Question

functools.wraps在python 2.4中不可用。在python 2.4中是否还有其他模块可用于代替？

Answer 1

您可以从Python2.5复制functools代码并使其在Python2.4中工作，只需进行少量更改（用 lambda 代替 partial ）：{{ 3}}

以下是partial的简单替代：

def partial(func, *args, **kwds):
    "Emulate Python2.6's functools.partial"
    return lambda *fargs, **fkwds: func(*(args+fargs), **dict(kwds, **fkwds))

Answer 2

阐述Raymond Hettinger的回答：

WRAPPER_ASSIGNMENTS = ('__module__', '__name__', '__doc__')
WRAPPER_UPDATES = ('__dict__',)
def wraps(wrapped, assigned=WRAPPER_ASSIGNMENTS, updated=WRAPPER_UPDATES):
  def partial(func, *args, **kwds):
    return lambda *fargs, **fkwds: func(*(args+fargs), **dict(kwds,**fkwds))
  def update_wrapper(wrapper, wrapped, assigned=WRAPPER_ASSIGNMENTS,
                     updated=WRAPPER_UPDATES):
    for attr in assigned:
      setattr(wrapper, attr, getattr(wrapped, attr))
    for attr in updated:
      getattr(wrapper, attr).update(getattr(wrapped, attr, {}))
    return wrapper
  return partial(update_wrapper, wrapped=wrapped,
                 assigned=assigned, updated=updated)

functools.wraps for python 2.4

2 个答案: