我有以下情节：
library(reshape)
library(ggplot2)
library(gridExtra)
require(ggplot2)
data2<-structure(list(IR = structure(c(4L, 3L, 2L, 1L, 4L, 3L, 2L, 1L
), .Label = c("0.13-0.16", "0.17-0.23", "0.24-0.27", "0.28-1"
), class = "factor"), variable = structure(c(1L, 1L, 1L, 1L,
2L, 2L, 2L, 2L), .Label = c("Real queens", "Simulated individuals"
), class = "factor"), value = c(15L, 11L, 29L, 42L, 0L, 5L, 21L,
22L), Legend = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c("Real queens",
"Simulated individuals"), class = "factor")), .Names = c("IR",
"variable", "value", "Legend"), row.names = c(NA, -8L), class = "data.frame")
p <- ggplot(data2, aes(x =factor(IR), y = value, fill = Legend, width=.15))
data3<-structure(list(IR = structure(c(4L, 3L, 2L, 1L, 4L, 3L, 2L, 1L
), .Label = c("0.13-0.16", "0.17-0.23", "0.24-0.27", "0.28-1"
), class = "factor"), variable = structure(c(1L, 1L, 1L, 1L,
2L, 2L, 2L, 2L), .Label = c("Real queens", "Simulated individuals"
), class = "factor"), value = c(2L, 2L, 6L, 10L, 0L, 1L, 4L,
4L), Legend = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c("Real queens",
"Simulated individuals"), class = "factor")), .Names = c("IR",
"variable", "value", "Legend"), row.names = c(NA, -8L), class = "data.frame")
q<- ggplot(data3, aes(x =factor(IR), y = value, fill = Legend, width=.15))
##the plot##
q + geom_bar(position='dodge', colour='black') + ylab('Frequency') + xlab('IR')+scale_fill_grey() +theme(axis.text.x=element_text(colour="black"), axis.text.y=element_text(colour="Black"))+ opts(title='', panel.grid.major = theme_blank(),panel.grid.minor = theme_blank(),panel.border = theme_blank(),panel.background = theme_blank(), axis.ticks.x = theme_blank())
我想y轴显示的整数。 这是否是通过倒圆或通过更巧妙的方法来实现是不是我真的很重要。
随着scale_y_continuous()和论证breaks=你可以为y轴的设定断裂点要显示的整数。
ggplot(data2, aes(x =factor(IR), y = value, fill = Legend, width=.15)) +
geom_bar(position='dodge', colour='black')+
scale_y_continuous(breaks=c(1,3,7,10))
如果你有scales包，你可以使用pretty_breaks()而无需手动指定休息。
q + geom_bar(position='dodge', colour='black') +
scale_y_continuous(breaks= pretty_breaks())
这是我使用：
ggplot(data3, aes(x = factor(IR), y = value, fill = Legend, width = .15)) +
geom_col(position = 'dodge', colour = 'black') +
scale_y_continuous(breaks = function(x) unique(floor(pretty(seq(0, (max(x) + 1) * 1.1)))))
您可以使用自定义的贴标机。 例如，该功能保证只产生整数符：
int_breaks <- function(x, n = 5) pretty(x, n)[pretty(x, n) %% 1 == 0]
用于
+ scale_y_continuous(breaks = int_breaks)
另一种选择是控制标签的格式与自定义格式
q + geom_bar(position='dodge', colour='black')+
scale_y_continuous(labels = function (x) floor(x))
见scales其他流行的格式化像包percent ， dollar ，...
这些解决方案并没有为我工作，并没有解释的解决方案。
的breaks参数的scale_*_continuous的功能可以与该需要的极限作为输入并返回场所作为输出一个自定义功能使用。 默认情况下，轴的限制将被5％的在每一侧上的连续数据（相对于数据的范围）扩大。 轴界限可能不会是整数值由于这种扩张。
我一直在寻找解决的办法是简单地四舍五入下限至最接近的整数，圆形的上限下调至最接近的整数，然后必须在这些端点之间的整数值休息。 因此，我用休息的功能：
brk <- function(x) seq(ceiling(x[1]), floor(x[2]), by = 1)
所需的代码片段是：
scale_y_continuous(breaks = function(x) seq(ceiling(x[1]), floor(x[2]), by = 1))
从原来的问题的可重复的例子是：
data3 <-
structure(
list(
IR = structure(
c(4L, 3L, 2L, 1L, 4L, 3L, 2L, 1L),
.Label = c("0.13-0.16", "0.17-0.23", "0.24-0.27", "0.28-1"),
class = "factor"
),
variable = structure(
c(1L, 1L, 1L, 1L,
2L, 2L, 2L, 2L),
.Label = c("Real queens", "Simulated individuals"),
class = "factor"
),
value = c(2L, 2L, 6L, 10L, 0L, 1L, 4L,
4L),
Legend = structure(
c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L),
.Label = c("Real queens",
"Simulated individuals"),
class = "factor"
)
),
row.names = c(NA,-8L),
class = "data.frame"
)
ggplot(data3, aes(
x = factor(IR),
y = value,
fill = Legend,
width = .15
)) +
geom_col(position = 'dodge', colour = 'black') + ylab('Frequency') + xlab('IR') +
scale_fill_grey() +
scale_y_continuous(
breaks = function(x) seq(ceiling(x[1]), floor(x[2]), by = 1),
expand = expand_scale(mult = c(0, 0.05))
) +
theme(axis.text.x=element_text(colour="black", angle = 45, hjust = 1),
axis.text.y=element_text(colour="Black"),
panel.grid.major = element_blank(),
panel.grid.minor = element_blank(),
panel.border = element_blank(),
panel.background = element_blank(),
axis.ticks.x = element_blank())
