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我正在努力在这张全球金融市场不同部门逐年变化的离散回报图表上找到正确的轴。我正在使用的代码是：

library(scatterplot3d)
data = c(32, 10.7, 37.5, -10, -50.2, 31.7, 20.1, -5.9, 29.8, 1.3,    24.7, 33.4, 32.2, 39.5, -53.2, 78.6, 19, -18.3, 18.2, -9.6,     22.8, 6.5, 15.2, 18.1, -5.2, 22.0, 15.7, -1.8, 15.1, -6.2,    20.8, 14.1, 27, 11.8, -43, 32.4, 8.4, -11.7, 17.9, 4.5,    11.1, 7.9, 14.7, 10.1, -23, 23.6, 10.5, -1, 11.8, 2.9,     11.1, 2.8, 11.8, 1.9, -26.2, 58.2, 15.1, 5, 15.8, 1.4,    10.9, 4.9, 15.8, 5.5, -37, 26.5, 15.1, 2.1, 16, 13.8,    9.1, 21.4, 2.1, 16.2, -35.6, 18.9, 16.8, -13.3, -1.1, -10.5,    9.1, -4.4, 6.6, 9.5, 4.8, 6.9, 5.5, 5.6, 4.3, -4.8,     8.3, 3.0, 0.4, 11.6, -2.4, 11.4, 6.3, 13.6, 7.0, -7.4,    5.4, 18.4, 23.0, 31.3, 5.5, 24.0, 29.7, 10.2, 7, -27,     5, 2.2, 4.3, 5.1, -3.1, 16, 8.5, 8.4, 9.4, -3.6,    4.5, 3.5, 4.8, 3.4, -2.5, 12.9, 2.4, 10.7, 6.8, -2.7,      3.3, 3, 3.1, 9, 13.7, -3.6, 5.9, 9.8, 2, -2.1,     1.2, 3, 4.8, 4.8, 1.8, 0.1, 0.1, 0.1, 0.1, 0) # should have length 150, check

my.datamat = matrix(data, nrow = 10)
rownames = c("2004", "2005", "2006", "2007", "2008", "2009", "2010", "2011", "2012", "YTD")
colnames = c("Global RE", "EM Equities", "EM Fixed", "Non US Eq", "60/40", "High Yield", "US Equities", "Nat. Res.", "Global Fixed", "TIPS", "Gold", "Inv. Grade", "Municipals", "Gov't Bonds", "Cash")
dimnames(my.datamat) = list(rownames, colnames)
my.datamat = t(my.datamat)


 plot(0,0, xlim=c(1,length(colnames(my.datamat)))
        ,ylim = range(my.datamat)
     , type='n'
     ,xaxt = "n", main = "Northern Trust Data (jmi4)", xlab  = "Year", ylab = "Return (%)")

sapply(1:length(colnames(my.datamat))
       ,function(i){lines(my.datamat[i,], col = i)} )

op = par(cex = .5)
legend("bottomright"
       , legend=rownames(my.datamat)
       ,lty=rep(1,length(rownames(my.datamat)))
       ,col=c(1:length(rownames(my.datamat)))
       )
par(op)

axis(1, at = 1:length(colnames(my.datamat))
      , labels=colnames(my.datamat))


 ### or do a heatmap
 quartz()
 library(gplots)
 my.heatmap = heatmap.2(my.datamat, Rowv = NA, Colv = NA, col = redblue(256), dendrogram = "none", scale = "column", main  = "Northern Trust Heatmap", key = TRUE, trace = "none", density.info = "none", keysize = 1)

当我绘制这个时，轴非常偏离。我怎样才能（a）固定轴和（b）在这些点之间线性插值，就像这里一样。谢谢你的帮助。

编辑

感谢 David Martin，这些是我能够制作的最终图表。我的代码已更新。以下是图表： 概率热图 叠加折线图

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1 回答 1

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您的轴标签重叠，因为您提供的标签数量是函数绘制“刻度”的两倍，但您告诉它在每个刻度处放置一个标签。您需要明确提供所需的刻度数，如下所示：

试试这个：

library(reshape2)
m = melt(my.datamat)

s3d = scatterplot3d(m, type = "h"
              , lwd = 5, pch = " "
              , x.ticklabs = rownames(my.datamat)
              , y.ticklabs = colnames(my.datamat)
              , color = grey(150:1 / 200), main = "Northern Trust Data"
              ,lab=c(length(rownames(my.datamat)),length(colnames(my.datamat))) # defines how many ticks should appear on each axis
                    )

您发布的链接中的表面并不完全是“在点之间插值”，它是通过点的回归平面。如果这对您的需求是可以接受的，那么您可以将其添加到情节中：

# regression plane
s3d$plane3d(lm(value~., data=nt.dat), lty.box="solid" )

不过老实说，我认为以 3D 形式绘制它会让人感到困惑，并且会使您的数据不必要地难以阅读。我建议您考虑改用叠加折线图：

plot(0,0, xlim=c(1,length(colnames(my.datamat)))
        ,ylim = range(my.datamat)
     , type='n'
     ,xaxt = "n")

sapply(1:length(rownames(my.datamat))
       ,function(i){lines(my.datamat[i,]
                          , col=i)} )

legend("topright"
       , legend=rownames(my.datamat)
       ,lty=rep(1,length(rownames(my.datamat)))
       ,col=c(1:length(rownames(my.datamat)))
       )

axis(1, at = 1:length(colnames(my.datamat))
      , labels=colnames(my.datamat))
于 2013-07-15T20:56:01.257 回答