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Julia 1.6.0 [d330b81b] PyPlot v2.10.0
Following https://docs.juliaplots.org/latest/generated/pyplot/#pyplot-ref27 ,
Θ = range(0, stop = 1.5π, length = 100) r = abs.(0.1 * randn(100) + sin.(3Θ)) plot(Θ, r, proj = :polar, m = 2)
fails with "Unknown property proj", as below. PyCall is a workaround:
using PyCall plt = pyimport("matplotlib.pyplot") plt.axes(polar=true) th = range(0, stop = 1.5π, length = 100) r = abs.(0.1 * randn(100) + sin.(3*th)) plt.plot(th, r) plt.show()
plot(Θ, r, proj = :polar, m = 2) ERROR: PyError ($(Expr(:escape, :(ccall(#= /home/ben/.julia/packages/PyCall/3fwVL/src/pyfncall.jl:43 =# @pysym(:PyObject_Call), PyPtr, (PyPtr, PyPtr, PyPtr), o, pyargsptr, kw))))) <class 'AttributeError'> AttributeError('Unknown property proj',) File "/usr/lib/python3/dist-packages/matplotlib/pyplot.py", line 3261, in plot ret = ax.plot(*args, **kwargs) File "/usr/lib/python3/dist-packages/matplotlib/__init__.py", line 1718, in inner return func(ax, *args, **kwargs) File "/usr/lib/python3/dist-packages/matplotlib/axes/_axes.py", line 1372, in plot for line in self._get_lines(*args, **kwargs): File "/usr/lib/python3/dist-packages/matplotlib/axes/_base.py", line 404, in _grab_next_args for seg in self._plot_args(this, kwargs): File "/usr/lib/python3/dist-packages/matplotlib/axes/_base.py", line 394, in _plot_args seg = func(x[:, j % ncx], y[:, j % ncy], kw, kwargs) File "/usr/lib/python3/dist-packages/matplotlib/axes/_base.py", line 301, in _makeline seg = mlines.Line2D(x, y, **kw) File "/usr/lib/python3/dist-packages/matplotlib/lines.py", line 426, in __init__ self.update(kwargs) File "/usr/lib/python3/dist-packages/matplotlib/artist.py", line 902, in update for k, v in props.items()] File "/usr/lib/python3/dist-packages/matplotlib/artist.py", line 902, in <listcomp> for k, v in props.items()] File "/usr/lib/python3/dist-packages/matplotlib/artist.py", line 895, in _update_property raise AttributeError('Unknown property %s' % k) Stacktrace: [1] pyerr_check @ ~/.julia/packages/PyCall/3fwVL/src/exception.jl:62 [inlined] [2] pyerr_check @ ~/.julia/packages/PyCall/3fwVL/src/exception.jl:66 [inlined] [3] _handle_error(msg::String) @ PyCall ~/.julia/packages/PyCall/3fwVL/src/exception.jl:83 [4] macro expansion @ ~/.julia/packages/PyCall/3fwVL/src/exception.jl:97 [inlined] [5] #107 @ ~/.julia/packages/PyCall/3fwVL/src/pyfncall.jl:43 [inlined] [6] disable_sigint @ ./c.jl:458 [inlined] [7] __pycall! @ ~/.julia/packages/PyCall/3fwVL/src/pyfncall.jl:42 [inlined] [8] _pycall!(ret::PyCall.PyObject, o::PyCall.PyObject, args::Tuple{StepRangeLen{Float64, Base.TwicePrecision{Float64}, Base.TwicePrecision{Float64}}, Vector{Float64}}, nargs::Int64, kw::PyCall.PyObject) @ PyCall ~/.julia/packages/PyCall/3fwVL/src/pyfncall.jl:29 [9] _pycall!(ret::PyCall.PyObject, o::PyCall.PyObject, args::Tuple{StepRangeLen{Float64, Base.TwicePrecision{Float64}, Base.TwicePrecision{Float64}}, Vector{Float64}}, kwargs::Base.Iterators.Pairs{Symbol, Any, Tuple{Symbol, Symbol}, NamedTuple{(:proj, :m), Tuple{Symbol, Int64}}}) @ PyCall ~/.julia/packages/PyCall/3fwVL/src/pyfncall.jl:11 [10] pycall(::PyCall.PyObject, ::Type{PyCall.PyAny}, ::StepRangeLen{Float64, Base.TwicePrecision{Float64}, Base.TwicePrecision{Float64}}, ::Vararg{Any, N} where N; kwargs::Base.Iterators.Pairs{Symbol, Any, Tuple{Symbol, Symbol}, NamedTuple{(:proj, :m), Tuple{Symbol, Int64}}}) @ PyCall ~/.julia/packages/PyCall/3fwVL/src/pyfncall.jl:83 [11] plot(::StepRangeLen{Float64, Base.TwicePrecision{Float64}, Base.TwicePrecision{Float64}}, ::Vararg{Any, N} where N; kws::Base.Iterators.Pairs{Symbol, Any, Tuple{Symbol, Symbol}, NamedTuple{(:proj, :m), Tuple{Symbol, Int64}}}) @ PyPlot ~/.julia/packages/PyPlot/XaELc/src/PyPlot.jl:177 [12] top-level scope @ REPL[5]:1
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Julia 1.6.0
[d330b81b] PyPlot v2.10.0
Following https://docs.juliaplots.org/latest/generated/pyplot/#pyplot-ref27 ,
fails with "Unknown property proj", as below. PyCall is a workaround:
The text was updated successfully, but these errors were encountered: