我有一个问题,但在此解决方案中找不到答案。我的意思是,我想ggplot在新函数中使用该函数,例如
library(ggplot2)
draw_point <- function(data, x, y ){
ggplot(data, aes_string(x, y)) +
geom_point()
}
Run Code Online (Sandbox Code Playgroud)
结果我必须使用引号:
draw_point(mtcars, "wt", "qsec")
Run Code Online (Sandbox Code Playgroud)
相反,我想使用某种lazyeval包来编写这个不带引号的函数:
draw_point(mtcars, wt, qsec)
Run Code Online (Sandbox Code Playgroud)
有可能做到吗?
我想radioButtons使用工具提示创建一个小部件shinyBS.我想要实现的是创建一个带有3个按钮的小部件,其中包含不同的信息tooltip.基于此解决方案, 它创建了3个具有不同id值的单独单选按钮.是否可以做同样的事情,但有一个带3个按钮的无线电小部件(即有一个id值)?
library(shiny)
library(shinyBS)
ui <- shinyUI(
fluidPage(
fluidRow(
column(3,
HTML("<div class='container'><br>
<h1>Test</h1>
<div>
<label id='radio_venue_1'>
<input type='radio' value='1' role='button'> button 1
</label>
</div>
<div>
<label id='radio_venue_2'>
<input type='radio' value='2' role='button'> button 2
</label>
</div>
<div>
<label id='radio_venue_3'>
<input type='radio' value='3' role='button'> button 3
</label>
</div>
</div>")),
bsTooltip(id = "radio_venue_1", title = "Button 1 Explanation", placement = "right", trigger = "hover"),
bsTooltip(id = "radio_venue_2", title = "Button 2 Explanation", placement = "right", trigger …Run Code Online (Sandbox Code Playgroud)
我想创建一个ggplot使用两个变量来填充不同的方式.基于这个解决方案我做了
x = c("Band 1", "Band 2", "Band 3")
y1 = c("1","2","3")
y2 = c("2","3","4")
to_plot <- data.frame(x=x,y1=y1,y2=y2)
melted<-melt(to_plot, id="x")
ggplot(melted,aes(x=x,y=value,fill=variable)) +
geom_bar(stat="identity",position = "identity", alpha=.3)
Run Code Online (Sandbox Code Playgroud)

但不是alpha参数,我想每个颜色Band以不同的方式和价值实现y1的颜色与边框的白色柱状给定的Band,并y2作为与给定的彩条Band.怎么做?
我想创建一个将使用summarise_和group_by_功能的简单函数.我知道非标准评估存在问题,但我可以解决它.
library(dplyr)
test_function <- function(.data, ..., variable){
dots <- quote(mean(substitute(variable)))
group_by_(.data,.dots = lazyeval::lazy_dots(...)) %>%
summarise_(.dots = dots)
}
Run Code Online (Sandbox Code Playgroud)
但是,当我尝试这个:
test_function(ggplot2::diamonds, cut, clarity, variable = price)
Run Code Online (Sandbox Code Playgroud)
这是行不通的.
谢谢,
添加图例和缩写说明时遇到问题.举个例子
df <- data.frame(value = c(seq(1,50,1),seq(21,70,1),seq(41,90,1)),
name = c(rep("A-B",50),rep("B-C",50),rep("A-C",50)))
ggplot(df, aes(name, value))+
geom_boxplot()
Run Code Online (Sandbox Code Playgroud)

我有data.frame和缩写的解释.
full_name <- data.frame(abb =c("A","B","C"),fname = c("Ananas","Banana","Cucumber"))
Run Code Online (Sandbox Code Playgroud)
我想添加带有每个字母含义的信息的图例.
我有一个data.frame:
df <- data.frame(id = rep(1:4, each = 3),x = c("A","B","C","D","E","A","A","C","D","A","C","E"))
Run Code Online (Sandbox Code Playgroud)
我想计算每个id内的连接:这是我想得到的输出:
connections |num. of connections
A - B | 1
B - C | 1
C - D | 1
A - C | 3
A - E | 2
A - D | 2
D - E | 1
C - E | 1
Run Code Online (Sandbox Code Playgroud)
怎么在dplyr中做到这一点?