R:如何更改facet_grid中每行的列数

时间:2016-08-21 02:21:19

标签: r ggplot2

通过ggplot,我有一个如下图,使用下面的代码。但是,如果我这样画,我根本看不到x轴。我想知道是否有任何方法可以解决这个问题,例如更改每行中的列数。我在ncol中尝试了facet_grid命令,但它不允许我这样做。

ggplot(derivative, aes(x = factor(move), fill = factor(move)), colour = black)+ 
geom_bar()+
facet_grid(Market~Season)+
 scale_fill_discrete(name="Relative Market Move",
                  breaks=c("neg.big", "neg.small", "pos.big", "pos.small"),
                  labels=c("Big Negative", "Small Negative", "Big Positive", "Small Positive"))+
 scale_x_discrete(labels=c("Large Negative", "Small Negative", "large Positive", "Small Positive"))+
labs( x = "") +ylab("Count") 

cramped faceted plot

1 个答案:

答案 0 :(得分:1)

你可能会更好地使用堆积条形图和"负面"指向下方。这将更有效地使用水平空间,并使您更容易看到时间趋势。例如:

    3,20
    2,40
    4,10
    1,50

首先创建一些假数据:

library(reshape2)

现在的情节:

set.seed(199)
dat = data.frame(index=rep(c("S&P 500","Shanghai","Hang Seng"), each=7),
                 year=rep(paste0(rep(2009:2015,each=2),rep(c("Sp","Au"),7)), 3),
                 replicate(3, sample(50:100,14*3)))
dat$big.neg = 300 - rowSums(dat[,3:5])
names(dat)[3:5] = c("big.pos","small.pos","small.neg")

# Set year order
dat$year = factor(dat$year, levels=dat$year[1:14])

# Melt to long format
dat = melt(dat, id.var=c("year","index"))

enter image description here