如何在 Julia 中更改散点图中的制造商

Question

我想根据向量中的值更改 2D 散点图的标记样式，最终在两个轴上显示 3 个维度。以下是我想做的事情以及我尝试时遇到的错误。

x = rand(1:30,100)
y = rand(20:30,100)
MyMarker = [fill("o",50);fill("x",50)]
scatter(x,y,marker=MyMarker,alpha=0.5)

LoadError: PyError (:PyObject_Call) <type 'exceptions.ValueError'>
ValueError(u"Unrecognized marker style ['o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'o', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x', 'x']",)
  File "/home/lara/.julia/v0.4/Conda/deps/usr/lib/python2.7/site-packages/matplotlib/pyplot.py", line 3251, in scatter
    edgecolors=edgecolors, data=data, **kwargs)
  File "/home/lara/.julia/v0.4/Conda/deps/usr/lib/python2.7/site-packages/matplotlib/__init__.py", line 1812, in inner
    return func(ax, *args, **kwargs)
  File "/home/lara/.julia/v0.4/Conda/deps/usr/lib/python2.7/site-packages/matplotlib/axes/_axes.py", line 3877, in scatter
    marker_obj = mmarkers.MarkerStyle(marker)
  File "/home/lara/.julia/v0.4/Conda/deps/usr/lib/python2.7/site-packages/matplotlib/markers.py", line 171, in __init__
    self.set_marker(marker)
  File "/home/lara/.julia/v0.4/Conda/deps/usr/lib/python2.7/site-packages/matplotlib/markers.py", line 252, in set_marker
    ' {0}'.format(marker))

while loading In[91], in expression starting on line 1

 in getindex at /home/lara/.julia/v0.4/PyCall/src/PyCall.jl:228
 in pysequence_query at /home/lara/.julia/v0.4/PyCall/src/conversions.jl:717
 [inlined code] from /home/lara/.julia/v0.4/PyCall/src/conversions.jl:733
 in pytype_query at /home/lara/.julia/v0.4/PyCall/src/conversions.jl:762
 in convert at /home/lara/.julia/v0.4/PyCall/src/conversions.jl:782
 in pycall at /home/lara/.julia/v0.4/PyCall/src/PyCall.jl:363
 in call at /home/lara/.julia/v0.4/PyCall/src/PyCall.jl:372
 in close_queued_figs at /home/lara/.julia/v0.4/PyPlot/src/PyPlot.jl:401

我尝试了以下方法，但这两种方法都不是我想要做的，但它们足够接近，而且都不起作用。

using Plots
pyplot(size=(400,200), legend=false)  # set backend and set some session defaults
scatter(rand(30),m = ColorGradient([:red,:green,:blue]),zcolor = repmat([0,0.5,1],10))

using Plots
pyplot(size=(400,200), legend=false)  # set backend and set some session defaults
scatter(rand(30),
        m = ColorGradient([:red,:green,:blue]),  # colors are defined by a gradient
        zcolor = repmat([0,0.5,1],10) # sample from the gradient, cycling through: 0, 0.5, 1
       )

谁能告诉我如何让ma​​rker接受向量？

更新 我有一个解决方法，但我不喜欢这样做：将数据分成单独的集合，然后将它们绘制在相同的轴上：

PyPlot.plot(spacerdataActualSpacer[:Length],spacerdataActualSpacer[:Curvature],marker="x",".")
PyPlot.plot(spacerdataNoActualSpacer[:Length],spacerdataNoActualSpacer[:Curvature],marker="o",".")

Answer 1

您可以尝试几种不同的方法（我是 Plots 开发人员，所以我不能确定开箱即用的方法对您有用）：

scatter(rand(30), group = repmat(1:3,10), marker=:auto)
scatter(rand(30), group = repmat(1:3,10), shape=[:+ :o :utri])
scatter(rand(30), group = repmat(1:3,10), shape=[:+ :o :utri], layout=(3,1))

如果您想要其中任何一个，请告诉我。老实说，我不完全确定我知道你理想的可视化会是什么样子。