我在ggplot2中制作一个躲闪的条形图,一个分组的数量为零,我想显示。我记得曾经HERE在{{3}}看到这一点,并认为scale_x_discrete(drop=F)
会起作用。它似乎不适用于躲避酒吧。如何显示零点计数?
例如,下图(下面的代码),type8~group4没有例子。我仍然希望绘图显示零计数的空白空间而不是消除栏。我怎么能这样做?
mtcars2 <- data.frame(type=factor(mtcars$cyl),
group=factor(mtcars$gear))
m2 <- ggplot(mtcars2, aes(x=type , fill=group))
p2 <- m2 + geom_bar(colour="black", position="dodge") +
scale_x_discrete(drop=F)
p2
答案 0 :(得分:15)
已更新 geom_bar()
需要stat = "identity"
它的价值:上面的计数表dat包含NA。有时候,有一个明确的0代替它是有用的;例如,如果下一步是将计数放在柱上方。下面的代码就是这样做的,虽然它可能并不比Joran简单。它涉及两个步骤:使用dcast
获取计数的交叉制表,然后使用melt
融合表格,然后像往常一样使用ggplot()
。
library(ggplot2)
library(reshape2)
mtcars2 = data.frame(type=factor(mtcars$cyl), group=factor(mtcars$gear))
dat = dcast(mtcars2, type ~ group, fun.aggregate = length)
dat.melt = melt(dat, id.vars = "type", measure.vars = c("3", "4", "5"))
dat.melt
ggplot(dat.melt, aes(x = type,y = value, fill = variable)) +
geom_bar(stat = "identity", colour = "black", position = position_dodge(width = .8), width = 0.7) +
ylim(0, 14) +
geom_text(aes(label = value), position = position_dodge(width = .8), vjust = -0.5)
答案 1 :(得分:12)
我知道的唯一方法是预先计算计数并添加一个虚拟行:
dat <- rbind(ddply(mtcars2,.(type,group),summarise,count = length(group)),c(8,4,NA))
ggplot(dat,aes(x = type,y = count,fill = group)) +
geom_bar(colour = "black",position = "dodge",stat = "identity")
我认为使用stat_bin(drop = FALSE,geom = "bar",...)
会起作用,但显然不会。
答案 2 :(得分:10)
这里是如何在不首先制作汇总表的情况下完成的 它在我的CRAN版本(2.2.1)中不起作用,但在ggplot(2.2.1.900)的最新开发版本中,我没有任何问题。
ggplot(mtcars, aes(factor(cyl), fill = factor(vs))) +
geom_bar(position = position_dodge(preserve = "single"))
答案 3 :(得分:7)
我问了同样的问题,但我只想使用data.table
,因为它是更大的数据集的更快解决方案。我包含了有关数据的注释,以便那些经验不足并且想要理解我为什么做我所做的事情的人可以轻松地做到这一点。以下是我操纵mtcars
数据集的方法:
library(data.table)
library(scales)
library(ggplot2)
mtcars <- data.table(mtcars)
mtcars$Cylinders <- as.factor(mtcars$cyl) # Creates new column with data from cyl called Cylinders as a factor. This allows ggplot2 to automatically use the name "Cylinders" and recognize that it's a factor
mtcars$Gears <- as.factor(mtcars$gear) # Just like above, but with gears to Gears
setkey(mtcars, Cylinders, Gears) # Set key for 2 different columns
mtcars <- mtcars[CJ(unique(Cylinders), unique(Gears)), .N, allow.cartesian = TRUE] # Uses CJ to create a completed list of all unique combinations of Cylinders and Gears. Then counts how many of each combination there are and reports it in a column called "N"
这是产生图形的调用
ggplot(mtcars, aes(x=Cylinders, y = N, fill = Gears)) +
geom_bar(position="dodge", stat="identity") +
ylab("Count") + theme(legend.position="top") +
scale_x_discrete(drop = FALSE)
它产生了这个图:
此外,如果有连续数据,例如diamonds
数据集中的数据(感谢mnel):
library(data.table)
library(scales)
library(ggplot2)
diamonds <- data.table(diamonds) # I modified the diamonds data set in order to create gaps for illustrative purposes
setkey(diamonds, color, cut)
diamonds[J("E",c("Fair","Good")), carat := 0]
diamonds[J("G",c("Premium","Good","Fair")), carat := 0]
diamonds[J("J",c("Very Good","Fair")), carat := 0]
diamonds <- diamonds[carat != 0]
然后使用CJ
也可以。
data <- data.table(diamonds)[,list(mean_carat = mean(carat)), keyby = c('cut', 'color')] # This step defines our data set as the combinations of cut and color that exist and their means. However, the problem with this is that it doesn't have all combinations possible
data <- data[CJ(unique(cut),unique(color))] # This functions exactly the same way as it did in the discrete example. It creates a complete list of all possible unique combinations of cut and color
ggplot(data, aes(color, mean_carat, fill=cut)) +
geom_bar(stat = "identity", position = "dodge") +
ylab("Mean Carat") + xlab("Color")
给我们这张图:
答案 4 :(得分:3)
使用count
中的complete
和dplyr
来执行此操作。
library(tidyverse)
mtcars %>%
mutate(
type = as.factor(cyl),
group = as.factor(gear)
) %>%
count(type, group) %>%
complete(type, group, fill = list(n = 0)) %>%
ggplot(aes(x = type, y = n, fill = group)) +
geom_bar(colour = "black", position = "dodge", stat = "identity")
答案 5 :(得分:0)
您可以利用gsub("[_]", " ", ctr_names)
函数的功能,该函数计算所有其级别的因子出现次数
table()