使用 interp 推断数据不会产生准确的图像

Question

我有一张图表，其中外推与初始内插不匹配。我希望热图填满整个图像。

一、插值代码：

library(akima)
library(reshape2)

xmin <- signif(min(CBLo2$MD1))
xmax <- signif(max(CBLo2$MD1))
ymin <- signif(min(CBLo2$MD2)) 
ymax <- signif(max(CBLo2$MD2))
gridint <- 100

fld <- with(CBLo2, interp(x = MD1, y = MD2, z = Abundance, 
            xo=seq(xmin, xmax, length=gridint), yo=seq(ymin, ymax, length=gridint) ))
df <- melt(fld$z, na.rm = TRUE)
names(df) <- c("MD1", "MD2", "Abundance")
df$MD1 <- fld$x[df$MD1]
df$MD2 <- fld$y[df$MD2]
contour(fld) # test plot

我不会 post 整个 ggplot 代码（用于下图），只是生成热图所需的代码：

ggplot() +
  geom_tile(inherit.aes=FALSE,data = df, aes(x = MD1, y = MD2,fill = Abundance)) +
  scale_fill_continuous(name = "Rain (mm)", low = "yellow", high = "green")

但是，当我尝试推断数据时（根据其他 post 的示例），我得到了以下图表，它与第一口井完全不匹配：

fld <- with(CBLo2, interp(x = MD1, y = MD2, z = Abundance, extrap=TRUE, linear=FALSE,
            xo=seq(xmin, xmax, length=gridint), yo=seq(ymin, ymax, length=gridint) ))

这是数据：

Abundance   MD1 MD2
9   -0.59042    0.76793119
42  -0.48544284 -0.09465043
13  0.51250586  -0.24599322
84  -0.30857525 -0.21529624
2   0.90449257  0.679926
16  0.24536209  0.24016424
52  -0.43144002 -0.75474149
4   1.23830339  -0.11985391
37  -1.10235817 0.33886773
79  0.01757236  -0.59635386

我做错了什么？我怎样才能使外推更准确？

Answer 1

TLDR 解决方案：

将linear = FALSE添加到所有interp()代码以保持一致性，并在scale_fill_continuous()中指定相同的限制。

解释:

这里有两个问题。

问题1：生成第一个fld的代码不包含参数linear = FALSE ，而用于第二个的代码。

让我们比较一下插值：

library(dplyr)

fld1 <- with(CBLo2, 
            interp(x = MD1, y = MD2, z = Abundance, 
                   xo=seq(xmin, xmax, length=gridint), 
                   yo=seq(ymin, ymax, length=gridint) ))
df1 <- melt(fld1$z, na.rm = TRUE) # 6426 obs

fld2 <- with(CBLo2, 
             interp(x = MD1, y = MD2, z = Abundance, 
                    extrap = TRUE, linear = FALSE,
                    xo=seq(xmin, xmax, length=gridint), 
                    yo=seq(ymin, ymax, length=gridint) ))
df2 <- melt(fld2$z, na.rm = TRUE) #1000 obs

df.combined <- left_join(df2, df1, by = c("Var1", "Var2"))
df.combined %>% 
  filter(!is.na(value.y)) %>%        # compare for the overlapping range
  mutate(diff = value.x - value.y) %>%
  select(diff) %>% 
  summary()

      diff         
 Min.   :-303.360  
 1st Qu.: -42.399  
 Median :   8.763  
 Mean   :  -7.552  
 3rd Qu.:  36.132  
 Max.   : 238.647

现在将linear = FALSE添加到第一个fld:

fld3 <- with(CBLo2, 
            interp(x = MD1, y = MD2, z = Abundance, 
                   linear = FALSE,
                   xo=seq(xmin, xmax, length=gridint), 
                   yo=seq(ymin, ymax, length=gridint) ))
df3 <- melt(fld3$z, na.rm = TRUE) # 6426 obs

df.combined <- left_join(df2, df3, by = c("Var1", "Var2"))
df.combined %>% 
  filter(!is.na(value.y)) %>%
  mutate(diff = value.x - value.y) %>%
  select(diff) %>% 
  summary()

      diff  
 Min.   :0  
 1st Qu.:0  
 Median :0  
 Mean   :0  
 3rd Qu.:0  
 Max.   :0

问题2：插值范围差别很大

# define column names
names(df2) <- c("MD1", "MD2", "Abundance")
names(df3) <- c("MD1", "MD2", "Abundance")

> range(df2$Abundance)
[1] -1136.341   420.369
> range(df3$Abundance)
[1] -297.9161  241.6618

我们可以看到，即使值在相同的 MD1/MD2 坐标处匹配，扩展的 df2 中的值范围也远远超过 df3 的范围。为了确保丰度值和颜色之间的映射相同，我们必须根据两者的组合范围来指定填充限制。

我将使用丑陋但视觉上明显不同的渐变来说明这一点：

library(gridExtra)

p <- ggplot() + 
  scale_fill_gradientn(name = "Rain (mm)", colours = rainbow(15),
                       limits = range(c(df2$Abundance, df3$Abundance)))

grid.arrange(p + geom_tile(data = df3, aes(x = MD1, y = MD2, fill = Abundance)),
             p + geom_tile(data = df2, aes(x = MD1, y = MD2, fill = Abundance)),
             nrow = 1)

如果我们叠加图，它们会完全重叠（调整透明度以显示 df3 的边缘）：

p + 
  geom_tile(data = df3, aes(x = MD1, y = MD2, fill = Abundance), alpha = 0.5) +
  geom_tile(data = df2, aes(x = MD1, y = MD2, fill = Abundance), alpha = 0.5)

使用 interp 推断数据不会产生准确的图像

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