[英]error with scatterPlot
我正在使用来自openair包的R的scatterPlot绘制轨迹(纬度与经度)。 使用“组”选项对轨迹进行分组时,不会绘制最后一个组。 这是一个示例代码:
df = data.frame(name = c(rep('C1',10),rep('C2',10),rep('C3',10),rep('C4',10)),
lat = seq(1,100,2.5),
lon = seq(101,200,2.5))
scatterPlot(df ,x = "lon", y = "lat", group = "name",map = TRUE )
scatterPlot(df ,x = "lon", y = "lat")
此外,我在尝试在后台绘制地图时出错:“使用数据包1时出错。长度为零的参数i”
谢谢伊利克
似乎是代码中的错误(基于lattice :: xyplot)。 openair::scatterPlot
的代码看起来很业余:
# ------segment where `group` parameter is passed to lattice code ---
id <- which(names(mydata) == group)
names(mydata)[id] <- "MyGroupVar"
plotType <- if (!Args$traj)
c("p", "g")
else "n"
if (method == "scatter") {
if (missing(k))
k <- NULL
Type <- type
xy.args <- list(x = myform, data = mydata, groups = mydata$MyGroupVar,
type = plotType, as.table = TRUE, scales = scales,
#---- end of extract
使用重命名分组变量的躲闪然后使用$与mydata$MyGroupVar
是一个黑客攻击。 应该用mydata[[group]]
做得更简单,更不容易出错。 建议你要求修复bug。
如果执行traceback()
您可以看到生成数据包错误时传递的参数:
traceback()4:xyplot.formula(x = lat~lon | default,data = list(lon = c(101,103.5,106,108.5,111,113.5,116,118.5,121,123.5,126,128.5,131 ,133.5,136,138.5,141,143.5,146,148.5,151,153.5,156,158.5,161,163.5,166,168.5,171,173.5,176,178.5,181,183.5,186,188.5,191,193.5 ,196,198.5),lat = c(1,3.5,6,8.5,11,13.5,16,18.5,21,23.5,26,28.5,31,33.5,36,38.5,41,43.5,46,48.5, 51,53.5,56,58.5,61,63.5,66,68.5,71,73.5,76,78.5,81,83.5,86,88.5,91,93.5,96,98.5),默认= c(1L,1L,1L ,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L ,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L),MyGroupVar = c(1L,1L,1L,1L,1L,1L,1L,1L,1L,1L, 1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,1L, 1L,1L,1L,1L,1L)),groups =“name”,type = c(“p”,“g”),as.table = TRUE,
(我认为)......“名称”周围的引号可能会导致错误。 它应该作为一个未引用的,真正的R名称/符号传递。
我认为你可以用普通的格子非常接近你想要的东西:
xyplot(data=df , lon~lat, groups = name, auto.key=TRUE, grid=TRUE)
如果您需要调整它,请参阅?xyplot
。
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