复制签名，转发来自包装器函数的所有参数 [英] Copy signature, forward all arguments from wrapper function

问题描述

我在一个类中有两个函数， plot()和 show(). show()作为方便的方法，除了在 plot()的代码中添加两行之外，没有做任何其他事情，例如

I have two functions in a class, plot() and show(). show(), as convenience method, does nothing else than to add two lines to the code of plot() like

def plot(
        self,
        show_this=True,
        show_that=True,
        color='k',
        boundary_color=None,
        other_color=[0.8, 0.8, 0.8],
        show_axes=True
        ):
    # lots of code
    return

def show(
        self,
        show_this=True,
        show_that=True,
        color='k',
        boundary_color=None,
        other_color=[0.8, 0.8, 0.8],
        show_axes=True
        ):
    from matplotlib import pyplot as plt
    self.plot(
        show_this=show_this,
        show_that=show_that,
        color=color,
        boundary_color=boundary_color,
        other_color=other_color,
        show_axes=show_axes
        )
    plt.show()
    return

这一切正常.

我遇到的问题是，在 show()包装程序中，这似乎途太多了.我真正想要的是:让 show()与 plot()具有相同的签名和默认参数，然后将所有参数转发给它.

The issue I have is that this seems way too much code in the show() wrapper. What I really want: Let show() have the same signature and default arguments as plot(), and forward all arguments to it.

有任何提示吗?

推荐答案

您可以使用参数打包/解包:

You can make use of argument packing/unpacking:

def show(self, *args, **kwargs):
    from matplotlib import pyplot as plt
    self.plot(*args, **kwargs)
    plt.show()

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