如何从Choroplethr映射中删除状态缩写

我正在使用一个

choroplethr

地图，如下所示。如何简单地删除州缩写

以下是复制代码：

library(choroplethr)
library(choroplethrMaps)

data(df_pop_state)
df_pop_state$value <- as.numeric(df_pop_state$value)

state_choropleth(df_pop_state, num_colors = 1,
                             title = "2012 State Population Estimates",
                             legend = "Population")

库（choroplethr）
图书馆（choroplethrMaps）
数据（df_pop_状态）
df_pop_state$value此函数返回一个ggplot对象，因此您可以手动检查图层并删除不需要的图层（此处您要删除GeomText图层）：

声明感谢您使用choroplethr。请注意，Choroplethr使用R6对象。事实上，
state\u choropleth
函数只是
statechropleth
R6对象的一个方便包装器：

> state_choropleth
function (df, title = "", legend = "", num_colors = 7, zoom = NULL, 
    reference_map = FALSE) 
{
    c = StateChoropleth$new(df)
    c$title = title
    c$legend = legend
    c$set_num_colors(num_colors)
    c$set_zoom(zoom)
    if (reference_map) {
        if (is.null(zoom)) {
            stop("Reference maps do not currently work with maps that have insets, such as maps of the 50 US States.")
        }
        c$render_with_reference_map()
    }
    else {
        c$render()
    }
}
<bytecode: 0x7fdda6aa3a10>
<environment: namespace:choroplethr>


我选择这种方法是因为，一般来说，我发现R中的许多函数都有大量的参数，这可能会令人困惑。缺点是函数比对象更容易编写文档（特别是在R中），因此经常会出现类似这样的问题

> state_choropleth
function (df, title = "", legend = "", num_colors = 7, zoom = NULL, 
    reference_map = FALSE) 
{
    c = StateChoropleth$new(df)
    c$title = title
    c$legend = legend
    c$set_num_colors(num_colors)
    c$set_zoom(zoom)
    if (reference_map) {
        if (is.null(zoom)) {
            stop("Reference maps do not currently work with maps that have insets, such as maps of the 50 US States.")
        }
        c$render_with_reference_map()
    }
    else {
        c$render()
    }
}
<bytecode: 0x7fdda6aa3a10>
<environment: namespace:choroplethr>

c = StateChoropleth$new(df_pop_state)
c$title = "2012 State Population Estimates"
c$legend = "Population"
c$set_num_colors(1)
c$show_labels = FALSE
c$render()