我的Shiny应用程序中有一个plotly
条形图,我想在结果条形图中为每一列设置特定的颜色。
#Here's some reproducible data
df=data.frame(Month=c("Jan","Feb","Mar","Apr","May","Jun"),Criteria1=c(10,15,20,15,7,6),Criteria2=c(3,8,5,7,9,10),Criteria3=c(11,18,14,9,3,1))
#Plot
colNames <- names(df)[-1] #Month is the first column
# Here is where I set the colors for each `Criteria`, assuming that the order of colors follows the same order as the factor levels of the `Criteria`.
p <- plotly::plot_ly(marker=list(colors=c('#CC1480', '#FF9673', '#E1C8B4')))
for(trace in colNames){
p <- p %>% plotly::add_trace(data = df, x = ~Month, y = as.formula(paste0("~`", trace, "`")), name = trace, type = "bar")
}
p %>%
layout(title = "Trend Over Time",showlegend = FALSE,
xaxis = list(title = ""),
yaxis = list (title = "Monthly Count of QoL Tweets"))
然而,结果图未显示我指定的任何颜色。
我做错了什么?任何指针都会非常感激。
答案 0 :(得分:1)
您可以将颜色指定给矢量:
colors <- c('#CC1480', '#FF9673', '#E1C8B4')
然后在略微修改的循环中添加跟踪。
p <- plotly::add_trace(p,
x = df$Month,
y = df[,trace],
marker = list(color = colors[[match(trace, colNames)]]),
name = trace,
type = "bar")
}
将为您提供以下图表
完整代码
library("plotly")
df=data.frame(Month=c("Jan", "Feb","Mar", "Apr", "May", "Jun"),
Criteria1 = c(10, 15,20,15,7,6),
Criteria2 = c(3, 8, 5, 7, 9, 10),
Criteria3 = c(11, 18, 14, 9, 3, 1))
colNames <- names(df)[-1] #Month is the first column
colors <- c('#CC1480', '#FF9673', '#E1C8B4')
p <- plotly::plot_ly()
#colNames = c('Criteria1')
for(trace in colNames){
p <- plotly::add_trace(p,
x = df$Month,
y = df[,trace],
marker = list(color = colors[[match(trace, colNames)]]),
name = trace,
type = "bar")
}
p
答案 1 :(得分:0)
我认为这里不需要循环,以下内容还提供了对df
熔化时特定级别(各个级别Criteria1
,Criteria2
)选择颜色的更多控制,Criteria3
library(plotly)
library(reshape2)
#Yout data.frame
df <- data.frame(Month = c("Jan","Feb","Mar","Apr","May","Jun"),
Criteria1 = c(10,15,20,15,7,6),
Criteria2 = c(3,8,5,7,9,10),
Criteria3 = c(11,18,14,9,3,1))
melt(df, id.vars = 'Month') %>% plot_ly(x = ~Month, y = ~value, type = 'bar',
color = ~variable,
colors = c(Criteria1 = '#CC1480', Criteria2 = '#FF9673', Criteria3 = '#E1C8B4'))