在Julia 1.5.2上使用animation.py时出错

问题描述

我使用this link的第一个示例制作视频。在我更换计算机之前的Julia版本中，我认为版本1.2可以正常工作。现在，一切正常，直到最后一行

myanim[:save]("testvid.mp4",writer=mywriter)

我收到错误消息的地方：

ERROR: PyError ($(Expr(:escape,:(ccall(#= /Users/XX/.julia/packages/PyCall/BcTLp/src/pyfncall.jl:43 =# @pysym(:PyObject_Call),PyPtr,(PyPtr,PyPtr),o,pyargsptr,kw))))) <class 'FileNotFoundError'>
FileNotFoundError(2,'No such file or directory')
  File "/Users/XX/.julia/conda/3/lib/python3.8/site-packages/matplotlib/animation.py",line 1123,in save
    with mpl.rc_context({'savefig.bBox': None}),\
  File "/Users/XX/.julia/conda/3/lib/python3.8/contextlib.py",line 113,in __enter__
    return next(self.gen)
  File "/Users/XX/.julia/conda/3/lib/python3.8/site-packages/matplotlib/animation.py",line 249,in saving
    self.setup(fig,outfile,dpi,*args,**kwargs)
  File "/Users/XX/.julia/conda/3/lib/python3.8/site-packages/matplotlib/animation.py",line 338,in setup
    self._run()
  File "/Users/XX/.julia/conda/3/lib/python3.8/site-packages/matplotlib/animation.py",line 348,in _run
    self._proc = subprocess.Popen(
  File "/Users/XX/.julia/conda/3/lib/python3.8/subprocess.py",line 854,in __init__
    self._execute_child(args,executable,preexec_fn,close_fds,File "/Users/XX/.julia/conda/3/lib/python3.8/subprocess.py",line 1702,in _execute_child
    raise child_exception_type(errno_num,err_msg,err_filename)

Stacktrace:
 [1] pyerr_check at /Users/XX/.julia/packages/PyCall/BcTLp/src/exception.jl:62 [inlined] 
 [2] pyerr_check at /Users/XX/.julia/packages/PyCall/BcTLp/src/exception.jl:66 [inlined]
 [3] _handle_error(::String) at /Users/XX/.julia/packages/PyCall/BcTLp/src/exception.jl:83
 [4] macro expansion at /Users/XX/.julia/packages/PyCall/BcTLp/src/exception.jl:97 [inlined]
 [5] #110 at /Users/XX/.julia/packages/PyCall/BcTLp/src/pyfncall.jl:43 [inlined]
 [6] disable_sigint at ./c.jl:446 [inlined]
 [7] __pycall! at /Users/XX/.julia/packages/PyCall/BcTLp/src/pyfncall.jl:42 [inlined]
 [8] _pycall!(::PyObject,::PyObject,::Tuple{String},::Int64,::PyObject) at /Users/XX/.julia/packages/PyCall/BcTLp/src/pyfncall.jl:29
 [9] _pycall!(::PyObject,::Base.Iterators.Pairs{Symbol,PyObject,Tuple{Symbol},NamedTuple{(:writer,),Tuple{PyObject}}}) at /Users/XX/.julia/packages/PyCall/BcTLp/src/pyfncall.jl:11
 [10] #_#117 at /Users/XX/.julia/packages/PyCall/BcTLp/src/pyfncall.jl:86 [inlined]
 [11] top-level scope at REPL[137]:1

有人知道发生了什么事吗？

完全相同的代码在以前的Julia版本中运行得很好。

解决方法

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