我正在尝试创建美国的chloropleth映射,该映射使用分类变量作为州色,但是我只得到一个空白映射。地物图与分类数据兼容吗?如果是这样,语法如何更改?
[对于我的数据,我只是上载了一张表格,其中包含状态,并随机包含“良好”,“不良”,“确定”之一。
我可以在下面的代码中进行哪些更改才能使其正常工作?我尝试了一种变通方法,该方法可以稍微改变状态的颜色,但是颜色栏会变色。 (value4是我的“良好”,“不良”,“确定”的分类变量)
很抱歉,如果我的问题不清楚或我的信息不好。如果有人有其他问题,我可以回答。在此先感谢
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$test],
z = ~ test
) %>%
layout(geo = list(scope = "usa"))
print(cw_map)
您需要以代码形式提供状态,所以让我们从此开始:
STATES <-c("AL", "AK", "AZ", "AR", "CA", "CO", "CT", "DE", "FL", "GA",
"HI", "ID", "IL", "IN", "IA", "KS", "KY", "LA", "ME", "MD", "MA",
"MI", "MN", "MS", "MO", "MT", "NE", "NV", "NH", "NJ", "NM", "NY",
"NC", "ND", "OH", "OK", "OR", "PA", "RI", "SC", "SD", "TN", "TX",
"UT", "VT", "VA", "WA", "WV", "WI", "WY")
[就像您一样,我们为每个州提供随机值4:
df = data.frame(state=STATES,
value4=sample(c("Good", "Bad", "OK."),length(STATES),replace=TRUE))
df <- mutate(df, test = ntile(x = value4, n = 3))
# or you can just do df$test = as.numeric(factor(df$value4))
我们可以绘制:
foo <- brewer.pal(n = 3,name = "Set1")
cw_map <- plot_ly(
data = df,
type = "choropleth",
locations = ~ state,
locationmode = "USA-states",
color = ~ test,
colors = foo,
z = ~ test
) %>%
layout(geo = list(scope = "usa"))
print(cw_map)