更新时间:2023-11-26 12:43:28
我认为我无法真正满足您的不制作新数据框"要求,但您可以即时创建新数据框:
I don't think I can really satisfy your "without making a new data frame" requirement, but you can create the new data frame on the fly:
ggplot(transform(iris,
Species=factor(Species,levels=c("virginica","setosa","versicolor")))) +
geom_histogram(aes(Petal.Width))+ facet_grid(Species~.)
或者,在 tidyverse 成语中:
or, in tidyverse idiom:
iris %>%
mutate(across(Species, factor, levels=c("virginica","setosa","versicolor"))) %>%
ggplot() +
geom_histogram(aes(Petal.Width))+
facet_grid(Species~.)
我同意如果有另一种方法来控制它会很好,但是 ggplot
已经是一个非常强大(且复杂)的引擎......
I agree it would be nice if there were another way to control this, but ggplot
is already a pretty powerful (and complicated) engine ...
请注意,(1) 数据集中的行的顺序与 (2) 因子的级别的顺序无关.#2 是 factor(...,levels=...)
变化的内容,以及 ggplot
用来确定构面顺序的内容.做#1(按指定顺序对数据框的行进行排序)是一个有趣的挑战.我想我实际上可以通过先做 #2 来实现这一点,然后使用 order()
或 arrange()
根据因子的数值进行排序:>
Note that the order of (1) the rows in the data set is independent of the order of (2) the levels of the factor. #2 is what factor(...,levels=...)
changes, and what ggplot
looks at to determine the order of the facets. Doing #1 (sorting the rows of the data frame in a specified order) is an interesting challenge. I think I would actually achieve this by doing #2 first, and then using order()
or arrange()
to sort according to the numeric values of the factor:
neworder <- c("virginica","setosa","versicolor")
library(plyr) ## or dplyr (transform -> mutate)
iris2 <- arrange(transform(iris,
Species=factor(Species,levels=neworder)),Species)
如果不改变因子水平的顺序,我无法立即看到一种快速的方法(你可以这样做,然后相应地重置因子水平的顺序).
I can't immediately see a quick way to do this without changing the order of the factor levels (you could do it and then reset the order of the factor levels accordingly).
一般来说,R 中依赖于分类变量级别顺序的函数基于因子级别顺序,而不是数据集中行的顺序:上述答案更普遍.
In general, functions in R that depend on the order of levels of a categorical variable are based on factor level order, not the order of the rows in the dataset: the answer above applies more generally.