在ggpplot2中绘制轮廓线。

I want to plot this figure created with filled.contour(), but in ggplot2, how do I do this?

我想用filled.contour()来绘制这个图形，但是在ggplot2中，我该怎么做呢?

I want to use ggplot2 because the graphing conventions are easier. The reason I want to use filled.contour() is because I tried geom_tile() and image.plot() and they both created very tile like outputs, and I need an output similar to filled.contour().

我想使用ggplot2，因为图形化的约定比较容易。我想使用filled.contour()的原因是我尝试了geo_tile()和image.plot()，它们都创建了非常像tile的输出，我需要一个类似filled.contour()的输出。

This is my figure:

这是我的图:

Code:

代码:

library(akima)

df <-read.table("Petra_phytoplankton+POM_xydata_minusNAs_noduplicates.txt",header=T)
attach(df)
names(df)
fld <- with(df, interp(x = longitude, y = latitude, z = d13C))

filled.contour.ungeoreferenced <- 
  (filled.contour(x = fld$x,
                  y = fld$y,
                  z = fld$z,
                  color.palette =
                    colorRampPalette(c("blue", "green", "yellow",
                                       "orange", "red")),
                  xlab = "Longitude",
                  ylab = "Latitude",
                  key.title = title(main = "d13C", 
                                    cex.main = 1)))

Snippet of data:

片段的数据:

latitude    longitude   d13C
-65 -70 -27.7
-61 150 -32.2
-61 150 -28.3
-60 116 -26.8
-60 116 -24.7
-47 38  -24.8
-38 150 -20.5
19  -65.7   -19.9
19  -65.5   -18.5
18  -60.7   -20
18  -58.5   -18.2
18  -57.8   -19
17  -55.4   -18.6
17  -50.8   -18
17  -47.1   -18.3
17  -45.5   -19.4
16  -43.3   -17.9
15  -40.7   -18.5
14  -39.3   -19.9
12  -36.7   -19.9
12  -36.2   -19.9
11  -34.4   -19.2
10  -32 -18.5
9   -30.3   -19.3
8   -29.2   -19.4
7   -26.6   -18.2
7   -25.5   -19.3
6   23.9    -20
3   -21.3   -20.4

3 个解决方案

#1

You can tweak the colors as you need:

你可以根据需要调整颜色:

gdat <- interp2xyz(fld, data.frame=TRUE)

ggplot(gdat) + 
  aes(x = x, y = y, z = z, fill = z) + 
  geom_tile() + 
  coord_equal() +
  geom_contour(color = "white", alpha = 0.5) + 
  scale_fill_distiller(palette="Spectral", na.value="white") + 
  theme_bw()

You can reduce the pixelation at the cost of some processing time by increasing the density of the interpolation:

你可以通过增加插值的密度来减少一些处理时间的像素化:

fld <- with(df, interp(x = longitude, 
                       y = latitude, 
                       z = d13C,
                       xo = seq(min(longitude), max(longitude), length=400),
                       duplicate="mean"))

and also reducing the bin width:

也减少了箱子的宽度:

ggplot(gdat) + 
  aes(x = x, y = y, z = z) + 
  geom_tile(aes(fill=z)) + 
  coord_equal() +
  stat_contour(aes(fill=..level..), geom="polygon", binwidth=0.005) + 
  geom_contour(color="white", alpha=0.5) +
  scale_fill_distiller(palette="Spectral", na.value="white") + 
  theme_bw()

NOTE: that is going to crunch for a noticeable few seconds on a decent desktop system. On my fairly beefy MacBook Pro it was:

注意:在一个不错的桌面系统上，这将会出现几秒钟的时间。在我相当结实的MacBook Pro上，它是:

   user  system elapsed 
  6.931   0.655   8.153 

#2

To follow up on @hrbrmstr's minimal example, you can also have ggplot2 compute "z" for you:

为了跟踪@hrbrmstr的最小示例，您还可以为您使用ggplot2计算“z”:

library(ggplot2)
ggplot(data = faithful, aes(x = eruptions, y = waiting)) +
  stat_density2d(aes(colour = ..level.., fill = ..level..), geom = "polygon")

#3

I took the example from the ggplot2 website.

我从ggplot2网站上拿了这个例子。

 # Generate data
 library(reshape2) # for melt
 volcano3d <- melt(volcano)
 names(volcano3d) <- c("x", "y", "z")

 # Basic plot
 v <- ggplot(volcano3d, aes(x, y, z = z)) +  
     stat_contour(geom="polygon", aes(fill=..level..))

Where x and y are your Long and Lat and z is d13C

x和y是你的长，Lat和z是d13C ?