proj=:polar fails

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](https://github.com/JuliaPy/PyCall.jl) 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)))))
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
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
```

Reproduce the documented example with Julia 1.6.0 and PyPlot v2.10.0, then inspect PyPlot.jl:177 where plot forwards keyword arguments to matplotlib. Compare that call with the matplotlib pyplot.py stack frames and the working PyCall example. Done means the documented proj=:polar example runs without the Unknown property proj error.

Do mô hình lập chỉ mục viết ra từ nội dung của issue.