R plotly:如何从跟踪名称中删除颜色映射变量
如果我将“颜色”映射到既不是x、y也不是名称的变量,则颜色映射变量将添加到悬停时的跟踪名称和图例中:R plotly:如何从跟踪名称中删除颜色映射变量,r,plotly,R,Plotly,如果我将“颜色”映射到既不是x、y也不是名称的变量,则颜色映射变量将添加到悬停时的跟踪名称和图例中: df <- data.frame(x = rnorm(5, 50, 1), y = letters[1:5], c = LETTERS[1:5], name = c("aa", "bb", "cc", "dd", "ee"), stringsAsFactors = FALSE) p
df <- data.frame(x = rnorm(5, 50, 1),
y = letters[1:5],
c = LETTERS[1:5],
name = c("aa", "bb", "cc", "dd", "ee"),
stringsAsFactors = FALSE)
p <- plot_ly(df, x = ~x, y = ~y,
type = "bar",
color = ~c,
name = ~name,
hoverinfo = "x+y+name")
p
df此代码设置color=~name
library(plotly)
library(RColorBrewer)
df <- data.frame(x = rnorm(5, 50, 1),
y = factor(letters[1:5],
levels = c("a", "e", "c", "d", "b")),
c = LETTERS[1:5],
name = c("qwe", "zxc", "sdf", "bnm", "ert"),
stringsAsFactors = FALSE)
p <- plot_ly(df, x = ~x, y = ~y,
type = "bar",
color = ~name,
colors = brewer.pal(8, "Blues")[4:8],
name = ~name,
hoverinfo = "x+y+name")
p
library(plotly)
图书馆(RColorBrewer)
df此代码设置color=~name
library(plotly)
library(RColorBrewer)
df <- data.frame(x = rnorm(5, 50, 1),
y = factor(letters[1:5],
levels = c("a", "e", "c", "d", "b")),
c = LETTERS[1:5],
name = c("qwe", "zxc", "sdf", "bnm", "ert"),
stringsAsFactors = FALSE)
p <- plot_ly(df, x = ~x, y = ~y,
type = "bar",
color = ~name,
colors = brewer.pal(8, "Blues")[4:8],
name = ~name,
hoverinfo = "x+y+name")
p
library(plotly)
图书馆(RColorBrewer)
df要从图例中删除颜色变量,请执行以下操作:name=~c
或更改:color=~name
,如果您希望图例的标签为“aa”、“bb”…..,谢谢@EnriquePérezHerrero。我注意到我的例子对于我遇到的现实生活问题来说太简单了,不能代表全部问题。我用一个更好更复杂的例子编辑了我的文章。您能帮忙吗?要从图例中删除颜色变量,请执行以下操作:name=~c
或更改:color=~name
,如果您希望图例的标签为“aa”、“bb”…..,谢谢@EnriquePérezHerrero。我注意到我的例子对于我遇到的现实生活问题来说太简单了,不能代表全部问题。我用一个更好更复杂的例子编辑了我的文章。你能帮忙吗?谢谢@Enrique,但是,我需要由变量c的特定序列映射的颜色。谢谢@Enrique,但是,我需要由变量c的特定序列映射的颜色。