热图颜色不能在暗地里工作

我试图用Shiny打印出一张使用情节图的热图。我想给他们一个自定义的色阶，但这并不是我想要的。当我使用colors =选项构建我的图表时，它似乎使用值的分布，而不是我给它指定颜色的zmin和zmax。热图颜色不能在暗地里工作

在我下面的示例代码中，您可以看到，我使用colors =选项为每个绘图提供相同的色彩比例（colorScale）。当我拥有一组分布良好的数据时，这种方式的工作方式如第一，第二和第四行图所示。

但是，在第三行中，图中的数据非常偏斜，您可以看到比例尺看起来不同于其他所有 - 它们具有蓝色和红色，但跳过中间的白色，而不是紫色。

在我的实际代码中，这对于在中间有很多值的图表造成了很大的问题，在两端都有一些极端值 - 我希望中间的值显示为白色，以表明那里没有变化，但是它们是紫色的，使得挑选重要的价值（极端的价值）变得更加困难。

有没有办法强制颜色分配行为如何我想要它？

感谢，崖

server.R

if (!require("pacman")) install.packages("pacman") 

pacman::p_load(shiny,data.table,plotly) 


colorScale <- colorRamp(c("darkblue","cornflowerblue","white","sandybrown","firebrick")) 

nCodeNames <- c("a","b","c","d","e","f","g","h","i","j","k","l") 
means = c(rnorm(600,0,2.5),runif(600,-5,5),runif(130,-4,-3.9),runif(70,4.5,5),rnorm(150,-3),rnorm(50,4),rnorm(180,-2.5),runif(20,4.93,4.98),runif(300,-4,3),rnorm(300,3.5)) 



dt <- data.table(age=rep(rep(c(11:20),times=20),times=12),composite=rep(rep(c(81:100),each=10),times=12),mean=means,n_code=rep(nCodeNames,each=200)) 

sub<-dt[n_code=="a"] 

shinyServer(function(input, output) { 

for(Ncode in nCodeNames){ 
    local({ 
    ncode = Ncode 
    output[[paste0("grid",ncode)]] <- renderPlotly({ 
     sub <- dt[n_code == ncode] 
     p <- plot_ly(data=sub, x=~age, y=~composite, z=~mean, type="heatmap", zmin=-5,zmax=5, 
        colors = colorScale, colorbar=list(thickness="15"))%>% 
      layout(title=ncode,xaxis=list(type="category",tickvals=c(11,15,20)),yaxis=list(title="",ticks="")) 
    }) 
    }) 
} 
})

ui.R

if (!require("pacman")) install.packages("pacman") 

pacman::p_load(shiny, plotly) 

nCodeNames <- c("a","b","c","d","e","f","g","h","i","j","k","l") 


shinyUI(navbarPage(
    "E-N Matrics: Proportion of E-Code Resulting in each N-Code", 

    tabPanel("Grid", 

      lapply(c(1:4), function(i) fluidRow(
       lapply(c(1:3), function(j) column(4, plotlyOutput(paste0("grid",nCodeNames[(i-1)*3+j])))) 
      )) 

      #fluidRow(column(4,plotlyOutput(paste0("grid",nCodeNames[(1-1)*3+1]))),column(4,plotly)) 
) 
))

来源

2016-11-10 Cliff

我遇到过类似的问题，在R上的colorscale plotly热图。当z参数的数据具有拧紧分布时，只有几个颜色中指定的颜色被绘制使用。

我发现一个解决方案，根据原始变量的分位数创建一个新变量，并将其传递给z参数。这是一般想法的R代码。您需要对其进行定制以使其适用于特定问题。

library(plotly) 
library(RColorBrewer) 

# create a dataframe where z has a skewed distribution 
set.seed(1) 
df = data.frame(x = rep(1:50, 20) , y = rep(1:20,each =50), z = rgamma(1000, 2, 0.5)) 

# check distribution of z 
plot_ly(data = df, x = ~z, type = "histogram")%>% 
    layout(title = "histogram of z") 

# original heatmap 
# pass the column z with screwed distribution to z argument 
plot_ly(data=df, x=~x, y=~y, z=~z, type="heatmap", 
     colors = "Spectral") %>% 
    layout(title = "original heatmap") 

# some data processing work 

# find unique quantiles of z 
quantiles = unique(quantile(df$z, seq(0,1,0.1))) 

# create a dummy column z1 of discrete values using the quantiles as cut off 
# the ideas is to arrage the data to subgroups of roughly the same size 
df$z1= cut(df$z, breaks = c(quantiles[1]-1,quantiles[-1]), right = TRUE, labels = FALSE) 

# check distribution of z1 
plot_ly(data = df, x = ~z1, type = "histogram")%>% 
    layout(title = "histogram of z1") 


# new heatmap 
# passes the new column z1 to z argument 
plot_ly(data=df, x=~x, y=~y, z=~z1, type="heatmap", 
     # make sure hovering over displays original z 
     text =~z, hoverinfo = "text", 
     # use the color palettes from RColorBrewer, 
     # or your customized colorscale 
     colors = "Spectral", 
     # map the label of the colorbar back to the quantiles 
     colorbar=list(tickmode="array", tickvals = 1:(length(quantiles)-1), ticktext = round(quantiles,2)[-1], title = "z")) %>% 
layout(title = "new heat map")

下面是原来的热图和plotly产生的新的热图。新的热图使用“光谱”调色板中的更多颜色区分较小的值。