使用 R 的美国地图上的热图

Question

我正在尝试在美国地图上生成热图。 该代码生成美国地图，但不显示任何数据。 它既不给出任何错误！ 我是 R 的新手，所以任何提示/想法都会很棒。

df <- structure(list(X = c(3L, 2L, 6L, 1L, 4L, 4L, 4L, 4L, 1L, 2L, 
5L, 3L, 0L, 4L, 4L, 3L, 3L, 4L, 2L, 4L, 3L, 4L, 4L, 5L, 3L, 4L, 
4L, 2L, 3L, 5L, 2L, 2L, 0L, 1L, 0L, 3L, 4L, 2L, 3L, 0L, 0L, 1L, 
3L, 3L, 1L, 2L, 0L, 0L, 3L, 4L, 3L, 4L, 4L, 3L, 2L, 0L, 2L, 4L, 
3L, 4L), Y = c(3L, 9L, 7L, 6L, 3L, 7L, 7L, 3L, 6L, 5L, 3L, 10L, 
3L, 3L, 4L, 4L, 4L, 3L, 5L, 5L, 4L, 5L, 4L, 4L, 4L, 5L, 6L, 5L, 
4L, 4L, 4L, 2L, 2L, 0L, 2L, 4L, 2L, 4L, 4L, 3L, 3L, 4L, 7L, 7L, 
2L, 3L, 2L, 4L, 3L, 4L, 3L, 3L, 7L, 3L, 4L, 3L, 3L, 3L, 3L, 3L
), Z = c(35L, 31L, 31L, 32L, 35L, 34L, 32L, 36L, 33L, 37L, 32L, 
30L, 39L, 35L, 33L, 35L, 35L, 0L, 35L, 30L, 35L, 33L, 31L, 33L, 
35L, 33L, 35L, 35L, 35L, 34L, 36L, 38L, 42L, 43L, 36L, 37L, 36L, 
39L, 35L, 38L, 40L, 39L, 33L, 33L, 41L, 38L, 38L, 41L, 39L, 35L, 
35L, 35L, 34L, 39L, 39L, 39L, 38L, 35L, 39L, 35L), type = structure(c(2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("BF", 
"DE", "TS", "ZN"), class = "factor"), lat = c(40.77486, 33.72621, 
30.38654, 39.21092, 42.56396, 35.25653, 38.98737, 39.17866, 41.42014, 
38.8756, 39.98261, 32.73808, 39.36327, 42.31465, 42.95051, 38.18899, 
26.05433, 43.00417, 45.21794, 36.13051, 30.00133, 40.7747, 35.48284, 
41.32453, 28.45942, 39.94898, 33.27457, 40.43147, 45.63667, 41.81694, 
32.73808, 39.36327, 42.31465, 42.95051, 35.9996, 34.83323, 41.76574, 
29.53539, 39.80981, 30.34521, 39.09211, 35.95858, 35.92756, 33.78383, 
38.18899, 36.13051, 30.00133, 40.7747, 35.48284, 41.32453, 28.45942, 
39.94898, 33.27457, 40.43147, 45.63667, 41.81694, 35.84396, 37.52477, 
34.03407, 38.60411), long = c(-75.94306, -85.44905, -97.78677, 
-77.6507, -72.23171, -81.42082, -101.89982, -84.7458, -82.23518, 
-104.89844, -83.27101, -98.0848, -106.07907, -83.24842, -85.86264, 
-85.81685, -80.99071, -88.04002, -94.50328, -87.32164, -90.19677, 
-74.68918, -97.61928, -96.59226, -81.96995, -75.87195, -112.45182, 
-80.05054, -123.21019, -71.45619, -98.0848, -106.07907, -83.24842, 
-85.86264, -80.0537, -82.39784, -72.7151, -96.11084, -86.41153, 
-81.82319, -94.8559, -83.99512, -115.53027, -119.30347, -85.81685, 
-87.32164, -90.19677, -74.68918, -97.61928, -96.59226, -81.96995, 
-75.87195, -112.45182, -80.05054, -123.21019, -71.45619, -78.78513, 
-77.5633, -117.63895, -121.41828)), class = "data.frame", row.names = c(NA, 
-60L))

我正在尝试为每种type (say, DE, TS, etc.)分别为X, Y and Z获取类似于下图（地图上没有文本type (say, DE, TS, etc.) 。 我试过的代码如下：-

library(ggplot2)
library(maps)

ggplot() + 
  geom_polygon(data=states, aes(x=long, y=lat, group=group), color="blue", fill="white")+
  geom_tile(data=df, aes(x=long, y=lat, fill = Y), alpha=0.3)+
  scale_fill_gradient(low = "blue", high = "red")

Answer 1

我无法运行您提供的示例，错误是：

object 'states' not found

但是，我可以运行这个稍微修改过的代码版本：

states <- map_data("state")
ggplot(data = states) +
    geom_polygon(aes(x = long, y = lat, group = group), color = "white") + 
    geom_point(data=df, aes(x=long, y=lat, color = Y), alpha=0.3)+
    scale_fill_gradient(low = "blue", high = "red")

有两个区别：

1. 我使用 ggplot2::map_data("state") （见上面的错误）
2. 使用 geom_point() 而不是 geom_tile()。 我不知道如何在“df”提供的数据上应用 geom_tile() ，因为您没有在规则间隔的“瓷砖集”上绘图。 geom_polygon() 可能更合适，但你有单点观察，即每个 Y 值一个“纬度”和“长”值，所以我不知道如何处理这些数据你想做的事情......

来自 geom_tile() 文档：

geom_rect 和 geom_tile 做同样的事情，但参数化不同：geom_rect 使用四个角的位置（xmin、xmax、ymin 和 ymax），而 geom_tile 使用平铺的中心及其大小（x、y、宽度、高度） ）。 geom_raster 是一种高性能特例，适用于所有切片大小相同的情况。