只需运行以下代码。我想显示一个坐标网格,但是什么也没发生:
df_runtime <- data.frame(x = c(0L, 20L),
y = c(0L, 10L),
z = c(0L, 50L), stringsAsFactors = FALSE)
car::scatter3d(x = df_runtime$x,
y = df_runtime$y,
z = df_runtime$z,
xlab = "x", ylab = "x", zlab = "z",
surface = FALSE, grid = TRUE)
从文档??car::scatter3d
中我意识到了
在回归曲面上绘制网格线(真或假)。
因此,grid
参数不是我想要的。 是否可以获取坐标网格?对我来说,这对于指导眼睛真的很有用。
在输入Carles之后进行编辑:
我想保留交互式图表-这就是为什么我在寻找car::scatter3d
解决方案。如果您不需要这样做,那么scatterplot3d
和FactoClass
的组合真的很不错。以下内容以非交互方式工作:
scatterplot3d::scatterplot3d(
df_runtime$x,
df_runtime$y,
df_runtime$z,
color = "blue", pch = 19, # filled blue circles
# type = "h", # lines to the horizontal plane
main = "Title",
xlab = "x",
ylab = "y",
zlab = "z",
angle = 35,
grid = FALSE)
FactoClass::addgrids3d(df_runtime$x,
df_runtime$y,
df_runtime$z,
angle = 35,
grid = c("xy", "xz", "yz"))
答案 0 :(得分:4)
如果要使用网格进行交互式绘图,则plotly
是另一种解决方案:
df_runtime <- data.frame(x = c(0L, 20L),
y = c(0L, 10L),
z = c(0L, 50L), stringsAsFactors = FALSE)
plotly::plot_ly(df_runtime,
x = ~x,
y = ~y,
z = ~z,
type = 'scatter3d',
mode = 'markers')
答案 1 :(得分:1)
对我来说,它不起作用。但是,您可以使用其他软件包,例如library("scatterplot3d")
。我刚刚提出了更多要点,它就可以了:
df_runtime <- structure(list(n_legs_array = rnorm(100,0,10),
n_vehicles_array = rnorm(100,0,10),
t = rnorm(100,0,10)),
.Names = c("n_legs_array", "n_vehicles_array", "t"),
row.names = c(1L, 2L),
class = "data.frame")
library("scatterplot3d")
scatterplot3d(x = df_runtime$n_legs_array,
y = df_runtime$t/60, # minutes
z = df_runtime$n_vehicles_array,
xlab = "n_legs", ylab = "time [min]", zlab = "n_vehicles",
grid = TRUE,box = FALSE,
color = "#56B4E9")