我想以粗体突出显示各个轴标签.我知道@MrFlick 的这个答案,但我无法弄清楚如何做这个a)多个项目,以及b)是否可以使用标签的名称而不是项目的编号列表(或表达式).
这是一个示例数据集:
require(ggplot2)
require(dplyr)
set.seed(36)
xx<-data.frame(YEAR=rep(c("X","Y"), each=20),
CLONE=rep(c("A","B","C","D","E"), each=4, 2),
TREAT=rep(c("T1","T2","T3","C"), 10),
VALUE=sample(c(1:10), 40, replace=T))
Run Code Online (Sandbox Code Playgroud)
然后我根据一个特定的因子组合对我的标签进行排序,然后应该在一个图的多个面板上进行维护.在这里查看我之前的问题.
clone_order <- xx %>% subset(TREAT == "C" & YEAR == "X") %>%
arrange(-VALUE) %>% select(CLONE) %>% unlist()
xx <- xx %>% mutate(CLONE = factor(CLONE, levels = clone_order))
ggplot(xx, aes(x=CLONE, y=VALUE, fill=YEAR)) +
geom_bar(stat="identity", position="dodge") +
facet_wrap(~TREAT)
Run Code Online (Sandbox Code Playgroud)
现在我想大胆Clone A,B并且E.我相信这会以某种方式起作用,但我无法弄清楚如何.理想情况下,这将是巨大的,知道如何做到这一点一)使用列表/表达项目的数量,以及B)通过使用标签,如A,B和E.
当错误消息本身被修改为以红色打印时,如何在RMarkdown中获得打印的错误消息以折叠成一个块?
在此示例中collapse = T按预期方式工作。
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE, collapse = TRUE)
```
```{r error=T}
x <- c(1,2,3,4,5)
x * 10
X * 10
```
Run Code Online (Sandbox Code Playgroud)
在此示例中,我将错误消息修改为红色(基于此答案)。但随后它不会随着其他崩溃而崩溃:
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE, collapse = TRUE)
knitr::knit_hooks$set(error = function(x, options) {
paste0("<pre style=\"color: red;\"><code>", x, "</code></pre>")
})
```
```{r error=T}
x <- c(1,2,3,4,5)
x * 10
X * 10
```
Run Code Online (Sandbox Code Playgroud)
我试图collapse = T在特定的代码块中再次指定,但这也不起作用:
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE, collapse = TRUE)
knitr::knit_hooks$set(error = …Run Code Online (Sandbox Code Playgroud) 我想过滤截止年之前最后一个测量年的所有观测值以及截止年之后所有年份的所有观测值。
这是一个例子:
d <- data.frame(group = c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2),
cut_off = c(2017,2017,2017,2017,2017,2017,2017,2017,2017,2016,2016,2016,2016,2016,2016,2016),
year = c(2000,2010,2010,2015,2015,2017,2017,2020,2024,2001,2009,2016,2017,2017,2021,2023),
value = c(10,20,30,40,50,60,70,80,90,100,110,120,130,140,150,160))
> d
group cut_off year value
1 1 2017 2000 10
2 1 2017 2010 20
3 1 2017 2010 30
4 1 2017 2015 40
5 1 2017 2015 50
6 1 2017 2017 60
7 1 2017 2017 70
8 1 2017 2020 80
9 1 2017 2024 90
10 1 2016 2001 100
11 2 2016 2009 110
12 …Run Code Online (Sandbox Code Playgroud) 我想在列总和不为零的数据帧中选择或子集变量,但也要保留其他因子变量.它应该相当简单,但我无法弄清楚如何select_if()使用dplyr以下方法对变量子集运行该函数:
df <- data.frame(
A = c("a", "a", "b", "c", "c", "d"),
B = c(0, 0, 0, 0, 0, 0),
C = c(3, 0, 0, 1, 1, 2),
D = c(0, 3, 2, 1, 4, 5)
)
require(dplyr)
df %>%
select_if(funs(sum(.) > 0))
#Error in Summary.factor(c(1L, 1L, 2L, 3L, 3L, 4L), na.rm = FALSE) :
# ‘sum’ not meaningful for factors
Run Code Online (Sandbox Code Playgroud)
然后我尝试只选择B, C, D,这有效,但我不会有变量A:
df %>%
select(-A) %>%
select_if(funs(sum(.) > …Run Code Online (Sandbox Code Playgroud)