如何基于分类变量在 R Plotly 中创建等值线图？

Question

我正在尝试创建美国的叶绿素地图，该地图使用州颜色的分类变量，但我只得到一张空白地图。 plotly maps 是否与分类数据兼容？如果是这样，语法有何变化？

对于我的数据，我只是上传 table 行，其中包含州和随机 "Good"、"Bad"、"OK."

我可以在下面的代码中更改什么以使其工作？我已经尝试了一种解决方法，它可以稍微改变状态的颜色，但颜色条变得不稳定。（value4 是我的 "Good"、"Bad"、"OK" 的分类变量）

如果我的问题不清楚或者我的信息不是很好，我们深表歉意。如果有人有进一步的问题，我可以回答。提前致谢

foo <- brewer.pal(n = 3,
                        name = "Set1")

df <- mutate(df, test = ntile(x = value4, n = 3))

cw_map <- plot_ly(
  data = df,
  type = "choropleth",
  locations = ~ state,
  locationmode = "USA-states",
  color = ~ test,
  colors = foo[df$test],
  z = ~ test
) %>%
  layout(geo = list(scope = "usa"))

print(cw_map)

Answer 1

您需要在代码形式中包含状态，所以让我们从这里开始：

STATES <-c("AL", "AK", "AZ", "AR", "CA", "CO", "CT", "DE", "FL", "GA", 
"HI", "ID", "IL", "IN", "IA", "KS", "KY", "LA", "ME", "MD", "MA", 
"MI", "MN", "MS", "MO", "MT", "NE", "NV", "NH", "NJ", "NM", "NY", 
"NC", "ND", "OH", "OK", "OR", "PA", "RI", "SC", "SD", "TN", "TX", 
"UT", "VT", "VA", "WA", "WV", "WI", "WY")

像您一样，我们为每个州提供随机值 4：

df = data.frame(state=STATES,
value4=sample(c("Good", "Bad", "OK."),length(STATES),replace=TRUE))

然后我们将您的 value4 作为因子和颜色等，就像您之前所做的那样：

df$value4 = factor(df$value4)
df$test = as.numeric(df$value4)
nfactor = length(levels(df$value4))
foo <- brewer.pal(n = nfactor,name = "Set1")
names(foo) = levels(df$value4)

要以离散形式显示颜色图例，您需要将其作为数据框提供，该数据框在 z 的 相对比例上定义中断。 R plotly 中没有很好地记录它，我使用 @emphet's plotly forum post and :

中的信息为 n 个因素编写了下面的解决方案
Z_Breaks = function(n){ CUTS = seq(0,1,length.out=n+1) rep(CUTS,ifelse(CUTS %in% 0:1,1,2)) } colorScale <- data.frame(z=Z_Breaks(nfactor), col=rep(foo,each=2),stringsAsFactors=FALSE) z col 1 0.0000000 #E41A1C 2 0.3333333 #E41A1C 3 0.3333333 #377EB8 4 0.6666667 #377EB8 5 0.6666667 #4DAF4A 6 1.0000000 #4DAF4A

然后我们绘制：

cw_map <- plot_ly( data = df, type = "choropleth", locations = ~ state, locationmode = "USA-states", z = df$test, colorscale=colorScale, colorbar=list(tickvals=1:nfactor, ticktext=names(foo)) ) %>% layout(geo = list(scope = "usa"))

如何基于分类变量在 R Plotly 中创建等值线图？

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