Label = c("FIXED", "FREE", "MOBILE"), class = "factor"), 
    value = c(6440.32, 6287.22, 9324, 7532, 7287.63, 6827.27, 
    6880.48, 7795.15, 7042.51, 2708.41, 1373.69, 6742.87, 7692.65, 
    7692.65, 8116.56, 7692.65, 7692.65, 7692.65, 7962.65, 8116.56, 
    5691.12, 2434, 8343, 7727.73, 7692.65, 7721.15, 1944.38, 
    6044.23, 8633.65, 7692.65, 7692.65, 8151.65, 7692.65, 7692.65, 
    2708.41, 3271.45, 3333.82, 1257.48, 6223.13, 7692.65, 6955.46, 
    7115.46, 7115.46, 7115.46, 7115.46, 6955.46, 7615.46, 2621.21, 
    2621.21, 445.61)), .Names = c("category", "profile", "value"
), class = "data.frame", row.names = c(NA, -50L))

library(ggplot2)

p = ggplot(d, aes(x=d$value, fill=d$category)) + geom_density(alpha=.3)
p + facet_grid(d$profile ~ .)

Answer 1

您的问题来自明确引用变量（即d$profile），而不是data调用中的ggplot参数。

当d$使用faceting或facet_grid时，您需要这样做。调用facet_wrap

也是一种好习惯

aes