Python：使用命名参数将函数传递给函数

我希望将一个函数传递给另一个函数，以及一个命名参数。这与类似，只是它不处理命名参数。评论中对此提出了一个问题，但没有答复

例如：

def printPath(path, displayNumber = False):
    pass

def explore(path, function, *args):
    contents = function(*args)

print explore(path, printPath, path, displayNumber = False)

这会产生以下错误：

TypeError: explore() got an unexpected keyword argument 'displayNumber'

---

您只需允许

explore

也接收命名参数：

def printPath(path, displayNumber = False):
    pass

def explore(path, function, *args, **kwargs):
    contents = function(*args, **kwargs)

print explore(path, printPath, path, displayNumber = False)

---

您只需允许

explore

也接收命名参数：

def printPath(path, displayNumber = False):
    pass

def explore(path, function, *args, **kwargs):
    contents = function(*args, **kwargs)

print explore(path, printPath, path, displayNumber = False)

---

您只需允许

explore

也接收命名参数：

def printPath(path, displayNumber = False):
    pass

def explore(path, function, *args, **kwargs):
    contents = function(*args, **kwargs)

print explore(path, printPath, path, displayNumber = False)

---

您只需允许

explore

也接收命名参数：

def printPath(path, displayNumber = False):
    pass

def explore(path, function, *args, **kwargs):
    contents = function(*args, **kwargs)

print explore(path, printPath, path, displayNumber = False)

---

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