Question

我正在使用oct2py与python运行octave函数。它正在工作，但当我尝试乘以2矩阵时出现错误。我该怎么做才能解决问题？

这是一个matlab函数示例

%% MATLAB
function lol = jk2(arg1,arg2)
    arg1 = arg1;
    arg2 = arg2;
    lol = arg1*arg2;

end

这是调用函数的代码

import oct2py
from oct2py import octave
a=3
b=4
octave.call("/MATLAB/jk2.m",a,b) # this call works
a=np.array([[1,2],[3,4]])
b=np.array([[5,6],[1,2]])
octave.call("/MATLAB/jk2.m",a,b) # this call report an errors

这是错误消息

>>> octave.call("/home/donbeo/Documents/MATLAB/jk2.m",a,b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/lib/python2.7/dist-packages/oct2py-1.2.0-py2.7.egg/oct2py/session.py", line 210, in call
    resp = self._eval(cmd, verbose=verbose)
  File "/usr/local/lib/python2.7/dist-packages/oct2py-1.2.0-py2.7.egg/oct2py/session.py", line 350, in _eval
    return self._session.evaluate(cmds, verbose, log, self.logger)
  File "/usr/local/lib/python2.7/dist-packages/oct2py-1.2.0-py2.7.egg/oct2py/session.py", line 523, in evaluate
    raise Oct2PyError(msg)
oct2py.utils.Oct2PyError: Oct2Py tried to run:
"""
[a__] = jk2(A__, B__)
"""
Octave returned:
binary operator '*' not implemented for 'int64 matrix' by 'int64 matrix' operations
>>>

Answer 1

以下是Python和Octave之间的界限变得模糊的示例。 Numpy将您的数组解释为整数类型（因为没有显式浮点数），但Octave会将数组视为双精度数。如果在数组定义中的任何位置添加句点，它都将起作用。

修正（测试）的例子：

from oct2py import octave
import numpy as np
a = np.array([[1, 2], [3, 4.]])  # notice the addition of the period
b = np.array([[5, 6], [1, 2], dtype=float])  # another way to specify floating point type
octave.call("/MATLAB/jk2.m", a, b)  # this call works just fine

矩阵乘法与oct2py

1 个答案: